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TnT

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CodeNet4Repair

like 0

s949440838

p03965

u827202523

Wrong Answer

AtCoDeer the deer and his friend TopCoDeer is playing a game. The game consists of N turns. In each turn, each player plays one of the two _gestures_ , _Rock_ and _Paper_ , as in Rock-paper-scissors, under the following condition: (※) After each turn, (the number of times the player has played Paper)≦(the number of times the player has played Rock). Each player's score is calculated by (the number of turns where the player wins) - (the number of turns where the player loses), where the outcome of each turn is determined by the rules of Rock-paper-scissors. _(For those who are not familiar with Rock-paper-scissors: If one player plays Rock and the other plays Paper, the latter player will win and the former player will lose. If both players play the same gesture, the round is a tie and neither player will win nor lose.)_ With his supernatural power, AtCoDeer was able to foresee the gesture that TopCoDeer will play in each of the N turns, before the game starts. Plan AtCoDeer's gesture in each turn to maximize AtCoDeer's score. The gesture that TopCoDeer will play in each turn is given by a string s. If the i-th (1≦i≦N) character in s is `g`, TopCoDeer will play Rock in the i-th turn. Similarly, if the i-th (1≦i≦N) character of s in `p`, TopCoDeer will play Paper in the i-th turn.

hands = input() ans = 0 for i, h in enumerate(hands): if i % 2 == 1: if h == "p": ans -= 1 else: if h == "g": ans += 1 print(ans)

s093455903

Accepted

hands = input() ans = 0 for i, h in enumerate(hands): if i % 2 == 0: if h == "p": ans -= 1 else: if h == "g": ans += 1 print(ans)

s680018459

p03400

u034128150

Wrong Answer

Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp.

N = int(input()) D, X = map(int, input().split()) As = [int(input()) for _ in range(N)] c = X for A in As: c += (D-1) // A print(c)

s028408751

Accepted

N = int(input()) D, X = map(int, input().split()) As = [int(input()) for _ in range(N)] c = X for A in As: c += ((D-1) // A) + 1 print(c)

s872288303

p03047

u089142196

Wrong Answer

Snuke has N integers: 1,2,\ldots,N. He will choose K of them and give those to Takahashi. How many ways are there to choose K consecutive integers?

N,K=map(int,input().split()) print(K-N+1)

s335606049

Accepted

N,K=map(int,input().split()) cnt=0 for i in range(1,N+1): if i+K-1<=N: cnt+=1 print(cnt)

s936745721

p03623

u088529499

Wrong Answer

Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.

list = list(map(int,input().split())) x=list[0] a=list[1] b=list[2] if a>x: s=a-x else: s=x-a if b>x: t=b-x else: t=x-b if s<t: print('B') else: print('A')

s901627818

Accepted

list = list(map(int,input().split())) x=list[0] a=list[1] b=list[2] if a>x: s=a-x else: s=x-a if b>x: t=b-x else: t=x-b if s<t: print('A') else: print('B')

s125773931

p03546

u698479721

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.

h,w=map(int,input().split()) grid = [] for i in range(10): array = list(map(int, input().strip().split(' '))) grid.append(array) wall = [] for i in range(h): array = list(map(int, input().strip().split(' '))) wall.append(array) def warshall_floyd(n): for k in range(n): for i in range(n): for j in range(n): grid[i][j] = min(grid[i][j], grid[i][k] + grid[k][j]) warshall_floyd(10) cost=[] for arrays in grid: cost.append(arrays[1]) ans = 0 for arrays in wall: for nums in arrays: ans += cost[nums] print(ans)

s343282298

Accepted

h,w=map(int,input().split()) grid = [] for i in range(10): array = list(map(int, input().strip().split(' '))) grid.append(array) wall = [] for i in range(h): array = list(map(int, input().strip().split(' '))) wall.append(array) def warshall_floyd(n): for k in range(n): for i in range(n): for j in range(n): grid[i][j] = min(grid[i][j], grid[i][k] + grid[k][j]) warshall_floyd(10) cost=[] for arrays in grid: cost.append(arrays[1]) ans = 0 for arrays in wall: for nums in arrays: if nums == -1: pass else: ans += cost[nums] print(ans)

s335223383

p03605

u961683150

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?

x=int(input()) if x==9 or x//10==9: print("Yes") else: print("No")

s863764223

Accepted

x=int(input()) if x==9 or x//10==9 or x%10==9: print("Yes") else: print("No")

s556784897

p03623

u263830634

Wrong Answer

Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.

S = list(input()) s = set(S) S = list(s) S.sort() if len(S) == 26: print (None) else: tmp = 0 for i in 'abcdefghijklmnopqrstuvwxyz': if tmp >= len(S): print (i) break if i == S[tmp]: tmp += 1 else: print (i) break

s717805276

Accepted

X, A, B = map(int, input().split()) if abs(X - A) < abs(X - B): print ('A') else: print ('B')

s393726218

p03610

u612635771

Wrong Answer

You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.

s = input() print(s[1::2])

s986303447

Accepted

s = input() print(s[::2])

s692869256

p04043

u683956577

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 = list(map(int, input().split())) print("Yes" if a[0]== a[2] == 5 and a[1] == 7 else "No")

s445957841

Accepted

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

s129676959

p02406

u967268722

Wrong Answer

In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }

n = int(input()) for i in range(3, n+1, 3): print(' '+str(i), end='')

s172711144

Accepted

n = int(input()) for i in range(1, n+1): if i%3 == 0:print(' '+str(i), end='') else: if '3' in str(i):print(' '+str(i), end='') print()

s721965584

p03401

u941753895

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()) if n==3: exit() al=list(map(int,input().split())) l1=[] l2=[] bl=[0]+al+[0] for i in range(len(bl)-1): l1.append(abs(bl[i]-bl[i+1])) if i!=len(bl)-2: l2.append(abs(bl[i]-bl[i+2])) for i in range(len(l2)): print(l2[i]+sum(l1[:i])+sum(l1[i+2:]))

s406832899

Accepted

import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time sys.setrecursionlimit(10**7) inf=10**20 mod=10**9+7 def LI(): return list(map(int,input().split())) def I(): return int(input()) def LS(): return input().split() def S(): return input() def main(): n=I() l=LI() l=[0]+l+[0] sm=sum([abs(l[i+1]-l[i]) for i in range(len(l)-1)]) for i in range(1,len(l)-1): a=abs(l[i]-l[i-1]) b=abs(l[i+1]-l[i]) c=abs(l[i+1]-l[i-1]) print(sm-(a+b-c)) main()

s782126503

p03659

u103520789

Wrong Answer

Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it. They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card. Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|.

N = int(input()) a_arr = list(map(int, input().split())) def main(arr): sum_sunuke = 0 sum_arai = sum(arr) min_val = 2*(10**5) for i in range(1,N): sum_sunuke += arr[i] sum_arai -= arr[i] val = abs(sum_sunuke - sum_arai) print(sum_sunuke) if min_val > val: min_val = val return min_val print(main(a_arr))

s527025308

Accepted

N = int(input()) a_arr = list(map(int, input().split())) def main(arr): sum_sunuke = 0 sum_arai = sum(arr) min_val = 10**10 # if N == 2: # return abs(arr[0]- arr[1]) for i in range(N-1): sum_sunuke += arr[i] sum_arai -= arr[i] val = abs(sum_sunuke - sum_arai) # print("arai: {}".format(sum_arai)) if min_val > val: min_val = val # print("min val: {}".format(min_val)) # print("*************") return min_val print(main(a_arr))

s250401797

p03155

u457460736

Wrong Answer

It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board. The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns. How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?

N=int(input()) H=int(input()) W=int(input()) print((N-H)*(N-W))

s338112452

Accepted

N=int(input()) H=int(input()) W=int(input()) print((N-H+1)*(N-W+1))

s271029715

p02694

u743164083

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?

m = 100 x = int(input()) y = 0 while x >= m: m = m + m // 100 y += 1 print(y)

s393673231

Accepted

m = 100 x = int(input()) y = 0 while x >= m: if x == m: break m = m + m // 100 y += 1 print(y)

s868589623

p03478

u167681750

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()) counter = 0 for i in range(1, n): str_i = str(i) sum_i = sum(map(int, str_i)) if a <= sum_i <= b: counter += 1 print(counter)

s002391470

Accepted

n, a, b = map(int, input().split()) answer = 0 for i in range(1, n + 1): sum_i = sum(map(int, list(str(i)))) if a <= sum_i <= b: answer += i print(answer)

s658876469

p03149

u652656291

Wrong Answer

You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".

