Datasets:

TnT

/

CodeNet4Repair

like 0

s981157252

p02747

u441694890

Wrong Answer

A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string.

hitachi = list(input()) if len(hitachi) % 2 == 1: print("NO") else: H = True for i in range(len(hitachi)): if i % 2 == 0: if hitachi[i] != "h": H = False else: if hitachi[i] != "i": H = False if H == True: print("YES") else: print("NO")

s000869398

Accepted

hitachi = list(input()) if len(hitachi) % 2 == 1: print("No") else: H = True for i in range(len(hitachi)): if i % 2 == 0: if hitachi[i] != "h": H = False else: if hitachi[i] != "i": H = False if H == True: print("Yes") else: print("No")

s787138982

p03860

u477114517

Wrong Answer

Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.

s = input() print('s[0]' + 's[8]' + 's[10]')

s364130835

Accepted

s = input() print('A' + s[8] + 'C')

s288479957

p03719

u137726327

Wrong Answer

You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.

# -*- coding: utf-8 -*- #a = int(input()) lists=list(map(int,input().split())) #s = input() #print("{} {}".format(a+b+c, s)) if lists[0]<=lists[2] and lists[1]>=lists[2]: print("YES") else: print("NO")

s263448868

Accepted

# -*- coding: utf-8 -*- #a = int(input()) lists=list(map(int,input().split())) #s = input() #print("{} {}".format(a+b+c, s)) if lists[0]<=lists[2] and lists[1]>=lists[2]: print("Yes") else: print("No")

s798408909

p02260

u144068724

Wrong Answer

Write a program of the Selection Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: SelectionSort(A) 1 for i = 0 to A.length-1 2 mini = i 3 for j = i to A.length-1 4 if A[j] < A[mini] 5 mini = j 6 swap A[i] and A[mini] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 6 of the pseudocode in the case where i ≠ mini.

def selectionSort(A,N): count = 0 for i in range(N): minj = i for j in range(i,N): if A[j] < A[minj]: minj = j A[i],A[minj] = A[minj],A[i] count += 1 return (count) if __name__ == '__main__': n = int(input()) data = [int(i) for i in range(n)] result = selectionSort(data,n) print(" ".join(map(str,data))) print(result)

s116487642

Accepted

def selection_Sort(A,N): count = 0 for i in range(N): minj = i for j in range(i,N): if A[j] < A[minj]: minj = j if i != minj: A[i],A[minj] = A[minj],A[i] count += 1 return (count) if __name__ == '__main__': n = int(input()) data = [int(i) for i in input().split()] result = selection_Sort(data,n) print(" ".join(map(str,data))) print(result)

s842473348

p02742

u572026348

Wrong Answer

We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:

def main(): h, w = list(map(int, input().split())) if (h==1 or w==1): print(1) elif (h%2==0 or w%2==0): print((h*w)/2) else: print((h*w)//2+1) if __name__ == '__main__': main()

s579693119

Accepted

def main(): h, w = list(map(int, input().split())) if (h==1 or w==1): print(1) elif (h%2==0 or w%2==0): print(int((h*w)/2)) else: print(int((h*w)//2+1)) if __name__ == '__main__': main()

s846390828

p03759

u051496905

Wrong Answer

Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.

a , b , c = map(int,input().split()) if b - a == c - b: print("Yes") else: print("No")

s350389061

Accepted

a , b , c = map(int,input().split()) if b - a == c - b: print("YES") else: print("NO")

s409995795

p03455

u922769680

Wrong Answer

AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.

a, b=map(int, input().split()) c=a*b # print(c) d=c%2 if d==0: print("EVEN") else: print("ODD")

s198662447

Accepted

a, b=map(int, input().split()) c=a*b # print(c) d=c%2 if d==0: print("Even") else: print("Odd")

s010934059

p03796

u075595666

Wrong Answer

Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.

N = int(input()) x = 1 for i in range(N): x = x*i if x >= 10**9+7: x = x-10**9-7 print(x%(10**+7))

s571388767

Accepted

N = int(input()) x = 1 for i in range(1,N+1): x = x*i x = x%(10**9+7) print(x%(10**9+7)) #457992974

s065377030

p03759

u277802731

Wrong Answer

Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.

#58a a,b,c = map(int,input().split()) print('Yes' if b-a==c-b else 'No')

s237790901

Accepted

#58a a,b,c = map(int,input().split()) print('YES' if b-a==c-b else 'NO')

s187702618

p03997

u667024514

Wrong Answer

You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.

a = int(input()) b = int(input()) c = int(input()) print((a+b)*c/2)

s321805022

Accepted

a = int(input()) b = int(input()) c = int(input()) ans = int((a+b)*c/2) print(ans)

s523274853

p03448

u100873497

Wrong Answer

You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.

a=int(input()) b=int(input()) c=int(input()) x=int(input()) ans=0 for i in range(0,a): for j in range(0,b): for k in range(0,c): if 500*i+100*j+50*k==x: ans+=1 print(ans)

s041284763

Accepted

a=int(input()) b=int(input()) c=int(input()) x=int(input()) ans=0 for i in range(a+1): for j in range(b+1): for k in range(c+1): if 500*i+100*j+50*k==x: ans+=1 print(ans)

s148271234

p03852

u101680358

Wrong Answer

Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.

S = input() S = S[::-1] T = ['dream'[::-1],'dreamer'[::-1],'erase'[::-1],'eraser'[::-1]] size = 0 while size < len(S): if S[size:size+5] == T[0] or S[size:size+5] == T[2]: size += 5 elif S[size:size+6] == T[3]: size += 6 elif S[size:size+7] == T[1]: size += 7 else : print('No') exit() print('Yes')

s255330110

Accepted

c = input() if c in 'aiueo': print('vowel') else: print('consonant')

s772915096

p02414

u656153606

Wrong Answer

Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula: \\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\] where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.

#Input n,m,l = map(int,input().split()) A = [list(map(int,input().split())) for i in range(n)] B = [list(map(int,input().split())) for i in range(m)] c = [list(sum([A[i][k] * B[k][j] for k in range(m)]) for j in range(l)) for i in range(n)] #Output for i in c: print(*c)

s813728885

Accepted

#Input n,m,l = map(int,input().split()) A = [list(map(int,input().split())) for i in range(n)] B = [list(map(int,input().split())) for i in range(m)] c = [list(sum([A[i][k] * B[k][j] for k in range(m)]) for j in range(l)) for i in range(n)] #Output for i in c: print(*i)

s446533441

p04043

u006425112

Wrong Answer

Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.

import sys a = map(int, sys.stdin.readline().split()) seven = 0 five = 0 for i in a: if i == seven: seven += 1 else: five += 1 if seven == 1 and five == 2: print("YES") else: print("NO")

s770648047

Accepted

import sys a = map(int, sys.stdin.readline().split()) seven = 0 five = 0 for i in a: if i == 7: seven += 1 else: five += 1 if seven == 1 and five == 2: print("YES") else: print("NO")

s488874730

p03197

u511379665

Wrong Answer

There is an apple tree that bears apples of N colors. The N colors of these apples are numbered 1 to N, and there are a_i apples of Color i. You and Lunlun the dachshund alternately perform the following operation (starting from you): * Choose one or more apples from the tree and eat them. Here, the apples chosen at the same time must all have different colors. The one who eats the last apple from the tree will be declared winner. If both you and Lunlun play optimally, which will win?

n=int(input()) cnt=float('inf') for i in range(n): a=int(input()) cnt=min(cnt,a) tm=n^cnt if tm%2==0: print("first") else: print("second")

s490862421

Accepted

n=int(input()) a=[int(input()) for i in range(n)] print( "second" if all(a[i]%2==0 for i in range(n)) else "first" )

s231122036

p03624

u750651325

Wrong Answer

You are given a string S consisting of lowercase English letters. Find the lexicographically (alphabetically) smallest lowercase English letter that does not occur in S. If every lowercase English letter occurs in S, print `None` instead.

S = input() lll = "abcdefghijklmnopqrstuvwxyz" for i in range(26): if lll[i] in S: print(lll[i]) exit() else: pass print("None")

s630725980

Accepted

S = input() lll = "abcdefghijklmnopqrstuvwxyz" for i in range(26): if lll[i] in S: pass else: print(lll[i]) exit() print("None")

s571096310

p03140

u432333240

Wrong Answer

You are given three strings A, B and C. Each of these is a string of length N consisting of lowercase English letters. Our objective is to make all these three strings equal. For that, you can repeatedly perform the following operation: * Operation: Choose one of the strings A, B and C, and specify an integer i between 1 and N (inclusive). Change the i-th character from the beginning of the chosen string to some other lowercase English letter. What is the minimum number of operations required to achieve the objective?