A = list(map(int,input().split())) a = A.count('1') b = A.count('7') c = A.count('9') d = A.count('4') if a == b == c == d == 1: print('YES') else: print('NO')

s370591587

Accepted

A = list(map(str,input().split())) a = A.count('1') b = A.count('7') c = A.count('9') d = A.count('4') if a == b == c == d == 1: print('YES') else: print('NO')

s115079733

p02257

u672822075

Wrong Answer

A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7. Write a program which reads a list of _N_ integers and prints the number of prime numbers in the list.

c = 0 n = int(input()) for _ in range(n): t = int(input()) for i in range(2,t): if t%i==0: break c += 1 print(c)

s119419711

Accepted

import math c = 0 n = int(input()) for _ in range(n): t = int(input()) for i in range(2,int(math.sqrt(t)+1)): if t%i==0: break else: c+=1 print(c)

s963508448

p02388

u100813820

Wrong Answer

Write a program which calculates the cube of a given integer x.

# x ??? 3 ??? # Input # Output # Sample Input 1 # 2 # Sample Output 1 # 8 x=2 print(x**3); # Sample Input 2 # 3 # Sample Output 2 # 27 x=3 print(x**3);

s096103038

Accepted

# x ??? 3 ??? # Input # Output # Sample Input 1 # 2 # Sample Output 1 # 8 x=int( input() ) print(x**3);

s509130318

p03637

u399721252

Wrong Answer

We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective.

n = int(input()) num = [ int(v) for v in input().split() ] a, b, c = 0, 0, 0 for i in num: if i % 2 != 0: a += 1 elif i % 4 != 0: b += 1 else: c += 1 print(a,c) if c >= a - 1: print("Yes") else: print("No")

s159954299

Accepted

n = int(input()) num = [ int(v) for v in input().split() ] a, b, c = 0, 0, 0 for i in num: if i % 2 != 0: a += 1 elif i % 4 != 0: b += 1 else: c += 1 if b == 0: if c >= a - 1: print("Yes") else: print("No") else: if c >= a: print("Yes") else: print("No")

s363485261

p03555

u162660367

Wrong Answer

You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.

C1=input().strip() C2=input().strip() print("No Yes".split()[C1==C2[::-1]])

s249907963

Accepted

C1=input().strip() C2=input().strip() print("NO YES".split()[int(C1==C2[::-1])])

s220472736

p03797

u870518235

Wrong Answer

Snuke loves puzzles. Today, he is working on a puzzle using `S`\- and `c`-shaped pieces. In this puzzle, you can combine two `c`-shaped pieces into one `S`-shaped piece, as shown in the figure below: Snuke decided to create as many `Scc` groups as possible by putting together one `S`-shaped piece and two `c`-shaped pieces. Find the maximum number of `Scc` groups that can be created when Snuke has N `S`-shaped pieces and M `c`-shaped pieces.

S, C = map(int, input().split()) ans = 0 if S*2 <= C: C = C - S*2 print(C, S, C//4) ans += S ans += C // 4 else: print(C//2) ans += C // 2 print(ans)

s531563959

Accepted

S, C = map(int, input().split()) ans = 0 if S*2 <= C: C = C - S*2 ans += S ans += C // 4 else: ans += C // 2 print(int(ans))

s283806971

p02927

u905582793

Wrong Answer

Today is August 24, one of the five Product Days in a year. A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day): * d_1 \geq 2 * d_{10} \geq 2 * d_1 \times d_{10} = m Takahashi wants more Product Days, and he made a new calendar called Takahashi Calendar where a year consists of M month from Month 1 to Month M, and each month consists of D days from Day 1 to Day D. In Takahashi Calendar, how many Product Days does a year have?

m,d = map(int,input().split()) ans = 0 for i in range(m): for j in range(d): if "1" in str(j): continue j1 = j//10 j2 = j%10 if i == j1*j2: ans += 1 print(ans)

s360703695

Accepted

m,d = map(int,input().split()) ans = 0 for i in range(1,m+1): for j in range(10,d+1): if "1" in list(str(j)): continue j1 = j//10 j2 = j%10 if i == j1*j2: ans += 1 print(ans)

s853318137

p03359

u457957084

Wrong Answer

In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?

n, b = map(int,input().split()) if n <= b: print(n + 1) else: print(n)

s462453115

Accepted

n, b = map(int,input().split()) if n <= b: print(n) else: print(n - 1)

s015783353

p03998

u868600519

Wrong Answer

Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.

cards = dict((x, list(input())) for x in 'abc') turn = 'a' while len(cards[turn]): turn = cards[turn].pop(0) print(turn.upper())

s822007714

Accepted

cards = {x: list(input()) for x in 'abc'} turn = 'a' while len(cards[turn]): turn = cards[turn].pop(0) print(turn.upper())

s511711348

p03693

u220870679

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 (100*r + 10*g + b) % 4 == 0: print("Yes") else: print("No")

s228382716

Accepted

r, g, b = map(int, input().split()) if (100*r + 10*g + b) % 4 == 0: print("YES") else: print("NO")

s419839108

p02850

u532966492

Wrong Answer

Given is a tree G with N vertices. The vertices are numbered 1 through N, and the i-th edge connects Vertex a_i and Vertex b_i. Consider painting the edges in G with some number of colors. We want to paint them so that, for each vertex, the colors of the edges incident to that vertex are all different. Among the colorings satisfying the condition above, construct one that uses the minimum number of colors.

def main(): n=int(input()) ab=[tuple(map(int,input().split())) for _ in [0]*(n-1)] d={ab[i]:i for i in range(n-1)} g=[[] for _ in [0]*n] [g[a-1].append(b-1) for a,b in ab] [g[b-1].append(a-1) for a,b in ab] g2=[set() for _ in [0]*(n-1)] for i in range(n): for j in g[i]: for k in g[j]: p=min(i,j) q=max(i,j) r=min(j,k) s=max(j,k) x=d[(p+1,q+1)] y=d[(r+1,s+1)] g2[x].add(y) g2[y].add(x) for i in range(n-1): g2[i].remove(i) color=[-1]*(n-1) color[0]=1 q=[[0,0]] while q: qq=[] while q: i,root=q.pop() cnt=1 for j in g2[i]: if j==root: continue if cnt==color[i]: cnt+=1 color[j]=cnt cnt+=1 qq.append([j,i]) q=qq print(*color) main()

s296875586

Accepted

def main(): n=int(input()) ab=[list(map(int,input().split())) for _ in [0]*(n-1)] g=[[] for _ in [0]*n] [g[a-1].append(b-1) for a,b in ab] [g[b-1].append(a-1) for a,b in ab] color=[-1]*n color[0]=0 q=[[0,0]] while q: qq=[] while q: i,root=q.pop() cnt=1 for j in g[i]: if j==root: continue if cnt==color[i]: cnt+=1 color[j]=cnt cnt+=1 qq.append([j,i]) q=qq d=[-1]*n d[0]=0 q=[0] cnt=0 while q: qq=[] cnt+=1 while q: i=q.pop() for j in g[i]: if d[j]==-1: d[j]=cnt qq.append(j) q=qq print(max(color[1:])) for i,j in ab: if d[i-1]>d[j-1]: k=i else: k=j print(color[k-1]) main()

s776314874

p03129

u550943777

Wrong Answer

Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.

n,k = map(int,input().split()) if n > 2*(k-1) : print('Yes') else: print('No')

s174668101

Accepted

n,k = map(int,input().split()) if n > 2*(k-1) : print('YES') else: print('NO')

s744426687

p03779

u999503965

Wrong Answer

There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.

x=int(input()) num=0 cnt=0 for i in range(10**9): num+=i cnt+=1 if num>=x: break if num==x: print(cnt) else: print(cnt-1)

s287334803

Accepted

x=int(input()) num=0 cnt=0 for i in range(1,10**9): num+=i cnt+=1 if num>=x: break print(cnt)

s330188823

p02255

u318430977

Wrong Answer

Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.

def insertion_sort(a, n): for i in range(1, n): v = a[i] j = i - 1 while j >= 0 and a[j] > v: a[j + 1] = a[j] j -= 1 a[j + 1] = v print(a) n = int(input()) a = input().split() for i in range(n): a[i] = int(a[i]) print(a) insertion_sort(a, n)

s926523607

Accepted

def insertion_sort(a, n): for i in range(1, n): v = a[i] j = i - 1 while j >= 0 and a[j] > v: a[j + 1] = a[j] j -= 1 a[j + 1] = v print_list_split_whitespace(a) def print_list_split_whitespace(a): for x in a[:-1]: print(x, end=' ') print(a[-1]) n = int(input()) a = input().split() for i in range(n): a[i] = int(a[i]) print_list_split_whitespace(a) insertion_sort(a, n)

s900979521

p03992

u163320134

Wrong Answer

This contest is `CODE FESTIVAL`. However, Mr. Takahashi always writes it `CODEFESTIVAL`, omitting the single space between `CODE` and `FESTIVAL`. So he has decided to make a program that puts the single space he omitted. You are given a string s with 12 letters. Output the string putting a single space between the first 4 letters and last 8 letters in the string s.