N = int(input()) A = input() B = input() C = input() target_list = [] for a, b, c in zip(A, B, C): if a==b and b==c: target_list.append(3) elif a==b or a==c: target_list.append(2) else: target_list.append(1) print(N*3 - sum(target_list))

s835533391

Accepted

N = int(input()) A = input() B = input() C = input() target_list = [] for a, b, c in zip(A, B, C): if a==b and b==c and c==a: target_list.append(3) elif a==b or b==c or c==a: target_list.append(2) else: target_list.append(1) print(N*3 - sum(target_list))

s858146500

p03693

u189385406

Wrong Answer

AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?

r,g,b=map(int,input().split()) if(r+g+b)%4 == 0: print('YES') else: print('NO')

s607115630

Accepted

g=int(input().replace(' ', '')) if g%4 == 0: print('YES') else : print('NO')

s269130469

p03494

u842838534

Wrong Answer

There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.

def how_many_times_divisible(n): ans = 0 while n % 2 == 0: n = n / 2 ans += 1 return ans n = input() a = list(map(int,input().split())) a_divided = list(map(how_many_times_divisible,a)) ans = max(a_divided) print(ans)

s667128286

Accepted

def how_many_times_divisible(n): ans = 0 while n % 2 == 0: n = n / 2 ans += 1 return ans n = input() a = list(map(int,input().split())) a_divided = list(map(how_many_times_divisible,a)) ans = min(a_divided) print(ans)

s088147588

p02613

u219937318

Wrong Answer

Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.

N=int(input()) ac=0 wa=0 tle=0 re=0 S=[input() for i in range(N)] for s in S: if s =='AC': ac=ac+1 elif s == 'WA': wa = wa + 1 elif s == 'TLE': tle = tle + 1 else: re = re + 1 print("AC ×",ac) print("WA ×",wa) print("TLE ×",tle) print("RE ×",re)

s846882605

Accepted

N=int(input()) ac=0 wa=0 tle=0 re=0 S=[input() for i in range(N)] for s in S: if s =='AC': ac=ac+1 elif s == 'WA': wa = wa + 1 elif s == 'TLE': tle = tle + 1 else: re = re + 1 print("AC x",ac) print("WA x",wa) print("TLE x",tle) print("RE x",re)

s927827482

p03448

u144072139

Wrong Answer

You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.

A = int(input()) B = int(input()) C = int(input()) X = int(input()) counts = 0 for a in range(A+1): for b in range(B+1): for c in range(C+1): if (A*500 + B*100 + C*50)==X: counts += 1 print(counts)

s069107916

Accepted

A = int(input()) B = int(input()) C = int(input()) X = int(input()) counts = 0 for a in range(A+1): for b in range(B+1): for c in range(C+1): if (a*500 + b*100 + c*50)==X: counts += 1 print(counts)

s666398952

p03573

u353652911

Wrong Answer

You are given three integers, A, B and C. Among them, two are the same, but the remaining one is different from the rest. For example, when A=5,B=7,C=5, A and C are the same, but B is different. Find the one that is different from the rest among the given three integers.

a,b,c=map(int,input().split()) print("c" if a==b else "b" if a==c else "a")

s387363769

Accepted

a,b,c=map(int,input().split()) print(c if a==b else b if a==c else a)

s089647804

p02694

u158703648

Wrong Answer

Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?

X = int(input()) money = 100 count = 0 while money <= X: money = int(money*1.01) count +=1 print(count)

s369974691

Accepted

X = int(input()) money = 100 count = 0 while money < X: money = int(money*1.01) count +=1 print(count)

s798529880

p02972

u241190159

Wrong Answer

There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.

def main(): N = int(input()) a = {i: inp for i, inp in enumerate(map(int, input().split()), 1)} M = 0 b = {i : 0 for i in range(1, N+1)} for i in range(N, 0, -1): sum_multiple = 0 cursor = 2 while i * cursor <= N: sum_multiple += a[i * cursor] cursor += 1 if a[i] == 1: if sum_multiple % 2: b[i] = 0 else: b[i] = 1 M += 1 else: if sum_multiple % 2: b[i] = 1 M += 1 else: b[i] = 0 b = [str(value) for value in b.values()] b = ' '.join(b) print(M) print(b) if __name__=="__main__": main()

s088222757

Accepted

def main(): N = int(input()) a = {i: inp for i, inp in enumerate(map(int, input().split()), 1)} b = {i : 0 for i in range(1, N+1)} ret = [] for i in range(N, 0, -1): sum_multiple = 0 cursor = 2 while i * cursor <= N: sum_multiple += b[i * cursor] cursor += 1 if a[i] == 1: if sum_multiple % 2 == 0: b[i] = 1 ret.append(i) else: if sum_multiple % 2: b[i] = 1 ret.append(i) M = len(ret) print(M) if M: ret = ' '.join([str(i) for i in ret]) print(ret) if __name__=="__main__": main()

s508776757

p03338

u698479721

Time Limit Exceeded

You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.

N = int(input()) s = input() i = 1 li = [] charas = 'abcdefghijklmnopqrstuvwxyz' while i < N: m = 0 s1 = s[0:i] s2 = s[i:] for chara in charas: if chara in s1 and chara in s2: m += 1 li.append(m) print(max(li))

s023582407

Accepted

N = int(input()) s = input() i = 1 li = [] charas = 'abcdefghijklmnopqrstuvwxyz' while i < N: m = 0 s1 = s[0:i] s2 = s[i:] for chara in charas: if chara in s1 and chara in s2: m += 1 li.append(m) i += 1 print(max(li))

s678799296

p03360

u214434454

Wrong Answer

There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations?

num = list(map(int,input().split())) k = int(input()) num.sort() print(sum(num[:2])+num[-1]**k)

s272161587

Accepted

num = list(map(int,input().split())) k = int(input()) num.sort() print(sum(num[:2])+num[-1] * 2**k)

s638913292

p02796

u670180528

Wrong Answer

In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep.

import sys input=sys.stdin.buffer.readline L=[] for _ in range(int(input())): x,l=map(int,input().split()) L.append((x-l,x+l)) L.sort(key=lambda x:x[1]) ans=0 cur=-1 for a,b in L: if a>=cur: cur=b ans+=1 print(ans)

s641860198

Accepted

import sys input=sys.stdin.buffer.readline L=[] for _ in range(int(input())): x,l=map(int,input().split()) L.append((x-l,x+l)) L.sort(key=lambda x:x[1]) ans=0 cur=-11111111111111111 for a,b in L: if a>=cur: cur=b ans+=1 print(ans)

s200353787

p03719

u726439578

Wrong Answer

You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.

a,b,c=map(int,input().split()) if a<=c<=b: print("YES") else: print("NO")

s413615375

Accepted

a,b,c=map(int,input().split()) if a<=c<=b: print("Yes") else: print("No")

s306777942

p03377

u129978636

Wrong Answer

There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.

A, B, X = map( int, input().split()) D = A + B if( D <= X): print('Yes') if( X <= D): print('No')

s314726613

Accepted

A, B, X = map( int, input().split()) D = A + B if( X <= D): if( A <= X): print('YES') else: print('NO') else: print('NO')

s131064436

p03470

u257541375

Wrong Answer

An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?

n = int(input()) a = [] for i in range(n): getn = int(input()) a.append(getn) b = set(a) print(len(a)-len(b)+1)

s906904309

Accepted

n = int(input()) a = [] for i in range(n): getn = int(input()) a.append(getn) b = set(a) print(len(b))

s515560482

p02613

u729272006

Wrong Answer

Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.

n = input() n = int(n) ac = 0 wa = 0 tle = 0 re = 0 for i in range(n): wk = input() if wk == "AC": ac +=1 elif wk == "WA": wa += 1 elif wk == "TLE": tle += 1 else : re += 1 print("AC × {0}".format(ac)) print("WA × {0}".format(wa)) print("TLE × {0}".format(tle)) print("RE × {0}".format(re))

s404755353

Accepted

n = input() n = int(n) ac = 0 wa = 0 tle = 0 re = 0 for i in range(n): wk = input() if wk == "AC": ac +=1 elif wk == "WA": wa += 1 elif wk == "TLE": tle += 1 else : re += 1 print("AC x {0}".format(ac)) print("WA x {0}".format(wa)) print("TLE x {0}".format(tle)) print("RE x {0}".format(re))

s666834220

p04044

u334222621

Wrong Answer

Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.

N, L=map(int, input().split()) lst = [] ans = str() for i in range(N): lst.append(input()) lst.sort for i in range(len(lst)): ans += lst[i] print(ans)

s407173672

Accepted

N, L=map(int, input().split()) lst = [] ans = "" for i in range(N): lst.append(input()) lst.sort() for i in range(len(lst)): ans += lst[i] print(ans)

s636458326

p03578

u085717502

Wrong Answer

Rng is preparing a problem set for a qualification round of CODEFESTIVAL. He has N candidates of problems. The difficulty of the i-th candidate is D_i. There must be M problems in the problem set, and the difficulty of the i-th problem must be T_i. Here, one candidate of a problem cannot be used as multiple problems. Determine whether Rng can complete the problem set without creating new candidates of problems.

#!/usr/bin/env python # coding: utf-8 # In[7]: N = int(input()) D = list(map(int, input().split())) M = int(input()) T = list(map(int, input().split())) # In[8]: for i,t in enumerate(T): if t in D: D.pop(D.index(t)) else: print("No") break else: print("Yes") # In[ ]:

s794570265

Accepted

#!/usr/bin/env python # coding: utf-8 # In[9]: import collections # In[32]: N = int(input()) D = list(map(int, input().split())) M = int(input()) T = list(map(int, input().split())) # In[34]: t_count = collections.Counter(T) d_count = collections.Counter(D) for key,c in t_count.items(): if key in d_count.keys(): if d_count[key] < c: print("NO") break else: print("NO") break else: print("YES") # In[ ]:

s195481500

p02697

u117541450

Wrong Answer

You are going to hold a competition of one-to-one game called AtCoder Janken. _(Janken is the Japanese name for Rock-paper-scissors.)_ N players will participate in this competition, and they are given distinct integers from 1 through N. The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive). You cannot assign the same integer to multiple playing fields. The competition consists of N rounds, each of which proceeds as follows: * For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there. * Then, each player adds 1 to its integer. If it becomes N+1, change it to 1. You want to ensure that no player fights the same opponent more than once during the N rounds. Print an assignment of integers to the playing fields satisfying this condition. It can be proved that such an assignment always exists under the constraints given.

N,M = map(int, input().split()) for i in range(M): a=i+1 b=2*M-i print(a,b)

s994746389

Accepted

N,M = map(int, input().split()) M1 = M//2 M2 = M-M1 for i in range(M1): a=i+1 b=2*M1+1-i print(a,b) for i in range(M2): a=i+1 b=2*M2-i print(2*M1+1 + a,2*M1+1 + b)

s063300600

p03657

u667084803

Wrong Answer

Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.

a,b=map(int,input().split()) if a%3==0 or b%3==0 or a+b%3==0: print("Possible") else: print("Impossible")

s868566094

Accepted

a,b=map(int,input().split()) if a%3==0 or b%3==0 or (a+b)%3==0: print("Possible") else: print("Impossible")

s916576940

p03546

u639343026

Wrong Answer

Joisino the magical girl has decided to turn every single digit that exists on this world into 1. Rewriting a digit i with j (0≤i,j≤9) costs c_{i,j} MP (Magic Points). She is now standing before a wall. The wall is divided into HW squares in H rows and W columns, and at least one square contains a digit between 0 and 9 (inclusive). You are given A_{i,j} that describes the square at the i-th row from the top and j-th column from the left, as follows: * If A_{i,j}≠-1, the square contains a digit A_{i,j}. * If A_{i,j}=-1, the square does not contain a digit. Find the minimum total amount of MP required to turn every digit on this wall into 1 in the end.

import sys input=sys.stdin.readline import numpy as np from scipy.sparse.csgraph import shortest_path h,w=map(int,input().split()) c=[list(map(int,input().split())) for i in range(10)] a=[list(map(int,input().split())) for i in range(h)] graph = np.zeros((10,10)) for i in range(10): for j in range(10): graph[i,j] = c[i][j] dist = shortest_path(graph,directed=True).astype(int) print(dist) res=0 for i in range(h): #print(a[i]) for j in range(w): if a[i][j]==-1:continue res+=dist[a[i][j]][1] #print(dist[a[i][j]][1]) print(res)

s263194223

Accepted

import sys input=sys.stdin.readline import numpy as np from scipy.sparse.csgraph import shortest_path h,w=map(int,input().split()) c=[list(map(int,input().split())) for i in range(10)] a=[list(map(int,input().split())) for i in range(h)] graph = np.zeros((10,10)) for i in range(10): for j in range(10): graph[i,j] = c[i][j] dist = shortest_path(graph,directed=True).astype(int) #print(dist) res=0 for i in range(h): #print(a[i]) for j in range(w): if a[i][j]==-1:continue res+=dist[a[i][j]][1] #print(dist[a[i][j]][1]) print(res)

s841114623

p02419

u823030818

Wrong Answer

Write a program which reads a word W and a text T, and prints the number of word W which appears in text T T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.

target = input() count = 0 while True: line = input() if line == 'END_OF_TEXT': break for word in line.split(): if word.lower == target.lower: count += 1 print(count)

s680689104

Accepted

target = input() count = 0 while True: line = input() if line == 'END_OF_TEXT': break for word in line.split(): if word.lower() == target.lower(): count += 1 print(count)

s228256040

p03069

u994988729

Wrong Answer

There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.

n=int(input()) s=input() ndot=0 nx=0 mx=10000000 for i in range(n+1): if i==0: ndot=sum([1 for i in s if i=="."]) nx=0 else: if s[i-1]==".": ndot-=1 elif s[i-1]=="#": nx+=1

s605276996

Accepted

n=int(input()) s=input() ndot=0 nx=0 mx=10000000 for i in range(n+1): if i==0: ndot=sum([1 for i in s if i=="."]) nx=0 else: if s[i-1]==".": ndot-=1 elif s[i-1]=="#": nx+=1 mx=min(mx,nx+ndot) print(mx)

s750651483

p03369

u180704972

Wrong Answer

In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.

S = input() Ramen = 700 if S[0] == '○': Ramen += 100 if S[1] == '○': Ramen += 100 if S[2] == '○': Ramen += 100 print(Ramen)

s795348735

Accepted

S = input() Ramen = 700 if S[0] == 'o': Ramen += 100 if S[1] == 'o': Ramen += 100 if S[2] == 'o': Ramen += 100 print(Ramen)

s652167519

p03760

u324549724

Wrong Answer

Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.

a = input() b = input() ans = [] count = 0 while True: if count == len(a): print(ans) exit() ans += a[count] if count == len(b): print(ans) exit() ans += b[count] count += 1

s192944187

Accepted

a = input() b = input() ans = "" count = 0 while True: if count == len(a): print(ans) exit() ans += a[count] if count == len(b): print(ans) exit() ans += b[count] count += 1

s183415964

p03407

u779728630

Wrong Answer

An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.

a, b, c = map(int, input().split()) print("Yes") if a + b <= c else print("No")

s864904732

Accepted

a, b, c = map(int, input().split()) print("Yes") if a + b >= c else print("No")

s710420700

p03889

u426964396

Wrong Answer

You are given a string S consisting of letters `b`, `d`, `p` and `q`. Determine whether S is a _mirror string_. Here, a mirror string is a string S such that the following sequence of operations on S results in the same string S: 1. Reverse the order of the characters in S. 2. Replace each occurrence of `b` by `d`, `d` by `b`, `p` by `q`, and `q` by `p`, simultaneously.

s=input().split() if s[0]=='d' and s[3]=='b': if s[1]=='p' and s[2]=='q': print('Yes') elif s[1]=='q' and s[2]=='p': print('Yes') else: print('No') elif s[0]=='b' and s[3]=='d': if s[1]=='p' and s[2]=='q': print('Yes') elif s[1]=='q' and s[2]=='p': print('Yes') else: print('No') else: print('No')

s489941984

Accepted

dict={'b':'d','d':'b','p':'q','q':'p'} s= input() t=''.join(map(lambda x: dict[x],list(s[::-1]))) if s==t: print("Yes") else : print("No")

s947098052

p03457

u933622697

Wrong Answer

AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.

n = int(input()) for i in range(n): t, x, y = map(int, input().split()) # input() when you access each trial if x + y < t or (x + y + t) % 2: print("No") exit() print("Yes")

s785750589

Accepted

n = int(input()) for i in range(n): t, x, y = map(int, input().split()) if t < x + y or t % 2 != (x + y) % 2: print("No") exit() print("Yes")

s832854624

p02694

u827553608

Wrong Answer

Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?

try: x=int(input()) o=100 count=0 while int(o)!=x: o=o+(o*(1/100)) print(o) count+=1 print(count) except: pass

s321023539

Accepted

try: x=int(input()) o=100 count=0 while o<x: o=int(o+(o*(1/100))) count+=1 print(count) except: pass

s449676898

p03962

u917558625

Wrong Answer

AtCoDeer the deer recently bought three paint cans. The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c. Here, the color of each paint can is represented by an integer between 1 and 100, inclusive. Since he is forgetful, he might have bought more than one paint can in the same color. Count the number of different kinds of colors of these paint cans and tell him.

a=list(map(int,input().split())) a.sort() if a[0]==a[1]==a[2]: print(1) else: if a[0]==a[1] or a[1]==a[2] or a[2]==a[0]: print(2) else: print(1)

s286476646

Accepted

a=list(map(int,input().split())) a.sort() if a[0]==a[1]==a[2]: print(1) else: if a[0]==a[1] or a[1]==a[2] or a[2]==a[0]: print(2) else: print(3)

s352499178

p02613

u689710606

Wrong Answer

Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.