s=input() print('{} {}'.format(s[:4],s[5:]))

s403985479

Accepted

s=input() print('{} {}'.format(s[:4],s[4:]))

s705040967

p03693

u779830746

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(str, input().split()) number = int(r + g + b) if number / 4 >= 1: print('Yes') else: print('No')

s166033046

Accepted

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

s123075020

p03998

u030626972

Wrong Answer

Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.

def main(): sa = list(input()) sb = list(input()) sc = list(input()) turn = 'a' while True: if turn == 'a': if len(sa)>0: turn = sa.pop(0) else: print('a') return elif turn == 'b': if len(sb)>0: turn = sb.pop(0) else: print('b') return elif turn == 'c': if len(sc)>0: turn = sc.pop(0) else: print('c') return if __name__ == "__main__": main()

s908246051

Accepted

def main(): sa = list(input()) sb = list(input()) sc = list(input()) turn = 'a' while True: if turn == 'a': if len(sa)>0: turn = sa.pop(0) else: print('A') return elif turn == 'b': if len(sb)>0: turn = sb.pop(0) else: print('B') return elif turn == 'c': if len(sc)>0: turn = sc.pop(0) else: print('C') return if __name__ == "__main__": main()

s861945281

p02612

u091307273

Wrong Answer

We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.

def main(): R, C, K = map(int, input().split()) grid = [] for r in range(R): grid.append([1 if c == '#' else 0 for c in input()]) n_configs = 0 for rm in range(2**R): for cm in range(2**C): nb = 0 for r in range(R): for c in range(C): if ((rm >> r) & 1) == 0 and ((cm >> c) & 1) == 0 and grid[r][c] == 1: nb += 1 if nb == K: n_configs += 1 print(n_configs, flush=True) return

s530717555

Accepted

price = int(input()) change = 1000 - (price % 1000) if change == 1000: change = 0 print(change)

s705127718

p02256

u566311709

Wrong Answer

Write a program which finds the greatest common divisor of two natural numbers _a_ and _b_

x, y = sorted(list(map(int, input().split())), reverse=True) while 1: if x % y == 0: print(y) break else: a = y b = x % y x = a y = b print(x, y)

s728046980

Accepted

a, b = map(int, input().split()) while b: a, b = b, a % b print(a)

s233242644

p04031

u633450100

Wrong Answer

Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective.

import numpy as np N = int(input()) a = [int(i) for i in input().split()] oppai = np.array(a) max = np.max(oppai) min = np.min(oppai) x = (max+min)//2 sum = [0] * 3 for i in range(x-1,x+2,1): oppai = np.array(a) oppai -= i oppai = oppai**2 sum[i-x+1] = np.sum(oppai) print('sum[{}]'.format(i-x+1),sum[i-x+1]) chinko = np.array(sum) #print(np.argmin(chinko)+x-1) print(np.min(chinko))

s508802844

Accepted

import numpy as np N = int(input()) a = [int(i) for i in input().split()] oppai = np.array(a) max = np.max(oppai) min = np.min(oppai) x = (max+min)//2 sum = [0] * 201 for i in range(-100,101): oppai = np.array(a) oppai -= i oppai = oppai**2 sum[i+100] = np.sum(oppai) #print('sum[{}]'.format(i+100),sum[i+100]) chinko = np.array(sum) #print(np.argmin(chinko)+x-1) print(np.min(chinko))

s870858695

p03545

u112623731

Wrong Answer

Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.

numbers = input() a,b,c,d = numbers i = len(numbers)-1 op = ['+','-'] for j in range(2 ** i): o = [op[int((j>>k)&1)] for k in range(i)] f=f"{a}{o[0]}{b}{o[1]}{c}{o[2]}{d}" print(f) if eval(f) == 7 else ''

s004569861

Accepted

numbers = input() a,b,c,d = numbers i = len(numbers)-1 op = ['+','-'] for j in range(2 ** i): o = [op[int((j>>k)&1)] for k in range(i)] f=f"{a}{o[0]}{b}{o[1]}{c}{o[2]}{d}" if eval(f) == 7: print(f+'=7') break

s863304063

p02388

u309095676

Wrong Answer

Write a program which calculates the cube of a given integer x.

x=3 print(x**3)

s088329476

Accepted

import sys x=int(sys.stdin.readline()) print(x**3)

s878448065

p00513

u662488567

Wrong Answer

IOI 不動産ではマンションの賃貸を行なっている． この会社が取り扱うマンションの部屋は 1LDK で， 下図のように面積が 2xy+x+y になっている． ただし， x, y は正整数である． IOI 不動産のカタログにはマンションの面積が昇順に（狭いものから順番に）書かれているが，この中にいくつか間違い（ありえない面積のもの）が混じっていることがわかった． カタログ（入力ファイル）は N+1 行で，最初の行に部屋数が書かれていて， 続く N 行に，１行に１部屋ずつ面積が昇順に書かれている． ただし， 部屋数は 100,000 以下， 面積は (2の31乗)-1 = 2,147,483,647 以下である． ５つの入力データのうち３つまでは， 部屋数 1000 以下，面積 30000 以下である． 間違っている行数（ありえない部屋の数）を出力しなさい． 出力ファイルにおいては， 出力の最後の行にも改行コードを入れること．

prime = dict() prime_nums = list() MAXIMUM = 2**16 def get_prime(): for i in range(2, MAXIMUM): if prime.get(i, True): prime_nums.append(i) for j in range(i * 2, MAXIMUM, i): prime[j] = False def check(x): x = 2*x + 1 for prime in prime_nums: if prime **2 > x: break if x % prime == 0: return False print(prime, x) return True get_prime() n = int(input()) count = 0 for _ in range(n): if check(int(input())): count += 1 print(count)

s816125111

Accepted

prime = dict() prime_nums = list() MAXIMUM = 2**16 def get_prime(): for i in range(2, MAXIMUM): if prime.get(i, True): prime_nums.append(i) for j in range(i * 2, MAXIMUM, i): prime[j] = False def check(x): x = 2*x + 1 for prime in prime_nums: if prime **2 > x: break if x % prime == 0: return False return True get_prime() n = int(input()) count = 0 for _ in range(n): if check(int(input())): count += 1 print(count)

s672205061

p04044

u881816188

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()) s=[str(input()) for _ in range(n)] sort_s=sorted(s) print(''.join(s))

s830177267

Accepted

n,l=map(int,input().split()) s=[str(input()) for _ in range(n)] sort_s=sorted(s) print(''.join(sort_s))

s716226271

p02396

u244790947

Wrong Answer

In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.

inputs = [] for i in range(10000) : inputraw = input() if inputraw == '0' : break inputs.append(inputraw) for i, input in enumerate(inputs) : print("Case "+str(i)+": "+ str(input))

s622778505

Accepted

inputs = [] for i in range(10000) : inputraw = input() if inputraw == '0' : break inputs.append(inputraw) for i, input in enumerate(inputs) : print("Case "+str(i+1)+": "+ str(input))

s905308964

p02659

u084491185

Wrong Answer

Compute A \times B, truncate its fractional part, and print the result as an integer.

A, B = input().split() a = float(A) b = float(B) S = a * b ans = S // 1 print(ans)

s280444648

Accepted

A, B = input().split() a = int(A) b = int(B.replace('.', '')) S = a * b ans = S // 100 print(int(ans))

s666243675

p02578

u298976461

Wrong Answer

N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal.

from sys import stdin input = stdin.readline n = input().rstrip() a = list(map(int, input().split())) ans = 0 for i in range(len(a)-1): if a[i] > a[i+1]: diff = a[i] - a[i+1] ans += diff print(ans)

s999495991

Accepted

from sys import stdin input = stdin.readline n = input().rstrip() a = list(map(int, input().split())) ans = 0 for i in range(len(a)-1): if a[i] > a[i+1]: diff = a[i] - a[i+1] a[i+1] += diff ans += diff print(ans)

s272714615

p03139

u088488125

Wrong Answer

We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.