n = int(input()) c0 = 0 c1 = 0 c2 = 0 c3 = 0 for i in range(n): s = input() if s == "AC": c0 += 1 elif s == "WA": c1 += 1 elif s == "TLE": c2 += 1 elif s == "RE": c3 += 1 else: exit() print(f"AC × {c0}") print(f"WA × {c1}") print(f"TLE × {c2}") print(f"RE × {c3}")

s065802632

Accepted

n = int(input()) c0 = 0 c1 = 0 c2 = 0 c3 = 0 for i in range(n): s = input() if s == "AC": c0 += 1 elif s == "WA": c1 += 1 elif s == "TLE": c2 += 1 elif s == "RE": c3 += 1 else: exit() print(f"AC x {c0}") print(f"WA x {c1}") print(f"TLE x {c2}") print(f"RE x {c3}")

s183415211

p03470

u830462928

Wrong Answer

An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?

n = int(input()) l = [] for i in range(0, n): l.append(int(input())) len(set(l))

s865170047

Accepted

n = int(input()) l = [] for i in range(0, n): l.append(int(input())) print(len(set(l)))

s827771762

p03544

u518455500

Wrong Answer

It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)

n = int(input()) l0 = 2 l1 = 1 li = 0 for i in range(2,n): li = l0 + l1 l0 = l1 l1 = li print(li)

s728350404

Accepted

n = int(input()) l0 = 2 l1 = 1 li = 0 if n == 0: print(l0) elif n == 1: print(l1) else: for i in range(2,n+1): li = l0 + l1 l0 = l1 l1 = li print(li)

s536941921

p03163

u437727817

Wrong Answer

There are N items, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), Item i has a weight of w_i and a value of v_i. Taro has decided to choose some of the N items and carry them home in a knapsack. The capacity of the knapsack is W, which means that the sum of the weights of items taken must be at most W. Find the maximum possible sum of the values of items that Taro takes home.

import numpy as np n,w = map(int,input().split()) WV = [list(map(int,input().split()))for _ in range(n)] dp = np.zeros(w+1,int) for i in range(n): W,V = WV[i] #print(str(dp)+":"+str(dp[W:])+":"+str(dp[:-W])) dp[W:] = np.maximum(dp[W:],dp[:-W]+V) print(str(dp)) print(dp[-1])

s802561193

Accepted

import numpy as np n,w = map(int,input().split()) WV = [list(map(int,input().split()))for _ in range(n)] dp = np.zeros(w+1,int) for i in range(n): W,V = WV[i] #print(str(dp)+":"+str(dp[W:])+":"+str(dp[:-W])) dp[W:] = np.maximum(dp[W:],dp[:-W]+V) #print(str(dp)) print(dp[-1])

s133846910

p03854

u797798686

Wrong Answer

You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.

from sys import stdin data = stdin.readline().rstrip() def shorten_phrase(x): l = len(x) if x.startswith("eraser"): x = x[6:] elif x.startswith("erase"): x = x[5:] elif x.startswith("dreamera"): x = x[5:] elif x.startswith("dreamer"): x = x[7:] elif x.startswith("dream"): x = x[5:] m = len(x) return l != m while shorten_phrase(data): if data.startswith("eraser"): data = data[6:] elif data.startswith("erase"): data = data[5:] elif data.startswith("dreamera"): data = data[5:] elif data.startswith("dreamer"): data = data[7:] elif data.startswith("dream"): data = data[5:] if len(data) == 0: print("YES") else: print("NO") print(data)

s533610972

Accepted

from sys import stdin data = stdin.readline().rstrip() def shorten_phrase(x): l = len(x) if x.startswith("eraser"): x = x[6:] elif x.startswith("erase"): x = x[5:] elif x.startswith("dreamera"): x = x[5:] elif x.startswith("dreamer"): x = x[7:] elif x.startswith("dream"): x = x[5:] m = len(x) return l != m while shorten_phrase(data): if data.startswith("eraser"): data = data[6:] elif data.startswith("erase"): data = data[5:] elif data.startswith("dreamera"): data = data[5:] elif data.startswith("dreamer"): data = data[7:] elif data.startswith("dream"): data = data[5:] if len(data) == 0: print("YES") else: print("NO")

s691094972

p01085

u105296105

Wrong Answer

The International Competitive Programming College (ICPC) is famous for its research on competitive programming. Applicants to the college are required to take its entrance examination. The successful applicants of the examination are chosen as follows. * The score of any successful applicant is higher than that of any unsuccessful applicant. * The number of successful applicants _n_ must be between _n_ min and _n_ max, inclusive. We choose _n_ within the specified range that maximizes the _gap._ Here, the _gap_ means the difference between the lowest score of successful applicants and the highest score of unsuccessful applicants. * When two or more candidates for _n_ make exactly the same _gap,_ use the greatest _n_ among them. Let's see the first couple of examples given in Sample Input below. In the first example, _n_ min and _n_ max are two and four, respectively, and there are five applicants whose scores are 100, 90, 82, 70, and 65. For _n_ of two, three and four, the gaps will be 8, 12, and 5, respectively. We must choose three as _n_ , because it maximizes the gap. In the second example, _n_ min and _n_ max are two and four, respectively, and there are five applicants whose scores are 100, 90, 80, 75, and 65. For _n_ of two, three and four, the gap will be 10, 5, and 10, respectively. Both two and four maximize the gap, and we must choose the greatest number, four. You are requested to write a program that computes the number of successful applicants that satisfies the conditions.

import sys while True: m, nmin, nmax = [int(i) for i in input().split()] if m == 0: sys.exit() p = [int(input()) for i in range(m)] ansnum = 1000000 ans = 0 for i in range(nmin-1,nmax): if ansnum >= p[i]-p[i+1]: ans = i ansnum = min(ansnum,p[i]-p[i+1]) print(ans)

s954885269

Accepted

import sys anslist = [] while True: m, nmin, nmax = [int(i) for i in input().split()] if m == 0: for i in anslist: print(i) sys.exit() p = [int(input()) for i in range(m)] ansnum = 0 ans = 0 for i in range(nmin-1,nmax): if ansnum <= p[i]-p[i+1] and not p[i]-p[i+1]==0: ans = i ansnum = p[i]-p[i+1] anslist.append(ans+1)

s604062312

p04043

u396858476

Wrong Answer

Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.

a, b, c = map(int, input().split()) if a == b and a == 5: if c == 7: print("Yes") else: print("No") elif b == c and b == 5: if a == 7: print("Yes") else: print("No") elif a == c and c == 5: if b == 7: print("Yes") else: print("No") else: print("No")

s759324128

Accepted

a, b, c = map(int, input().split()) if a == b and a == 5: if c == 7: print("YES") else: print("NO") elif b == c and b == 5: if a == 7: print("YES") else: print("NO") elif a == c and c == 5: if b == 7: print("YES") else: print("NO") else: print("NO")

s705297988

p03130

u782098901

Wrong Answer

There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads. Determine if we can visit all the towns by traversing each of the roads exactly once.

ab = [[], [], [], [], []] for i in range(3): a, b = map(int, input().split()) ab[a].append(b) ab[b].append(a) ways = [False] * 4 def f(now, count): if count > 4: return False for n in ab[now]: if ways[n - 1]: continue ways[n - 1] = True if all(ways): print("Yes") exit() f(n, count + 1) ways[n - 1] = False for i in range(1, 5): ways = [False] * 4 ways[i - 1] = True f(i, 1) ways[i - 1] = False print("NO")

s567491016

Accepted

c = [0] * 5 for _ in range(3): a, b = map(int, input().split()) c[a] += 1 c[b] += 1 for i in range(1, 5): if c[i] > 2: print("NO") exit() print("YES")

s316782358

p03006

u945418216

Wrong Answer

There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i). We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation: * Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1. Find the minimum total cost required to collect all the balls when we optimally choose p and q.

N = int(input()) XY = sorted([list(map(int,input().split())) for _ in range(N)]) PQ = [] for i in range(N): prev = XY[i] for j in range(N): if i==j: continue next = XY[j] p = next[0] - prev[0] q = next[1] - prev[1] PQ.append([p,q]) ans = 0 for pq in PQ: ans = max(ans, N-PQ.count(pq)-1) print(ans)

s224702804

Accepted

N = int(input()) XY = sorted([list(map(int,input().split())) for _ in range(N)]) PQ = [] for i in range(N): prev = XY[i] for j in range(N): if i!=j: next = XY[j] p = next[0] - prev[0] q = next[1] - prev[1] PQ.append((p,q)) ans = 1 for pq in PQ: ans = max(ans, PQ.count(pq)) print(N - ans if N>1 else 1)

s179511175

p03575

u887207211

Wrong Answer

You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges. The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i. An edge whose removal disconnects the graph is called a _bridge_. Find the number of the edges that are bridges among the M edges.