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

s012328160

Accepted

n,a,b=map(int, input().split()) print(min(a,b),max(0,a+b-n))

s472433075

p03711

u268516119

Wrong Answer

Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.

x,y=map(int,input().split()) a=[4,6,9,11] if x==2 or y==2: ans='No' elif x in a==y in a: ans='Yes' else:ans='No' print(ans)

s372980655

Accepted

x,y=map(int,input().split()) a=[4,6,9,11] if ((x in a)==(y in a)) and x!=2 and y!=2: ans='Yes' else:ans='No' print(ans)

s006679200

p03711

u682730715

Wrong Answer

Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.

a = [1,3,5,7,8,10,12] b = [4,6,9,11] x, y = map(int,input().split()) for i in a: if x == i: x = "a" if y == i: y == "a" for i in b: if x == i: x = "b" if y == i: y == "b" if x == y: print("Yes") else: print("No")

s566165673

Accepted

a = [1,3,5,7,8,10,12] b = [4,6,9,11] x, y = map(int,input().split()) for i in a: if x == i: x = "a" if y == i: y = "a" for i in b: if x == i: x = "b" if y == i: y = "b" if x == y: print("Yes") else: print("No")

s979261084

p03795

u820315626

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.

import math n = int(input()) print(math.factorial(n) / (pow(10, 9) + 7))

s793533195

Accepted

n = int(input()) print(800 *n - ( 200 * int(n / 15)))

s326989942

p01225

u104171359

Wrong Answer

あなたの友達は最近 UT-Rummy というカードゲームを思いついた． このゲームで使うカードには赤・緑・青のいずれかの色と1から9までのいずれかの番号が つけられている． このゲームのプレイヤーはそれぞれ9枚の手札を持ち， 自分のターンに手札から1枚選んで捨てて， 代わりに山札から1枚引いてくるということを繰り返す． このように順番にターンを進めていき， 最初に手持ちのカードに3枚ずつ3つの「セット」を作ったプレイヤーが勝ちとなる． セットとは，同じ色の3枚のカードからなる組で，すべて同じ数を持っているか 連番をなしているもののことを言う． 連番に関しては，番号の巡回は認められない． 例えば，7, 8, 9は連番であるが 9, 1, 2は連番ではない． あなたの友達はこのゲームをコンピュータゲームとして売り出すという計画を立てて， その一環としてあなたに勝利条件の判定部分を作成して欲しいと頼んできた． あなたの仕事は，手札が勝利条件を満たしているかどうかを判定する プログラムを書くことである．

#!usr/bin/env python3 import sys def main(): WINNING_HANDS = [ '111', '222', '333', '444', '555', '666', '777', '888', '999', '123', '234', '345', '456', '567', '678', '789' ] num_of_datasets = int(sys.stdin.readline().strip('\n')) datasets = [{'R': [], 'G': [], 'B': []} for _ in range(num_of_datasets)] results = list() for dataset in range(num_of_datasets): n_set = [int(num) for num in sys.stdin.readline().strip('\n').split()] c_set = [colour for colour in sys.stdin.readline().strip('\n').split()] for idx, colour in enumerate(c_set): if colour == 'R': datasets[dataset]['R'].append(n_set[idx]) elif colour == 'G': datasets[dataset]['G'].append(n_set[idx]) elif colour == 'B': datasets[dataset]['B'].append(n_set[idx]) match_count = int() for rgb_key in datasets[dataset]: nums = ''.join(str(num) for num in sorted(datasets[dataset][rgb_key])) for hand in WINNING_HANDS: if hand in nums: match_count += 1 if match_count == 3: results.append(1) continue if dataset < num_of_datasets-1: results.append(0) for result in results: print(result) if __name__ == '__main__': main()

s674013433

Accepted

#!usr/bin/env python3 import sys def main(): WINNING_HANDS = [ '111', '222', '333', '444', '555', '666', '777', '888', '999', '123', '234', '345', '456', '567', '678', '789' ] num_of_datasets = int(sys.stdin.readline().strip('\n')) datasets = [{'R': [], 'G': [], 'B': []} for _ in range(num_of_datasets)] results = list() for dataset in range(num_of_datasets): n_set = [num for num in sys.stdin.readline().strip('\n').split()] c_set = [colour for colour in sys.stdin.readline().strip('\n').split()] for idx, colour in enumerate(c_set): if colour == 'R': datasets[dataset]['R'].append(n_set[idx]) elif colour == 'G': datasets[dataset]['G'].append(n_set[idx]) elif colour == 'B': datasets[dataset]['B'].append(n_set[idx]) match_count = int() for rgb_key in datasets[dataset]: if len(datasets[dataset][rgb_key]) > 1: nums = sorted(datasets[dataset][rgb_key]) else: continue for hand in WINNING_HANDS: while hand[0] in nums and hand[1] in nums and hand[2] in nums: tmp = nums.copy() if hand[0] in tmp: tmp.pop(tmp.index(hand[0])) if hand[1] in tmp: tmp.pop(tmp.index(hand[1])) if hand[2] in tmp: tmp.pop(tmp.index(hand[2])) match_count += 1 nums = tmp else: break else: break else: break if match_count == 3: results.append(1) break if dataset < num_of_datasets and match_count != 3: results.append(0) for result in results: print(result) if __name__ == '__main__': main()

s257962862

p04031

u175590965

Wrong Answer

Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective.

n = int(input()) a = list(map(int,input().split())) ans = 10**10 for i in range(-100,101): t = 0 for j in a: t += (j-i)**2 if t < ans: ans =t print(ans)

s580802146

Accepted

n = int(input()) a = list(map(int,input().split())) ans = 10**10 for i in range(-100,101): t = 0 for j in a: t += (j-i)**2 if t < ans: ans =t print(ans)

s926917039

p03826

u611090896

Wrong Answer

There are two rectangles. The lengths of the vertical sides of the first rectangle are A, and the lengths of the horizontal sides of the first rectangle are B. The lengths of the vertical sides of the second rectangle are C, and the lengths of the horizontal sides of the second rectangle are D. Print the area of the rectangle with the larger area. If the two rectangles have equal areas, print that area.

a,b,c,d = map(int,input().split()) print(a*b,c*d)

s862737407

Accepted

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

s556484680

p02744

u309141201

Wrong Answer

In this problem, we only consider strings consisting of lowercase English letters. Strings s and t are said to be **isomorphic** when the following conditions are satisfied: * |s| = |t| holds. * For every pair i, j, one of the following holds: * s_i = s_j and t_i = t_j. * s_i \neq s_j and t_i \neq t_j. For example, `abcac` and `zyxzx` are isomorphic, while `abcac` and `ppppp` are not. A string s is said to be in **normal form** when the following condition is satisfied: * For every string t that is isomorphic to s, s \leq t holds. Here \leq denotes lexicographic comparison. For example, `abcac` is in normal form, but `zyxzx` is not since it is isomorphic to `abcac`, which is lexicographically smaller than `zyxzx`. You are given an integer N. Print all strings of length N that are in normal form, in lexicographically ascending order.

n = int(input()) pw = ['a', 'b', 'c'] ans = ['' for _ in range(n)] # print(ord('b')) def dfs(i): if i == n: print(''.join(ans)) return for j in range(97, 100): # print(j) ans[i] = chr(j) dfs(i+1) dfs(0)

s224853777

Accepted

n = int(input()) ans = ['' for _ in range(n)] def dfs(i, mx): if i == n: print(''.join(ans)) return for j in range(mx+1): ans[i] = chr(ord('a') + j) dfs(i+1, max(j+1, mx)) dfs(0, 0)

s117989465

p03796

u597455618

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()) ans = 0 for i in range(n+1): ans *= i print(ans % (10 **9 + 7))

s854291542

Accepted

n = int(input()) ans = 1 for i in range(1,n+1): ans *= i ans %= (10**9 + 7) print(ans)

s353565729

p03739

u936985471

Wrong Answer

You are given an integer sequence of length N. The i-th term in the sequence is a_i. In one operation, you can select a term and either increment or decrement it by one. At least how many operations are necessary to satisfy the following conditions? * For every i (1≤i≤n), the sum of the terms from the 1-st through i-th term is not zero. * For every i (1≤i≤n-1), the sign of the sum of the terms from the 1-st through i-th term, is different from the sign of the sum of the terms from the 1-st through (i+1)-th term.

N=int(input()) A=list(map(int,input().split())) cur=A[0] ans=0 isplus=True ind=1 if cur<0: isplus=False if cur==0: allzero=True firstNotZero=0 firstNotZeroInd=0 for i in range(N): if A[i]!=0: firstNotZero=A[i] firstNotZeroInd=i allzero=False break if allzero: print(N) exit(0) if firstNotZero>0: cur=-1 ind=firstNotZeroInd isplus=False print("ind",ind,"isplus",isplus) else: cur=1 ind=firstNotZeroInd isplus=True print("ind",ind,"isplus",isplus) for i in range(ind,N): print("i",i,"cur",cur,"A[i]",A[i]) if isplus: if cur+A[i]>=0: diff=abs((cur+A[i])-(-1)) print("cur",cur,"cur+A[i]",cur+A[i]," -> -1 diff",diff) ans+=diff print("ans",ans) cur=-1 else: print("no problem") cur+=A[i] isplus=False else: if cur+A[i]<=0: diff=abs((cur+A[i])-1) print("cur",cur,"cur+A[i]",cur+A[i]," -> 1 diff",diff) ans+=diff print("ans",ans) cur=1 else: print("no problem") cur+=A[i] isplus=True print(ans)

s109731590

Accepted

N=int(input()) A=list(map(int,input().split())) ans=[0]*2 for pat in range(2): cur=0 isplus=pat for i in range(N): cur+=A[i] if isplus: if cur<=0: ans[pat]+=abs(cur-1) cur=1 else: if cur>=0: ans[pat]+=abs(cur-(-1)) cur=-1 isplus^=1 print(min(ans))

s112239166

p03386

u394794741

Wrong Answer

Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.

A,B,K = map(int, input().split()) if (B-A)>2*K: for i in range(A,A+K,1): print(i) for j in range(B-K,B,1): print(j) else: for k in range(A,B+1,1): print(k)

s157996680

Accepted

A,B,K = map(int, input().split()) if (B-A)>=2*K: for i in range(A,A+K,1): print(i) for j in range(B-K+1,B+1,1): print(j) else: for k in range(A,B+1,1): print(k)

s873193017

p03599

u094191970

Wrong Answer

Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the beaker. * Operation 3: Put C grams of sugar into the beaker. * Operation 4: Put D grams of sugar into the beaker. In our experimental environment, E grams of sugar can dissolve into 100 grams of water. Snuke will make sugar water with the highest possible density. The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker. Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it. If there is more than one candidate, any of them will be accepted. We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent. Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water.

a,b,c,d,e,f=map(int,input().split()) def calc(w,s): sugar=0 # while sugar/(w+sugar)<=e/(100+e) and w+sugar<=f: while sugar/w <= e/100 and w+sugar<=f: sugar+=s return w,sugar-s w1,s1=calc(100*a,c) w2,s2=calc(100*a,d) w3,s3=calc(100*b,c) w4,s4=calc(100*b,d) print(w1,s1,w2,s2,w3,s3,w4,s4) ansx=0 ansy=0 mol=0 for i in range(30): for j in range(30): for k in range(30): for l in range(30): sugar=s1*i+s2*j+s3*k+s4*l solvent=sugar+w1*i+w2*j+w3*k+w4*l # print(sugar,solvent) if solvent!=0 and sugar/solvent<=e/(100+e) and solvent<=f: if mol < (100*sugar)/solvent: ansx=solvent ansy=sugar mol=(100*sugar)/solvent print(ansx,ansy)

s290953549

Accepted

a,b,c,d,e,f=map(int,input().split()) water_g=[] for i in range(f//100+1): for j in range(f//100+1): # print(i,j) w=100*a*i + 100*b*j if w<=f: water_g.append(w) else: break water_g=sorted(list(set(water_g)),reverse=True) sugar_g=[] for i in range(f*e//100): for j in range(f*e//100): s=c*i + d*j if s<=f: sugar_g.append(s) else: break sugar_g=sorted(list(set(sugar_g)),reverse=True) #print(water_g, sugar_g) ans_water=100*a ans_sugar=0 for water in water_g: if water==0: break for sugar in sugar_g: if water+sugar<=f and sugar/water<=e/100: if sugar/(sugar+water)>ans_sugar/(ans_sugar+ans_water): ans_water=water ans_sugar=sugar print(ans_water+ans_sugar,ans_sugar)

s464009415

p03846

u036104576

Wrong Answer

There are N people, conveniently numbered 1 through N. They were standing in a row yesterday, but now they are unsure of the order in which they were standing. However, each person remembered the following fact: the absolute difference of the number of the people who were standing to the left of that person, and the number of the people who were standing to the right of that person. According to their reports, the difference above for person i is A_i. Based on these reports, find the number of the possible orders in which they were standing. Since it can be extremely large, print the answer modulo 10^9+7. Note that the reports may be incorrect and thus there may be no consistent order. In such a case, print 0.

import sys import itertools # import numpy as np import time import math import heapq from collections import defaultdict sys.setrecursionlimit(10 ** 7) INF = 10 ** 18 MOD = 10 ** 9 + 7 read = sys.stdin.buffer.read readline = sys.stdin.buffer.readline readlines = sys.stdin.buffer.readlines # map(int, input().split()) N = int(input()) A = list(map(int, input().split())) X = [0] * N for i, a in enumerate(A): j = (N - 1 - a) // 2 X[j] += 1 for i, x in enumerate(X): if x == 1 and not (i == (N + 1) // 2): print(0) exit() if x > 2: print(0) exit() ans = 1 for x in X: if x == 0: break ans *= x ans %= MOD print(ans)

s715394220

Accepted

import sys import itertools # import numpy as np import time import math import heapq from collections import defaultdict sys.setrecursionlimit(10 ** 7) INF = 10 ** 18 MOD = 10 ** 9 + 7 read = sys.stdin.buffer.read readline = sys.stdin.buffer.readline readlines = sys.stdin.buffer.readlines # map(int, input().split()) N = int(input()) A = list(map(int, input().split())) X = [0] * N for i, a in enumerate(A): j = (N - 1 - a) // 2 X[j] += 1 for i, x in enumerate(X): if x == 1 and not (i == (N + 1) // 2 - 1): print(0) exit() if x > 2: print(0) exit() ans = 1 for x in X: if x == 0: break ans *= x ans %= MOD print(ans)

s525425333

p03457

u362560965

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.

import sys N = int(input()) t = [0 for i in range(N)] x = [0 for i in range(N)] y = [0 for i in range(N)] for i in range(N): t[i], x[i], y[i] = map(int, input().split()) for i in range(N-1): time = t[i+1] - t[i] dist = abs(x[i+1] - x[i]) + abs(y[i+1] - y[i]) if (time - dist) >= 0 and (time - dist) % 2 == 0: pass else: print("No") sys.exit()

s765590970

Accepted

import sys N = int(input()) t = [0 for i in range(N+1)] x = [0 for i in range(N+1)] y = [0 for i in range(N+1)] for i in range(1, N+1): t[i], x[i], y[i] = map(int, input().split()) for i in range(N): time = t[i+1] - t[i] dist = abs(x[i+1] - x[i]) + abs(y[i+1] - y[i]) if (time - dist) >= 0 and (time - dist) % 2 == 0: pass else: print("No") sys.exit() print("Yes")

s745053512

p03993

u548624367

Wrong Answer

There are N rabbits, numbered 1 through N. The i-th (1≤i≤N) rabbit likes rabbit a_i. Note that no rabbit can like itself, that is, a_i≠i. For a pair of rabbits i and j (i＜j), we call the pair (i，j) a _friendly pair_ if the following condition is met. * Rabbit i likes rabbit j and rabbit j likes rabbit i. Calculate the number of the friendly pairs.

N = int(input()) a = list(map(lambda x:int(x)-1,input().split())) ans = 0 for i in range(N): if a[i] == i: ans += 1 print(ans//2)

s466176747

Accepted

N = int(input()) a = list(map(lambda x:int(x)-1,input().split())) ans = 0 for i in range(N): if a[a[i]] == i: ans += 1 print(ans//2)

s945862364

p02842

u991619971

Wrong Answer

Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.

N=int(input()) x=N//1.08 if int(x*1.08) == N: print(x) elif int((x+1)*1.08) == N: print(x+1) else: print(':(')

s271426830

Accepted

#N,a,b = map(int,input().split()) N = int(input()) X=int(N/1.08) if int(X*1.08) == N: print(X) elif int((X+1)*1.08) == N: print(X+1) else: print(':(')

s315825219

p03854

u279718559

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

s862016273

Accepted

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

s989904302

p02406

u343251190

Wrong Answer

In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }

import sys n = int(input()) sys.stdout.write(" ") answer = [] for i in range(n + 1): if i%3 == 0: answer.append(i) elif '3' in str(i): answer.append(i) print(*answer)

s536969289

Accepted

import sys n = int(input()) sys.stdout.write(" ") answer = [] for i in range(3, n + 1): if i%3 == 0: answer.append(i) elif '3' in str(i): answer.append(i) print(*answer)

s926348020

p02694

u894172792

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()) n = 100 c = 0 while n -100< x: n = int(n *1.01) c += 1 print(c)

s533618410

Accepted

x = int(input()) n = 100 c = 0 while n < x: n += int(n / 100) c += 1 #print(n,c) print(c)

s002720889

p03371

u193598069

Wrong Answer

"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.