N, M = map(int,input().split()) v = [i for i in range(1,N+1)] e = [[]]*(N+1) for i in range(M): a, b = map(int,input().split()) e[a] = e[a] + [b] e[b] = e[b] + [a] def dfs(v, e): for num in v: if(len(e[num]) == 1): tmp_num = e[num][0] v.remove(num) e[num] = [] e[tmp_num].remove(num) print(e,v) return dfs(v, e)+1 return 0 print(dfs(v, e))

s506435017

Accepted

N, M = map(int,input().split()) v = list(range(1,N+1)) edges = [[] for _ in range(N+1)] for _ in range(M): a, b = map(int,input().split()) edges[a] += [b] edges[b] += [a] def dfs(v, e): for x in v: if(len(e[x]) == 1): t = e[x][0] e[x] = [] e[t].remove(x) v.remove(x) return dfs(v, e)+1 return 0 print(dfs(v, edges))

s855840218

p03401

u422552722

Wrong Answer

There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis. You planned a trip along the axis. In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0. However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i. You will visit the remaining spots as planned in the order they are numbered. You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned. For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.

n = int(input()) a = list(map(int, input().split(" "))) print(n, a) if n == 2: print(abs(a[0]) + abs(a[0]-a[1]) + abs(a[1])) else: for i in range(len(a)): b = a[:] b.remove(b[i]) #print(b) cost = abs(b[0]) for j in range(len(b)-1): #print(b[j],b[j+1]) cost += abs(b[j] - b[j+1]) cost += abs(b[-1]) print(cost)

s490068071

Accepted

n = int(input()) a = list(map(int, input().split(" "))) a.insert(0,0) a.append(0) total = 0 for i in range(len(a)-1): total += abs(a[i] - a[i+1]) for j in range(1, len(a)-1): print(total - abs(a[j-1] - a[j]) - abs(a[j] - a[j+1]) + abs(a[j+1]- a[j-1]))

s894090473

p03607

u513900925

Wrong Answer

You are playing the following game with Joisino. * Initially, you have a blank sheet of paper. * Joisino announces a number. If that number is written on the sheet, erase the number from the sheet; if not, write the number on the sheet. This process is repeated N times. * Then, you are asked a question: How many numbers are written on the sheet now? The numbers announced by Joisino are given as A_1, ... ,A_N in the order she announces them. How many numbers will be written on the sheet at the end of the game?

N = int(input()) A = list(int(input()) for _ in range(N)) A.sort() print(A) count = 0 test = -1 stack = 0 for i in range(N): if test != A[i]: count += 1 stack = 1 test = A[i] else: if stack == 1: count = count - 1 stack = 0 else: stack = 1 count += 1 print(count)

s074456891

Accepted

N = int(input()) A = list(int(input()) for _ in range(N)) A.sort() count = 0 test = -1 stack = 0 for i in range(N): if test != A[i]: count += 1 stack = 1 test = A[i] else: if stack == 1: count = count - 1 stack = 0 else: stack = 1 count += 1 print(count)

s878437593

p03854

u502721867

Wrong Answer

You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.

s = input() s = s.replace("erase","").replace("eraser","").replace("dream","").replace("dreamer","") print(s) if s !="": print("No") else: print("Yes")

s218118702

Accepted

import re s = input() p = re.compile(r'^(dream|dreamer|erase|eraser)*(dream|dreamer|erase|eraser)$') x = p.search(s) if x : print("YES") else: print("NO")

s087926605

p03855

u677523557

Wrong Answer

There are N cities. There are also K roads and L railways, extending between the cities. The i-th road bidirectionally connects the p_i-th and q_i-th cities, and the i-th railway bidirectionally connects the r_i-th and s_i-th cities. No two roads connect the same pair of cities. Similarly, no two railways connect the same pair of cities. We will say city A and B are _connected by roads_ if city B is reachable from city A by traversing some number of roads. Here, any city is considered to be connected to itself by roads. We will also define _connectivity by railways_ similarly. For each city, find the number of the cities connected to that city by both roads and railways.

from operator import itemgetter import sys input = sys.stdin.readline N, K, L = map(int, input().split()) PQ = [list(map(int, input().split())) for _ in range(K)] RS = [list(map(int, input().split())) for _ in range(L)] class UnionFind(): def __init__(self, n): self.n = n self.root = [-1]*(n+1) self.rnk = [0]*(n+1) def Find_Root(self, x): if(self.root[x] < 0): return x else: self.root[x] = self.Find_Root(self.root[x]) return self.root[x] def Unite(self, x, y): x = self.Find_Root(x) y = self.Find_Root(y) if(x == y): return elif(self.rnk[x] > self.rnk[y]): self.root[x] += self.root[y] self.root[y] = x else: self.root[y] += self.root[x] self.root[x] = y if(self.rnk[x] == self.rnk[y]): self.rnk[y] += 1 def isSameGroup(self, x, y): return self.Find_Root(x) == self.Find_Root(y) def Count(self, x): return -self.root[self.Find_Root(x)] A = UnionFind(N) for p, q in PQ: A.Unite(p, q) B = UnionFind(N) for r,s in RS: B.Unite(r, s) C = [] for i in range(1, N+1): a = A.root[i] if A.root[i] > 0 else i b = B.root[i] if B.root[i] > 0 else i C.append([i, a, b]) C.sort(key=itemgetter(2)) C.sort(key=itemgetter(1)) D = [1 for _ in range(N)] pa, pb = 0, 0 c = 0 P = [] for p, a, b in C: P.append(p) if pa == a and pb == b: c += 1 D[p-1] += c else: c = 0 pa, pb = a, b #print(D) #print(P) Q = P[::-1] #print(Q) for i, p in enumerate(Q): a = D[p-1] #print(i, p, a, pa) if i == 0: l = a pa = a continue if pa != 1 or a != 1: l = max(l, a) D[p-1] = l else: l = 1 pa = a for a in D: print(a, end=' ') print()

s733956797

Accepted

class UnionFind(): def __init__(self, n): self.n = n self.root = [-1]*(n+1) self.rnk = [0]*(n+1) def Find_Root(self, x): if(self.root[x] < 0): return x else: self.root[x] = self.Find_Root(self.root[x]) return self.root[x] def Unite(self, x, y): x = self.Find_Root(x) y = self.Find_Root(y) if(x == y): return elif(self.rnk[x] > self.rnk[y]): self.root[x] += self.root[y] self.root[y] = x else: self.root[y] += self.root[x] self.root[x] = y if(self.rnk[x] == self.rnk[y]): self.rnk[y] += 1 def isSameGroup(self, x, y): return self.Find_Root(x) == self.Find_Root(y) def Count(self, x): return -self.root[self.Find_Root(x)] N, K, L = map(int, input().split()) U1 = UnionFind(N) U2 = UnionFind(N) for _ in range(K): a, b = map(int, input().split()) U1.Unite(a, b) for _ in range(L): a, b = map(int, input().split()) U2.Unite(a, b) Roots = [0] dic = {} for n in range(1, N+1): r1 = U1.Find_Root(n) r2 = U2.Find_Root(n) r = r1*2*N + r2 Roots.append(r) if not r in dic.keys(): dic[r] = 1 else: dic[r] += 1 ans = [] for n in range(1, N+1): ans.append(dic[Roots[n]]) print(" ".join([str(a) for a in ans]))

s066686017

p03110

u211160392

Wrong Answer

Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?

N = int(input()) x = [] u = [] Y = 0 for i in range(N): s = [i for i in input().split()] print(s) x.append(float(s[0])) u.append(s[1]) if u[i] == 'BTC': Y += x[i]*380000 else: Y += x[i] print(Y)

s008883363

Accepted

N = int(input()) x = [] u = [] Y = 0 for i in range(N): s = [i for i in input().split()] x.append(float(s[0])) u.append(s[1]) if u[i] == 'BTC': Y += x[i]*380000 else: Y += x[i] print(Y)

s405617852

p03563

u045408189

Wrong Answer

Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it.

r=float(input()) g=float(input()) print(2*g-r)

s835519223

Accepted

r=float(input()) g=float(input()) print(int(2*g-r))

s080051567

p03854

u060736237

Wrong Answer

You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.

def main(): s = input() dream = 0 erase = 1 work = [] temp1 = 0 temp2 = 0 kind = -1 append = work.append for c in s: if c == 'd': if kind != -1: append([kind, temp2-temp1]) kind = dream temp1 = temp2 elif c == 's': if kind != -1: append([kind, temp2-temp1-3]) kind = erase temp1 = temp2-3 temp2 += 1 append([kind, len(s)-temp1]) result = [] append = result.append for i, j in work: if i == dream: if j <= 5: append("dream") else: append("dreamer") else: if j <= 5: append("erase") else: append("eraser") result = "Yes" if "".join(result)==s else "No" print(result) main()

s010213438

Accepted

def main(): s = input() dream = 0 erase = 1 work = [] temp1 = 0 temp2 = 0 kind = -1 append = work.append for c in s: if c == 'd': if kind != -1: append([kind, temp2-temp1]) kind = dream temp1 = temp2 elif c == 's': if kind != -1: append([kind, temp2-temp1-3]) kind = erase temp1 = temp2-3 temp2 += 1 append([kind, len(s)-temp1]) result = [] append = result.append for i, j in work: if i == dream: if j <= 5: append("dream") else: append("dreamer") else: if j <= 5: append("erase") else: append("eraser") result = "YES" if "".join(result)==s else "NO" print(result) main()

s044386462

p03693

u757324869

Wrong Answer

AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?

p = input().split() s = int(p[0] + p[1] + p[2]) print(s) if s % 4 == 0: print("YES") else: print("NO")

s721491366

Accepted

p = input().split() s = int(p[0])*100 + int(p[1])*10 + int(p[2]) if s % 4 == 0: print("YES") else: print("NO")

s863919733

p02613

u437351386

Wrong Answer

Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.

n=int(input()) s=[] for i in range(n): s.append(input()) print("AC"+" × "+str(s.count("AC"))) print("WA"+" × "+str(s.count("WA"))) print("TLE"+" × "+str(s.count("TLE"))) print("RE"+" × "+str(s.count("RE")))

s929212396

Accepted

n=int(input()) s=[] for i in range(n): s.append(input()) print("AC"+" x "+str(s.count("AC"))) print("WA"+" x "+str(s.count("WA"))) print("TLE"+" x "+str(s.count("TLE"))) print("RE"+" x "+str(s.count("RE")))

s615493984

p03228

u167908302

Wrong Answer

In the beginning, Takahashi has A cookies, and Aoki has B cookies. They will perform the following operation alternately, starting from Takahashi: * If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other person. Find the numbers of cookies Takahashi and Aoki respectively have after performing K operations in total.

#coding:utf-8 a, b, k = map(int, input().split()) for i in range(0, k): if i % 2 == 0: b += a // 2 a /= 2 else: a += b // 2 b /= 2 print(a, b)

s039091479

Accepted

#coding:utf-8 a, b, k = map(int, input().split()) for i in range(0, k): if i % 2 == 0: b += a // 2 a = a // 2 else: a += b // 2 b = b // 2 print(a, b)

s759317519

p03385

u513081876

Wrong Answer

You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.

S = list(map(str, input())) if S.sort() == ['a', 'b', 'c']: print('Yes') else: print('No')

s726925201

Accepted

S = list(input()) if len(S) == len(set(S)): print('Yes') else: print('No')

s458621942

p03448

u787059958

Wrong Answer

You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.

A = int(input()) B = int(input()) C = int(input()) x = int(input()) la=[] for a in range(A+1): la.append(a) lb=[] for b in range(B+1): lb.append(b) lc=[] for c in range(C+1): lc.append(c) count=0 for i in range(A): for j in range(B): for p in range(C): if(500*la[i]+100*lb[j]+50*lc[p]==x): count+=1 print(count)

s064538804

Accepted

A = int(input()) B = int(input()) C = int(input()) x = int(input()) count=0 for i in range(A+1): for j in range(B+1): for p in range(C+1): if(10*i+2*j+p==x/50): count+=1 print(count)

s309836470

p03401

u583010173

Wrong Answer

There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis. You planned a trip along the axis. In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0. However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i. You will visit the remaining spots as planned in the order they are numbered. You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned. For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.

# -*- coding: utf-8 -*- n = int(input()) spot = [int(x) for x in input().split()] spot = [0] + spot + [0] print(spot) cost = 0 for i in range(len(spot)-1): cost += abs(spot[i] - spot[i+1]) #print(cost) change = 0 for k in range(len(spot)-2): if (spot[k+1]-spot[k])*(spot[k+2]-spot[k+1]) >= 0: change = 0 else: change = 2*min([abs(spot[k+1]-spot[k]),abs(spot[k+2]-spot[k+1])]) print(cost-change)

s172857646

Accepted

# -*- coding: utf-8 -*- n = int(input()) spot = [int(x) for x in input().split()] spot = [0] + spot + [0] cost = 0 for i in range(len(spot)-1): cost += abs(spot[i] - spot[i+1]) change = 0 for k in range(len(spot)-2): if (spot[k+1]-spot[k])*(spot[k+2]-spot[k+1]) >= 0: change = 0 else: change = 2*min([abs(spot[k+1]-spot[k]),abs(spot[k+2]-spot[k+1])]) print(cost-change)

s029716988

p03645

u193264896

Wrong Answer

In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him.

import sys readline = sys.stdin.buffer.readline sys.setrecursionlimit(10 ** 8) INF = float('inf') MOD = 10 ** 9 + 7 def main(): N, M = map(int, readline().split()) path = [[] for _ in range(N+1)] for _ in range(M): a,b = map(int, readline().split()) path[a].append(b) path[b].append(a) print(path[1]) print(path[N]) if set(path[1])&set(path[N]): print('POSSIBLE') else: print('IMPOSSIBLE') if __name__ == '__main__': main()

s330013978

Accepted

import sys readline = sys.stdin.buffer.readline sys.setrecursionlimit(10 ** 8) INF = float('inf') MOD = 10 ** 9 + 7 def main(): N, M = map(int, readline().split()) path = [[] for _ in range(N+1)] for _ in range(M): a,b = map(int, readline().split()) path[a].append(b) path[b].append(a) if set(path[1])&set(path[N]): print('POSSIBLE') else: print('IMPOSSIBLE') if __name__ == '__main__': main()

s402091359

p02261

u963402991

Wrong Answer

Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).

# -*- coding:utf-8 -*- def Selection_Sort(A,n): for i in range(n): mini = i for j in range(i,n): if int(A[j][1]) < int(A[mini][1]): mini = j if A[i] != A[mini]: A[i], A[mini] = A[mini], A[i] return A def Bubble_Sort(A, n): for i in range(n): for j in range(n-1,i,-1): if int(A[j][1]) < int(A[j-1][1]): A[j], A[j-1] = A[j-1], A[j] return A n = int(input()) A = input().strip().split() print (A[1][1]) B = A[:] A = Bubble_Sort(A,n) print (' '.join(A)) print ("Stable") B =Selection_Sort(B,n) print (' '.join(B)) if A == B: print ("Stable") else: print ("Not stable")

s318521388

Accepted

# -*- coding:utf-8 -*- def Selection_Sort(A,n): for i in range(n): mini = i for j in range(i,n): if int(A[j][1]) < int(A[mini][1]): mini = j if A[i] != A[mini]: A[i], A[mini] = A[mini], A[i] return A def Bubble_Sort(A, n): for i in range(n): for j in range(n-1,i,-1): if int(A[j][1]) < int(A[j-1][1]): A[j], A[j-1] = A[j-1], A[j] return A n = int(input()) A = input().strip().split() B = A[:] A = Bubble_Sort(A,n) print (' '.join(A)) print ("Stable") B =Selection_Sort(B,n) print (' '.join(B)) if A == B: print ("Stable") else: print ("Not stable")

s873497185

p03671

u703890795

Wrong Answer

Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately. He has very high awareness of safety, and decides to buy two bells, one for each hand. The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the minimum total price of two different bells.

a, b, c = map(int, input().split()) print(max(a+b, b+c, c+a))

s004110066

Accepted

a, b, c = map(int, input().split()) print(min(a+b, b+c, c+a))

s733230351

p03730

u103099441

Wrong Answer

We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.

a, b, c = map(int, input().split()) z = float('inf') n = int(a / b) + 1 flag = False while True: x, y = divmod(n * b + c, a) if not x: flag = True break elif y >= z: break else: n += 1 z = min(z, y) if flag: print('YES') else: print('NO')

s673465606

Accepted

a, b, c = map(int, input().split()) n = max(int(b / a), 1) p = 'NO' s = set() r = n * a % b while r not in s: if r == c: p = 'YES' break else: s.add(r) n += 1 r = n * a % b print(p)

s003968299

p03695

u405779580

Wrong Answer

In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users.

N =int(input()) scores = [int(i)//400 for i in input().split()] d = [0]*10 for s in scores: if s > 7: s = 8 d[s] = d[s] + 1 print(d) if any(d[:8]): m = [bool(i) for i in d[:8]].count(True) else: m = 1 M = min(8, m+d[8]) print(m, M)

s965418984

Accepted

from collections import Counter input() c = Counter([min(int(i)//400,8) for i in input().split()]) lt3200 = [bool(c[i]) for i in range(8)].count(True) print(max(1, lt3200), lt3200+c[8])

s273065193

p03547

u614734359

Wrong Answer

In programming, hexadecimal notation is often used. In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively. In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`. When X and Y are seen as hexadecimal numbers, which is larger?

x,y = input().split() if x>y: print(x) else: print(y)

s264970798

Accepted

x,y = input().split() if x>y: print('>') elif x<y: print('<') else: print('=')

s016981737

p03795

u454866339

Wrong Answer

Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.