A, B, C, X, Y = map(int, input().split()) ans = 0 if A + B <= C: ans = A*X + B*Y else: ans = 2*C*max(X, Y) print (ans)

s183826545

Accepted

A, B, C, X, Y = map(int, input().split()) ans = 0 if A + B <= 2*C: ans = A*X + B*Y else: ans += 2*C*min(X, Y) if X >= Y: ans += min(A, 2*C) * (X-Y) else: ans += min(B, 2*C) * (Y-X) print (ans)

s844334165

p02842

u624075921

Wrong Answer

Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.

x=int(input()) val = x//1.08 if val*1.08 == x: print(val) else: print(":(")

s208735358

Accepted

from math import floor n = int(input()) val = round(n/1.08) if floor(val*1.08) == n: print(val) elif floor((val+1)*1.08) ==n: print(val+1) else: print(":(")

s371299678

p03836

u623814058

Wrong Answer

Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.

SX,SY,TX,TY=map(int,input().split()) vx=TX-SX vy=TY-SY print('U'*vy+'R'*vx+'L'*vx+'D'*vy+'L'+'U'*(vy+1)+'R'*(vx+1)+'D'+ 'R'+'D'*(vy+1)+'L'*(vx+1)+'U')

s775873384

Accepted

p,q,x,y=map(int,input().split()) x-=p y-=q u='U'*y+'R'*x d='D'*y+'L'*x+'LU' print(u+d+u+'RDRD'+d)

s984833458

p03361

u459150945

Wrong Answer

We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j). Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`. However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black. Determine if square1001 can achieve his objective.

import sys H, W = map(int, input().split()) s = [input() for _ in range(H)] print(s) dx = [1, 0, -1, 0] dy = [0, 1, 0, -1] for i in range(H): for j in range(W): if s[i][j] == '#': for k in range(4): try: if s[i + dy[k]][j + dx[k]] == '#': flag = False break else: flag = True except IndexError: pass if flag: print('No') sys.exit() print('Yes')

s146679806

Accepted

import sys H, W = map(int, input().split()) s = [input() for _ in range(H)] dx = [1, 0, -1, 0] dy = [0, 1, 0, -1] for i in range(H): for j in range(W): if s[i][j] == '#': for k in range(4): try: if s[i + dy[k]][j + dx[k]] == '#': flag = False break else: flag = True except IndexError: pass if flag: print('No') sys.exit() print('Yes')

s442101004

p03854

u179987763

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() if s.find("1") != -1: print("NO") elif s.find("2") != -1: print("NO") else: a = s.replace("dream", "11111").replace("erase", "22222").replace("11111er", "1111111").replace("22222r", "111111") print(a) if a.count("1") == len(a): print("YES") else: print("NO")

s433854653

Accepted

s = input() if s.find("1") != -1: print("NO") elif s.find("2") != -1: print("NO") else: a = s.replace("dream", "11111").replace("erase", "22222").replace("11111er", "1111111").replace("22222r", "111111").replace("22222", "11111") if a.count("1") == len(a): print("YES") else: print("NO")

s980145470

p02406

u498462680

Wrong Answer

In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }

inData = int(input()) seqData = list(range(1,inData+1,1)) outData = [0] * inData for thisData in seqData: if thisData == inData: print(thisData) else: if thisData%3 == 0: print(thisData, end = "") else: pass

s510336343

Accepted

inData = int(input()) seqData = list(range(1,inData+1,1)) outData = [0] * inData for thisData in seqData: if thisData == inData: if thisData%3 == 0 or thisData%10==3: print(" "+str(thisData)) else: for i in range(5): i=i+1 over = 10 ** i tmp = thisData/over if int(tmp % 10) == 3: print(" "+str(thisData)) break else: pass print("") else: if thisData%3 == 0 or thisData%10==3: print(" "+str(thisData), end = "") else: for i in range(5): i=i+1 over = 10 ** i tmp = thisData/over if int(tmp % 10) == 3: print(" "+str(thisData), end = "") break else: pass

s034841135

p02383

u216804574

Wrong Answer

Write a program to simulate rolling a dice, which can be constructed by the following net. As shown in the figures, each face is identified by a different label from 1 to 6. Write a program which reads integers assigned to each face identified by the label and a sequence of commands to roll the dice, and prints the integer on the top face. At the initial state, the dice is located as shown in the above figures.

d = [_ for _ in input().split()] commands = input() for c in commands: if c == 'E': d[0], d[1], d[3], d[5] = d[1], d[5], d[0], d[ 3] elif c == 'W': d[0], d[1], d[3], d[5] = d[3], d[0], d[5], d[ 1] elif c == 'N': d[0], d[2], d[4], d[5] = d[2], d[5], d[0], d[ 4] elif c == 'S': d[0], d[2], d[4], d[5] = d[4], d[0], d[5], d[ 2] print(d[0])

s804616009

Accepted

class dice: def __init__(self, dice): self.d = dice def roll(self, c): d = self.d if c == 'E': d[0], d[2], d[3], d[5] = d[3], d[0], d[5], d[2] elif c == 'W': d[0], d[2], d[3], d[5] = d[2], d[5], d[0], d[3] elif c == 'N': d[0], d[1], d[4], d[5] = d[1], d[5], d[0], d[4] elif c == 'S': d[0], d[1], d[4], d[5] = d[4], d[0], d[5], d[1] self.d = d def print(self): print(self.d[0]) d = dice([_ for _ in input().split()]) commands = input() for c in commands: d.roll(c) d.print()

s592881866

p02417

u782850499

Wrong Answer

Write a program which counts and reports the number of each alphabetical letter. Ignore the case of characters.

text = input() count={} for i in range(97,97+26): count[chr(i)] = 0 for letter in text: if letter in count: count[letter] += 1 for x,y in sorted(count.items()): print("{0} : {1}".format(x,y))

s086014793

Accepted

count={} for i in range(97,97+26): count[chr(i)] = 0 text=[] while True: try: text.append(input()) except EOFError: break text_out="".join(text) text_lower=text_out.lower() for letter in text_lower: if letter in count: count[letter] += 1 for x,y in sorted(count.items()): print("{0} : {1}".format(x,y))

s151070711

p03436

u358859892

Wrong Answer

We have an H \times W grid whose squares are painted black or white. The square at the i-th row from the top and the j-th column from the left is denoted as (i, j). Snuke would like to play the following game on this grid. At the beginning of the game, there is a character called Kenus at square (1, 1). The player repeatedly moves Kenus up, down, left or right by one square. The game is completed when Kenus reaches square (H, W) passing only white squares. Before Snuke starts the game, he can change the color of some of the white squares to black. However, he cannot change the color of square (1, 1) and (H, W). Also, changes of color must all be carried out before the beginning of the game. When the game is completed, Snuke's score will be the number of times he changed the color of a square before the beginning of the game. Find the maximum possible score that Snuke can achieve, or print -1 if the game cannot be completed, that is, Kenus can never reach square (H, W) regardless of how Snuke changes the color of the squares. The color of the squares are given to you as characters s_{i, j}. If square (i, j) is initially painted by white, s_{i, j} is `.`; if square (i, j) is initially painted by black, s_{i, j} is `#`.

from collections import deque H, W = map(int, input().split()) S = [input() for _ in range(H)] s = 0 for c in S: s += c.count('.') que = deque() dist = [[-1 for j in range(W)] for i in range(H)] dist[0][0] = 0 st = (0, 0) que.append(st) dydx = [(1, 0), (0, 1), (-1, 0), (0, -1)] while que: p, q = que.popleft() for dy, dx in dydx: print(dy, dx) if 0 <= p+dy and p+dy < H and 0 <= q+dx and q+dx < W: if dist[p+dy][q+dx] != -1 or S[p+dy][q+dx] != '.': continue dist[p+dy][q+dx] = dist[p][q] + 1 que.append((p+dy, q+dx)) if dist[H-1][W-1] == -1: print(-1) else: print(s-dist[H-1][W-1]-1)

s565488903

Accepted

from collections import deque H, W = map(int, input().split()) S = [input() for _ in range(H)] s = 0 for c in S: s += c.count('.') que = deque() dist = [[-1 for j in range(W)] for i in range(H)] dist[0][0] = 0 st = (0, 0) que.append(st) dydx = [(1, 0), (0, 1), (-1, 0), (0, -1)] while que: p, q = que.popleft() for dy, dx in dydx: if 0 <= p+dy and p+dy < H and 0 <= q+dx and q+dx < W: if dist[p+dy][q+dx] != -1 or S[p+dy][q+dx] != '.': continue dist[p+dy][q+dx] = dist[p][q] + 1 que.append((p+dy, q+dx)) if dist[H-1][W-1] == -1: print(-1) else: print(s-dist[H-1][W-1]-1)

s035665166

p02694

u629540524

Time Limit Exceeded

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()) n, y = 100, 0 while n < x: n = n % 1.01 y = y + 1 print(n) print(y)

s557224429

Accepted

import math x = int(input()) n, y = 100, 0 while n < x: n = math.floor(n * 1.01) y = y + 1 print(y)

s359846682

p02614

u536560967

Wrong Answer

We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.