N = 60 x = 800 * N y = 200 * (N // 15) print(x - y)

s953093466

Accepted

N = int(input()) x = int(800 * N) y = int(200 * (N // 15)) print(int(x - y))

s069613034

p04045

u632369368

Wrong Answer

Iroha is very particular about numbers. There are K digits that she dislikes: D_1, D_2, ..., D_K. She is shopping, and now paying at the cashier. Her total is N yen (the currency of Japan), thus she has to hand at least N yen to the cashier (and possibly receive the change). However, as mentioned before, she is very particular about numbers. When she hands money to the cashier, the decimal notation of the amount must not contain any digits that she dislikes. Under this condition, she will hand the minimum amount of money. Find the amount of money that she will hand to the cashier.

import itertools N, K = [int(s) for s in input().split()] D = [int(s) for s in input().split()] U = [str(n) for n in range(10) if n not in D] ret = N * 10 for v in [int(''.join(T)) for T in itertools.product(U, repeat=4)]: print(v) if N <= v: ret = min(ret, v) print(ret)

s628513182

Accepted

from functools import reduce N, _ = [int(s) for s in input().split()] D = [s for s in input().split()] R = N while True: if reduce(lambda x, y: x and y, [s not in D for s in list(str(R))]): break R += 1 print(R)

s356534286

p03712

u403551852

Wrong Answer

You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.

h,w = map(int,input().split()) aa = [input() for _ in range(h)] print(aa) for i in range(h+2): if i == 0 or i == h+1: print('#'*(w+2)) else: print('#'+aa[i-1]+'#')

s702678584

Accepted

h,w = map(int,input().split()) aa = [input() for _ in range(h)] for i in range(h+2): if i == 0 or i == h+1: print('#'*(w+2)) else: print('#'+aa[i-1]+'#')

s663757566

p02413

u216425054

Wrong Answer

Your task is to perform a simple table calculation. Write a program which reads the number of rows r, columns c and a table of r × c elements, and prints a new table, which includes the total sum for each row and column.

r,c=[int(x) for x in input().split()] s_row=[0 for x in range(c+1)] for i in range(r): row=[int(x) for x in input().split()] row.append(sum(row)) print(" ".join([str(x) for x in row])) s_row=[x+y for x,y in zip(s_row,row)]

s297476583

Accepted

r,c=[int(x) for x in input().split()] s_row=[0 for x in range(c+1)] for i in range(r): row=[int(x) for x in input().split()] row.append(sum(row)) print(" ".join([str(x) for x in row])) s_row=[x+y for x,y in zip(s_row,row)] print(" ".join([str(x) for x in s_row]))

s491044473

p03377

u816376170

Wrong Answer

There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.

a = input().split() if int(a[0])<int(a[2]) and int(a[0])+int(a[1])>int(a[2]): print("yes") else: print("no")

s056235433

Accepted

a = input().split() if int(a[0])<=int(a[2]) and int(a[0])+int(a[1])>=int(a[2]): print("YES") else: print("NO")

s015910872

p02261

u209358977

Wrong Answer

Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).

# -*- coding: utf-8 -*- def conv_input(arr): ret = [] for e in arr: ret.append((e[0], int(e[1]))) return ret def conv_output(conv_arr): ret = [] for e in conv_arr: ret.append(e[0] + str(e[1])) return ret def is_stable(arr1, arr2): ret = 'Stable' for a1, a2 in zip(arr1, arr2): if a1 == a2: continue else: ret = 'Not Stable' break return ret def bubble_sort(n, arr): flag = True while flag: flag = False for i in reversed(range(n)): if i is 0: break if arr[i][1] < arr[i - 1][1]: tmp = arr[i] arr[i] = arr[i - 1] arr[i - 1] = tmp flag = True return arr def selection_sort(n, arr): for i in range(n): minj = i for j in range(i, n): if arr[j][1] < arr[minj][1]: minj = j if i != minj: tmp = arr[minj] arr[minj] = arr[i] arr[i] = tmp return arr if __name__ == '__main__': n = int(input()) arr1 = arr2 = [str(s) for s in input().split()] arr1 = conv_output(bubble_sort(n, conv_input(arr1))) print(' '.join(map(str, arr1))) print('Stable') arr2 = conv_output(selection_sort(n, conv_input(arr2))) print(' '.join(map(str, arr2))) print(is_stable(arr1, arr2))

s315645120

Accepted

# -*- coding: utf-8 -*- def conv_input(arr): ret = [] for e in arr: ret.append((e[0], int(e[1]))) return ret def conv_output(conv_arr): ret = [] for e in conv_arr: ret.append(e[0] + str(e[1])) return ret def is_stable(arr1, arr2): ret = 'Stable' for a1, a2 in zip(arr1, arr2): if a1 == a2: continue else: ret = 'Not stable' break return ret def bubble_sort(n, arr): flag = True while flag: flag = False for i in reversed(range(n)): if i is 0: break if arr[i][1] < arr[i - 1][1]: tmp = arr[i] arr[i] = arr[i - 1] arr[i - 1] = tmp flag = True return arr def selection_sort(n, arr): for i in range(n): minj = i for j in range(i, n): if arr[j][1] < arr[minj][1]: minj = j if i != minj: tmp = arr[minj] arr[minj] = arr[i] arr[i] = tmp return arr if __name__ == '__main__': n = int(input()) arr1 = arr2 = [str(s) for s in input().split()] arr1 = conv_output(bubble_sort(n, conv_input(arr1))) print(' '.join(map(str, arr1))) print('Stable') arr2 = conv_output(selection_sort(n, conv_input(arr2))) print(' '.join(map(str, arr2))) print(is_stable(arr1, arr2))

s777274673

p03478

u825440127

Wrong Answer

Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).

n, a, b= map(int, input().split()) ans = 0 for i in range(1, n+1): if a <= sum([int(c) for c in str(i)]) <= b: ans += 1 print(ans)

s800420435

Accepted

n, a, b= map(int, input().split()) ans = 0 for i in range(1, n+1): if a <= sum([int(c) for c in str(i)]) <= b: ans += i print(ans)

s955401192

p03680

u626337957

Wrong Answer

Takahashi wants to gain muscle, and decides to work out at AtCoder Gym. The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it. Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up. Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.

N = int(input()) nums = [0] for _ in range(N): nums.append(int(input())) check_cnt = 1 _next = nums[1] while True: if check_cnt >= N: break else: if next == 2: print(check_cnt) exit() _next = nums[_next] check_cnt += 1 print(-1)

s443432025

Accepted

N = int(input()) nums = [0] for _ in range(N): nums.append(int(input())) check_cnt = 1 _next = nums[1] while True: if check_cnt >= N: break else: if _next == 2: print(check_cnt) exit() _next = nums[_next] check_cnt += 1 print(-1)

s234504789

p03694

u672898046

Wrong Answer

It is only six months until Christmas, and AtCoDeer the reindeer is now planning his travel to deliver gifts. There are N houses along _TopCoDeer street_. The i-th house is located at coordinate a_i. He has decided to deliver gifts to all these houses. Find the minimum distance to be traveled when AtCoDeer can start and end his travel at any positions.

s = list(map(int, input().split())) print(max(s)-min(s))

s623879675

Accepted

n = int(input()) s = list(map(int, input().split())) print(max(s)-min(s))

s124491327

p03131

u349444371

Wrong Answer

Snuke has one biscuit and zero Japanese yen (the currency) in his pocket. He will perform the following operations exactly K times in total, in the order he likes: * Hit his pocket, which magically increases the number of biscuits by one. * Exchange A biscuits to 1 yen. * Exchange 1 yen to B biscuits. Find the maximum possible number of biscuits in Snuke's pocket after K operations.

k,a,b=map(int,input().split()) if b<a+2: print(1+k) else: if k<=a: print(1+k) else: if (k-a-1)%2!=0: print(b+(b-a)*(k-a-1)//2+1) else: print(b+(b-a)*(k-a-1)//2)

s728322944

Accepted

k,a,b=map(int,input().split()) if b<a+2: print(1+k) else: if k<=a: print(1+k) else: if (k-a-1)%2==0: print(b+(b-a)*((k-a-1)//2)) else: print(b+(b-a)*((k-a-1)//2)+1)

s976783226

p02613

u166057331

Wrong Answer

Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.

N = int(input()) C0 = 0 C1 = 0 C2 = 0 C3 = 0 for i in range(0,N): S = str(input()) if S == 'AC': C0 += 1 elif S == 'WA': C1 += 1 elif S == 'TLE': C2 += 1 elif S == 'RE': C3 += 1 print('AC ×',C0) print('WA ×',C1) print('TLE ×',C2) print('RE ×',C3)

s973211983

Accepted

N = int(input()) C0 = 0 C1 = 0 C2 = 0 C3 = 0 for i in range(0,N): S = str(input()) if S == 'AC': C0 += 1 elif S == 'WA': C1 += 1 elif S == 'TLE': C2 += 1 elif S == 'RE': C3 += 1 print('AC x',C0) print('WA x',C1) print('TLE x',C2) print('RE x',C3)

s701764160

p04043

u433559751

Wrong Answer

Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.

N = list(map(int, input().split(" "))) if N.count(5) == 2 and N.count(7) == 1: print('Yes') else: print('No')

s097102202

Accepted

N = list(map(int, input().split(" "))) if N.count(5) == 2 and N.count(7) == 1: print('YES') else: print('NO')

s224728240

p00007

u298999032

Wrong Answer

Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week. Write a program which computes the amount of the debt in n weeks.

x=100000 n=int(input()) for i in range(n): x*=1.05 if x%1000!=0: x+=1000-x%1000 print(x)

s970391000

Accepted

x=100000 for i in range(int(input())): x*=1.05 if x%1000!=0: x+=1000-x%1000 print(int(x))

s003157769

p03643

u268792407

Wrong Answer

This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer.

n=int(input()) from numpy import log print(pow(2,int(log(n)/log(2))))

s719359982

Accepted

print("ABC"+input())

s187722405

p03544

u698567423

Wrong Answer

It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)

n = int(input()) lf = 2 ls = 1 for _ in range(n-1): lf, ls = ls, lf + ls print(ls)

s212829865

Accepted

n = int(input()) lf = 2 ls = 1 for _ in range(n-1): lf, ls = ls, lf + ls print(ls)

s825003527

p03643

u886655280

Wrong Answer

This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer.

input = int(input()) input_list = list(range(1, input + 1)) count_list = [] for e in input_list: count = 0 e_tmp = e while True: if e_tmp % 2 == 0: e_tmp = e_tmp/2 count += 1 else: break count_list.append([e, count]) max_value = 0 max_key = 0 for i in range(input): curr_value = int(count_list[i][1]) if curr_value > max_value: max_value = curr_value max_key = count_list[i][0] print(max_key)

s032271490

Accepted

N = input() ans = 'ABC' + N print(ans)

s229669238

p02613

u048013400

Wrong Answer

Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.

from collections import Counter N = int(input()) S = [input() for i in range(N)] s = Counter(S) print('AC×'+ str(s['AC'])) print('WA×' + str(s['WA'])) print('TLE×' + str(s['TLE'])) print('RE×' + str(s['RE']))

s153503446

Accepted

from collections import Counter N = int(input()) S = [input() for i in range(N)] s = Counter(S) print('AC x ',end='') print(s['AC']) print('WA x ', end='') print(s['WA']) print('TLE x ',end='') print(s['TLE']) print('RE x ', end='') print(s['RE'])

s480401316

p03472

u356608129

Wrong Answer

You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, …, Katana N, and can perform the following two kinds of attacks in any order: * Wield one of the katana you have. When you wield Katana i (1 ≤ i ≤ N), the monster receives a_i points of damage. The same katana can be wielded any number of times. * Throw one of the katana you have. When you throw Katana i (1 ≤ i ≤ N) at the monster, it receives b_i points of damage, and you lose the katana. That is, you can no longer wield or throw that katana. The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total?

import sys (N, H) = tuple(map(int, input().split(" "))) a = [] b = [] result = -1 for i in range(N): (ai,bi) = tuple(map(int, input().split(" "))) a.append(ai) b.append(bi) a_max = max(a) sorted(b).reverse() b_length = len(b) for i, bi in enumerate(b): if bi < a_max: b_length = i S = 0 for l in range(b_length): S += b[l] if S >= H: result = l + 1 print(result) sys.exit() j = b_length result = b_length + int((H - S)/a_max) print(result)

s153656361

Accepted

import sys import math (N, H) = tuple(map(int, input().split(" "))) a = [] b = [] result = 0 for i in range(N): (ai,bi) = tuple(map(int, input().split(" "))) a.append(ai) b.append(bi) a_max = max(a) b.sort() b.reverse() b_length = len(b) for i, bi in enumerate(b): if bi < a_max: b_length = i break S = 0 for l in range(b_length): S += b[l] if S >= H: result = l + 1 print(result) sys.exit() result = b_length + int(math.ceil((H - S)/a_max)) print(result)

s283160489

p03379

u764956288

Wrong Answer

When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N.

n = int(input()) As = list(map(int, input().split())) As.sort() left = As[n//2 - 1] right = As[n//2] for x in As: if x <= left: print(right) else: print(left)

s040512014

Accepted

n = int(input()) As = list(map(int, input().split())) Xs = sorted(As) left = Xs[n//2 - 1] right = Xs[n//2] for x in As: if x <= left: print(right) else: print(left)

s181072385

p03798

u432042540

Wrong Answer

Snuke, who loves animals, built a zoo. There are N animals in this zoo. They are conveniently numbered 1 through N, and arranged in a circle. The animal numbered i (2≤i≤N-1) is adjacent to the animals numbered i-1 and i+1. Also, the animal numbered 1 is adjacent to the animals numbered 2 and N, and the animal numbered N is adjacent to the animals numbered N-1 and 1. There are two kinds of animals in this zoo: honest sheep that only speak the truth, and lying wolves that only tell lies. Snuke cannot tell the difference between these two species, and asked each animal the following question: "Are your neighbors of the same species?" The animal numbered i answered s_i. Here, if s_i is `o`, the animal said that the two neighboring animals are of the same species, and if s_i is `x`, the animal said that the two neighboring animals are of different species. More formally, a sheep answered `o` if the two neighboring animals are both sheep or both wolves, and answered `x` otherwise. Similarly, a wolf answered `x` if the two neighboring animals are both sheep or both wolves, and answered `o` otherwise. Snuke is wondering whether there is a valid assignment of species to the animals that is consistent with these responses. If there is such an assignment, show one such assignment. Otherwise, print `-1`.

n = int(input()) s = input() a = False b = False sw = [None] * n sw[0] = False sw[1] = False def next(s,sw,i,j): if s[i] == 'o': return not (sw[j] ^ sw[i]) else: return sw[j] ^ sw[i] def printsw(sw,n): s = '' for i in range(n): if sw[i]: s += 'S' else: s += 'W' print(s) flag = True for j in range(4): for i in range(2,n): sw[i] = next(s,sw,i-2,i-1) zero = next(s,sw,n-2,n-1) one = next(s,sw,n-1,0) if zero == sw[0]: if one == sw[1]: printsw(sw,n) flag = False break sw[0] = not sw[0] if j == 1: sw[1] = True if flag: print(-1)

s710981000

Accepted

n = int(input()) s = input() a = False b = False sw = [None] * n sw[0] = True sw[1] = True def next(s,sw,j,i): if s[i] == 'o': return not (sw[j] ^ sw[i]) else: return sw[j] ^ sw[i] def printsw(sw,n): s = '' for i in range(n): if sw[i]: s += 'S' else: s += 'W' print(s) flag = True for j in range(4): for i in range(2,n): sw[i] = next(s,sw,i-2,i-1) zero = next(s,sw,n-2,n-1) one = next(s,sw,n-1,0) if zero == sw[0]: if one == sw[1]: printsw(sw,n) flag = False break sw[0] = not sw[0] if j == 1: sw[1] = False if flag: print(-1)

s522700119

p03493

u425177436

Wrong Answer

Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.

print(list(map(int, input().split())).count(1))

s290064136

Accepted

print(list(map(int, list(input()))).count(1))

s062041733

p03679

u063052907

Wrong Answer

Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.

#coding: utf-8 X, A, B = map(int, input().split()) print(["delicious", "safe", "dangerous"][(A > B) + (A + X < B) ])

s965355673

Accepted

#coding: utf-8 X, A, B = map(int, input().split()) print(["delicious", "safe", "dangerous"][(A < B) + (A + X < B)])

s460403208

p03486

u623687794

Wrong Answer

You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.

s=input() t=input() schar=[] tchar=[] for i in range(len(s)): schar.append(ord(s[i])) for i in range(len(t)): tchar.append(ord(t[i])) schar.sort() tchar.sort(reverse=True) flag=0 for i in range(min(len(s),len(t))): if schar[i]>tchar[i]: flag=1 break if flag==1: print("No") else: if len(s)>=len(t): print("No") else: print("Yes")

s574517849

Accepted

s=input() t=input() schar=[] tchar=[] for i in range(len(s)): schar.append(ord(s[i])) for i in range(len(t)): tchar.append(ord(t[i])) schar.sort() tchar.sort(reverse=True) flag=0 for i in range(min(len(s),len(t))): if schar[i]==tchar[i]: continue elif schar[i]>tchar[i]: flag=1 break else: flag=-1 break if flag==1: print("No") elif flag==-1: print("Yes") else: if len(s)<len(t): print("Yes") else: print("No")

s700154241

p02608

u119015607

Wrong Answer

Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).

n = int(input()) ans = [0 for i in range(10001)] for i in range(105): for j in range(105): for m in range(105): v = i**2+j**2+m**2+i*j+j*m+m*i #print(v) if v<10001: ans[v]+=1 #print(count) for i in range(n): print(ans[i+1])

s224486696

Accepted

n = int(input()) ans = [0 for i in range(10001)] for i in range(1,105): for j in range(1,105): for m in range(1,105): v = i**2+j**2+m**2+i*j+j*m+m*i #print(v) if v<10001: ans[v]+=1 #print(count) for i in range(n): print(ans[i+1])

s395699444

p03605

u806403461

Wrong Answer

It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?

N = str(input()) for a in N: if a == '9': print('Yes') else: print('No')

s004175409

Accepted

N = str(input()) ans = 'No' for a in N: if a == '9': ans = 'Yes' break else: ans = 'No' print(ans)

s182757548

p03854

u998082063

Wrong Answer

You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.

s = input().replace("eracer", "").replace("erace", "").replace("dreamer", "").replace("dream", "") if s: print("NO") else: print("YES")

s444812039

Accepted

s = input().replace("eraser", "").replace("erase", "").replace("dreamer", "").replace("dream", "") if s: print("NO") else: print("YES")