H, W, K=map(int,input().split()) F = [input() for _ in range(H)] print(F) ans = 0 for i in range(2**H): for j in range(2**W): cnt = 0 for ii in range(H): for jj in range(W): if (i>>ii)%2==0 or (j>>jj)%2==0 or F[ii][jj]==".": continue cnt+=1 if cnt==K: ans+=1 print(ans)

s766715673

Accepted

H, W, K=map(int,input().split()) F = [input() for _ in range(H)] ans = 0 for i in range(2**H): for j in range(2**W): cnt = 0 for ii in range(H): for jj in range(W): if (i>>ii)%2==0 or (j>>jj)%2==0 or F[ii][jj]==".": continue cnt+=1 if cnt==K: ans+=1 print(ans)

s757965259

p03636

u385825353

Wrong Answer

The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.

s = input() n = len(s) print(s[0]+str(n)+s[-1])

s135467792

Accepted

s = input() n = len(s) print(s[0]+str(n-2)+s[-1])

s115002494

p04000

u560301743

Wrong Answer

We have a grid with H rows and W columns. At first, all cells were painted white. Snuke painted N of these cells. The i-th ( 1 \leq i \leq N ) cell he painted is the cell at the a_i-th row and b_i-th column. Compute the following: * For each integer j ( 0 \leq j \leq 9 ), how many subrectangles of size 3×3 of the grid contains exactly j black cells, after Snuke painted N cells?

H, W, N = [int(v) for v in input().split()] points = (map(int, input().split()) for _ in range(N)) memory = {} for ai, bi in points: for i in range(max(2, ai - 1), min(H - 1, ai + 1) + 1): for j in range(max(2, bi - 1), min(W - 1, bi + 1) + 1): key = (i, j) if not key in memory: memory[key] = 0 memory[key] += 1 counter = {i: 0 for i in range(1, 9 + 1)} for v in memory.values(): counter[v] += 1 for i in range(1, 9 + 1): print(counter[i])

s883653680

Accepted

H, W, N = [int(v) for v in input().split()] points = (map(int, input().split()) for _ in range(N)) memory = {} for ai, bi in points: for i in range(max(2, ai - 1), min(H - 1, ai + 1) + 1): for j in range(max(2, bi - 1), min(W - 1, bi + 1) + 1): key = (i, j) if not key in memory: memory[key] = 0 memory[key] += 1 counter = {i: 0 for i in range(1, 10)} for v in memory.values(): counter[v] += 1 print((H - 2) * (W - 2) - sum(counter.values())) for i in range(1, 10): print(counter[i])

s834545163

p03729

u635141733

Wrong Answer

You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.

x, y, z = [i for i in input().split()] if x[-1] != y[0]: print('no') elif y[-1] != z[0]: print('no') else: print ('yes')

s070683076

Accepted

x, y, z = [i for i in input().split()] if x[-1] != y[0]: print('NO') elif y[-1] != z[0]: print('NO') else: print ('YES')

s658959021

p02388

u907607057

Wrong Answer

Write a program which calculates the cube of a given integer x.

import sys def main(): x = int(sys.stdin.readline()) print(x**2) return if __name__ == '__main__': main()

s842148726

Accepted

import sys def main(): x = int(sys.stdin.readline()) print(x**3) return if __name__ == '__main__': main()

s684144750

p03386

u672475305

Wrong Answer

Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.

a,b,k = map(int,input().split()) lst = [] for i in range(a,a+k): if a<=i<=b: lst.append(i) for i in range(b-k+1,b+1): if a<=i<=b: lst.append(i) ans = list(set(lst)) for i in range(len(ans)): print(ans[i])

s209590361

Accepted

a,b,k = map(int,input().split()) lst = [] for i in range(a,a+k): if a<=i<=b: lst.append(i) for i in range(b-k+1,b+1): if a<=i<=b: lst.append(i) ans = list(set(lst)) ans.sort() for i in range(len(ans)): print(ans[i])

s624011640

p02397

u914146430

Wrong Answer

Write a program which reads two integers x and y, and prints them in ascending order.

xy_data=[] while True: xy=list(map(int,input().split())) xy.sort() print(xy) if xy[0]==xy[1]==0: break else: xy_data.append(xy) for xy in xy_data: print(*xy)

s451332735

Accepted

xy_data=[] while True: xy=list(map(int,input().split())) xy.sort() if xy[0]==xy[1]==0: break else: xy_data.append(xy) for xy in xy_data: print(*xy)

s247461944

p02972

u815878613

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 = list(map(int, input().split()))[::-1] B = list(range(1, N + 1))[::-1] C = [] for a, b in zip(A, B): n = 1 cnt = 0 while b * n <= N: cnt += A[N - b * n] n += 1 C.append(cnt % 2) print(*C[::-1]) main()

s502452246

Accepted

import sys import numpy as np readline = sys.stdin.buffer.readline def main(): N = int(readline()) A = [0] + list(map(int, readline().split())) A = np.array(A) C = np.zeros(N + 1, dtype=np.int32) if A[-1] == 1: C[-1] == 1 for j in range(N, 0, -1): cnt = np.count_nonzero(C[j::j]) C[j] = (cnt % 2) ^ A[j] ans = np.sum(C) print(ans) if ans != 0: B = np.arange(N + 1) li = B[C == 1] print(*li, sep='\n') main()

s182830078

p04043

u456394640

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.

nums = [int(e) for e in input().split()] if nums.count(5) == 2 and nums.count(7) == 1: print('Yes') else: print('No')

s764794055

Accepted

nums = [int(e) for e in input().split()] if nums.count(5) == 2 and nums.count(7) == 1: print('YES') else: print('NO')

s896694709

p03719

u016393440

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 = (int(i) for i in input().split()) if a <= c and b >= c: print('YES') else: print('NO')

s366329962

Accepted

a, b, c = (int(i) for i in input().split()) if a <= c and b >= c: print('Yes') else: print('No')

s807465990

p03610

u137808818

Wrong Answer

You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.

n = input() print(n[0:3:2])

s460974113

Accepted

n = input() print(n[::2])

s644102748

p00434

u797673668

Wrong Answer

JOI大学のM教授はプログラミングの授業を担当している．クラスには30人の受講生がいて各受講生には1番から30番までの出席番号がふられている．この授業の課題を28人の学生が提出した．提出した28人の出席番号から提出していない2人の出席番号を求めるプログラムを作成せよ．

print(*sorted(set(range(1, 31)) ^ set(int(input()) for _ in range(28))))

s110055716

Accepted

for i in sorted(set(range(1, 31)) ^ set(int(input()) for _ in range(28))): print(i)

s841199939

p03673

u024340351

Wrong Answer

You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.

n = int(input()) l = list(map(int, input().split())) ogya = list() for i in range (0, n): if i%2==0: ogya.append(l[i]) else: ogya.insert(0, l[i]) if n%2==0: print(ogya, sep=" ") else: ogya.reverse() print(ogya, sep=" ")

s256628969

Accepted

n = int(input()) L = list(map(int, input().split())) l = [] if n == 1: print(*L) exit() if n%2 == 0: for i in range (0,n//2): l.append(L[n-1-2*i]) for i in range (0,n//2): l.append(L[2*i]) else: for i in range (0,n//2+1): l.append(L[n-1-2*i]) for i in range (0,n//2): l.append(L[2*i+1]) print(*l)

s148953251

p02841

u881100099

Wrong Answer

In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019. Integers M_1, D_1, M_2, and D_2 will be given as input. It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1. Determine whether the date 2019-M_1-D_1 is the last day of a month.

x=input().split() print(x)

s566194048

Accepted

x=input().split() y=input().split() if(x[0]!=y[0]): print("1") else: print("0")

s000544424

p03813

u419963262

Wrong Answer

Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise.

x=int(input()) a=x%11 b=x//11 if 0<a<6: ans=2*b+1 elif a==0: ans=2*b elif 5<a<11: ans=2*(b+1) print(ans)

s977887084

Accepted

N=int(input()) if N<1200: print("ABC") else: print("ARC")

s740361248

p03569

u560867850

Wrong Answer

We have a string s consisting of lowercase English letters. Snuke can perform the following operation repeatedly: * Insert a letter `x` to any position in s of his choice, including the beginning and end of s. Snuke's objective is to turn s into a palindrome. Determine whether the objective is achievable. If it is achievable, find the minimum number of operations required.

s = input() n = len(s) sr = s.replace("x", "") if sr != "".join(reversed(sr)): print(-1) else: ans = 0 l = 0 r = -1 while l < n and -r <= n: if s[l] == s[r]: l += 1 r -= 1 continue elif s[l] == "x" and s[r] != "x": ans += 1 l += 1 elif s[l] != "x" and s[r] == "x": ans += 1 r -= 1 print(ans)

s796922911

Accepted

s = input() n = len(s) sr = s.replace("x", "") if sr != "".join(reversed(sr)): print(-1) else: ans = 0 l = 0 r = -1 while l - r < n: if s[l] == s[r]: l += 1 r -= 1 continue elif s[l] == "x" and s[r] != "x": ans += 1 l += 1 elif s[l] != "x" and s[r] == "x": ans += 1 r -= 1 print(ans)

s749726134

p03719

u858436319

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<=b and b<=c: print('Yes') else: print('No')

s402625272

Accepted

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

s615303956

p03658

u747602774

Wrong Answer

Snuke has N sticks. The length of the i-th stick is l_i. Snuke is making a snake toy by joining K of the sticks together. The length of the toy is represented by the sum of the individual sticks that compose it. Find the maximum possible length of the toy.

N,K=map(int,input().split()) li=list(map(int,input().split())) li.sort(reverse=True) ans=0 for k in range(K): ans=ans+li[k] print(ans)

s794527094

Accepted

N,K=map(int,input().split()) li=list(map(int,input().split())) li.sort(reverse=True) ans=0 for k in range(K): ans=ans+li[k] print(ans)

s832096378

p03578

u125545880

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.

import collections def main(): N = int(input()) D = list(map(int, input().split())) M = int(input()) T = list(map(int, input().split())) Dcounter = collections.Counter(D) Tcounter = collections.Counter(T) for t in T: if Dcounter[t] - Tcounter[t] < 0: print('No') return print('Yes') if __name__ == "__main__": main()

s109801051

Accepted

import collections def main(): N = int(input()) D = list(map(int, input().split())) M = int(input()) T = list(map(int, input().split())) Dcounter = collections.Counter(D) Tcounter = collections.Counter(T) for t in T: if Dcounter[t] - Tcounter[t] < 0: print('NO') return print('YES') if __name__ == "__main__": main()

s382147843

p03371

u042113240

Wrong Answer

"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.

A, B, C, X, Y = map(int, input().split()) value = float('inf') D = max(A,B) for i in range(D+1): valueX = 2*i*C+(X-i)*A+(Y-i)*B if valueX < value: value = valueX print(str(value))

s401517905

Accepted

A, B, C, X, Y = map(int, input().split()) value = float('inf') D = max(X,Y) for i in range(D+1): E = max((X-i)*A, 0) F = max((Y-i)*B, 0) valueX = 2*i*C+E+F if valueX < value: value = valueX print(str(value))

s135852824

p02407

u482227082

Wrong Answer

Write a program which reads a sequence and prints it in the reverse order.

n = int(input()) l = list(map(int, input().split())) for i in l[::-1]: print(i)

s659811060

Accepted

# # 6a # def main(): n = int(input()) a = list(map(int, input().split())) a.reverse() print(*a) if __name__ == '__main__': main()

s305360732

p02390

u152353734

Wrong Answer

Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.

S = int(input()) h = 60 m = 60 * 60 s = 3600 * 60 print(h, m, s, sep=':')

s842841252

Accepted

S = int(input()) h = S // 3600 m = S % 3600 // 60 s = S % 60 print(h, m, s, sep=':')

s850157122

p03193

u223646582

Wrong Answer

There are N rectangular plate materials made of special metal called AtCoder Alloy. The dimensions of the i-th material are A_i \times B_i (A_i vertically and B_i horizontally). Takahashi wants a rectangular plate made of AtCoder Alloy whose dimensions are exactly H \times W. He is trying to obtain such a plate by choosing one of the N materials and cutting it if necessary. When cutting a material, the cuts must be parallel to one of the sides of the material. Also, the materials have fixed directions and cannot be rotated. For example, a 5 \times 3 material cannot be used as a 3 \times 5 plate. Out of the N materials, how many can produce an H \times W plate if properly cut?

N,H,W=map(int,input().split()) ans=0 for _ in range(N): A,B=map(int,input().split()) if A<=H and B<=W: ans+=1 print(ans)

s343300255

Accepted

N,H,W=map(int,input().split()) ans=0 for _ in range(N): A,B=map(int,input().split()) if A>=H and B>=W: ans+=1 print(ans)

s772099843

p03487

u497046426

Wrong Answer

You are given a sequence of positive integers of length N, a = (a_1, a_2, ..., a_N). Your objective is to remove some of the elements in a so that a will be a **good sequence**. Here, an sequence b is a **good sequence** when the following condition holds true: * For each element x in b, the value x occurs exactly x times in b. For example, (3, 3, 3), (4, 2, 4, 1, 4, 2, 4) and () (an empty sequence) are good sequences, while (3, 3, 3, 3) and (2, 4, 1, 4, 2) are not. Find the minimum number of elements that needs to be removed so that a will be a good sequence.

import collections N = int(input()) A = list(map(int, input().split())) count = collections.Counter() for a in A: count[a] += 1 ans = 0 for num, c in count.items(): if num < c: ans += c - num elif num > c: ans += c else: continue

s085557405

Accepted

import collections N = int(input()) A = list(map(int, input().split())) count = collections.Counter() for a in A: count[a] += 1 ans = 0 for num, c in count.items(): if num < c: ans += c - num elif num > c: ans += c else: continue print(ans)

s664008588

p03565

u987170100

Wrong Answer

E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.

S = input() T = input() len_T = len(T) lst = [] for x in range(len(S) - len_T + 1): flg = True s = S[x:len_T+x] for idx in range(len_T): if s[idx] != '?' and s[idx] != T[idx]: flg = False break if flg: tmp = S[:x] + T + S[x + len_T:] print(tmp) tmp = tmp.replace('?', 'a') lst.append(tmp) if len(lst) == 0: print('UNRESTORABLE') else: print(min(lst))

s153404058

Accepted

S = input() T = input() len_T = len(T) lst = [] for x in range(len(S) - len_T + 1): flg = True s = S[x:len_T+x] for idx in range(len_T): if s[idx] != '?' and s[idx] != T[idx]: flg = False break if flg: tmp = S[:x] + T + S[x + len_T:] tmp = tmp.replace('?', 'a') lst.append(tmp) if len(lst) == 0: print('UNRESTORABLE') else: print(min(lst))

s177662924

p03380

u437638594

Wrong Answer

Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.

import bisect n = int(input()) a_list = sorted(list(map(int, input().split()))) a_i = a_list.pop() half = a_i // 2 if len(a_list) == 1: print(*[a_i, a_list[0]]) exit() left_index = bisect.bisect_left(a_list, half) if left_index >= 1: if a_list[left_index] - half < half - a_list[left_index-1]: print(*[a_i, a_list[left_index]]) else: print(*[a_i, a_list[left_index-1]]) if left_index == 0: print(*[a_i, a_list[left_index]])

s364459942

Accepted

import bisect n = int(input()) a_list = sorted(list(map(int, input().split()))) a_i = a_list.pop() half = a_i / 2 if len(a_list) == 1: print(*[a_i, a_list[0]]) exit() left_index = bisect.bisect_left(a_list, half) if left_index >= 1: if a_list[left_index] - half < half - a_list[left_index-1]: print(*[a_i, a_list[left_index]]) else: print(*[a_i, a_list[left_index-1]]) if left_index == 0: print(*[a_i, a_list[left_index]])

s930470171

p03474

u469063372

Wrong Answer

The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.

A, B = map(int, input().split()) S = input() if A + B + 1 != len(S) or S[A] != '-': print('No') else: if '-' in S[:A] or '-' in S[1+1:]: print('No') else: print('Yes')

s262013930

Accepted

A, B = map(int, input().split()) S = input() if S[A] == '-': if S[:A].isdecimal() and S[A+1:].isdecimal(): print('Yes') exit() print('No')

s879639799

p03861

u353855427

Wrong Answer

You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?

A,B,C = list(map(int,input().split())) print((B-A+1)//C)

s924463771

Accepted

A,B,C = list(map(int,input().split())) print((B//C-(A-1)//C))

s613560609

p02844

u315485238

Wrong Answer

AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code. The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code. How many different PIN codes can he set this way? Both the lucky number and the PIN code may begin with a 0.

N = int(input()) S = input() from itertools import product p = product([str(i) for i in range(10)], repeat=3) ans = 0 for i,j,k in p: x = S.find(i) if x<0: continue y = x + S[x+1:].find(j) if y<0: continue z = x + y + S[y+1:].find(k) if z<0: continue ans += 1 print(ans)

s905132987

Accepted

N = int(input()) S = input() from itertools import product p = product([str(i) for i in range(10)], repeat=3) ans = 0 for i,j,k in p: x = S.find(i) if x<0: continue y = S[x+1:].find(j) if y<0: continue z = S[x+y+2:].find(k) if z<0: continue #print(i,j,k, S, S[x+1:],S[x+y+2:]) ans += 1 print(ans)