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TnT

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CodeNet4Repair

like 0

s851650423

p03160

u479638406

Wrong Answer

There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.

n = int(input()) h = list(map(int, input().split())) dp = [0]*(n+1) for i in range(n): dp[i+1] = min(dp[i] + abs(h[i]-h[i-1]), dp[i-1] + abs(h[i]-h[i-2])) print(dp[-1])

s278483940

Accepted

n = int(input()) h = list(map(int, input().split())) dp = [0]*n dp[1] = abs(h[1]-h[0]) for i in range(2, n): dp[i] = min(dp[i-1] + abs(h[i]-h[i-1]), dp[i-2] + abs(h[i]-h[i-2])) print(dp[-1])

s911236031

p03997

u993435350

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

s276407021

Accepted

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

s400816705

p03457

u076543276

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.

t = [] x = [] y = [] N = int(input()) s = [] for i in range(N): t1, x1, y1 = [int(i) for i in input().split()] t.append(t1) x.append(x1) y.append(y1) for i in range(N): if t[i] < x[i] + y[i]: s.append(0) else: if (t[i] % 2 == 1 and x[i] % 2 == 0 and y[i] % 2 == 0) or (t[i] % 2 == 1 and x[i] % 2 == 1 and y[i] % 2 == 1): s.append(0) elif (t[i] % 2 == 0 and x[i] % 2 == 1 and y[i] % 2 == 0) or (t[i] % 2 == 1 and x[i] % 2 == 0 and y[i] % 2 == 1): s.append(0) else: s.append(1) for i in s: if i == 0: print("NO") break else: print("YES")

s902745472

Accepted

t = [] x = [] y = [] N = int(input()) s = [] for i in range(N): t1, x1, y1 = [int(i) for i in input().split()] t.append(t1) x.append(x1) y.append(y1) for i in range(N): if i == 0 and t[i] < abs(x[i]) + abs(y[i]): s.append(0) elif i > 0 and t[i] - t[i - 1] < abs(x[i] - x[i - 1]) + abs(y[i] - y[i - 1]): s.append(0) else: if (t[i] % 2 == 1 and x[i] % 2 == 0 and y[i] % 2 == 0) or (t[i] % 2 == 1 and x[i] % 2 == 1 and y[i] % 2 == 1): s.append(0) elif (t[i] % 2 == 0 and x[i] % 2 == 1 and y[i] % 2 == 0) or (t[i] % 2 == 0 and x[i] % 2 == 0 and y[i] % 2 == 1): s.append(0) else: s.append(1) for i in s: if i == 0: print("No") break else: print("Yes")

s702226369

p02694

u406767170

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()) M = 100 for i in range(10): M = int(M*1.01) print(str(i) + ":" + str(M))

s060496403

Accepted

X = int(input()) M = 100 year = 0 while 1==1: year += 1 M = int(M*1.01) if M >= X: break print(year)

s738220740

p03478

u574189773

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()) some_sum = 0 for num in range(n+1): tmp_num = num sum_num = 0 while tmp_num > 0: sum_num += tmp_num % 10 tmp_num //= 10 if a <= sum_num and sum_num <= b: some_sum += num print(num) print(some_sum)

s515150276

Accepted

n, a, b = map(int, input().split()) some_sum = 0 for num in range(n+1): tmp_num = num sum_num = 0 while tmp_num > 0: sum_num += tmp_num % 10 tmp_num //= 10 if a <= sum_num and sum_num <= b: some_sum += num #print(num) print(some_sum)

s420878533

p03228

u010437136

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.

import sys def solve(A,B,K): ab = [A,B] for i in range(K): if ab[0]%2!=0: ab[0]-=1 ab[1]+=ab[0]//2 ab[0]-=ab[0]//2 ab.reverse() if K%2!=0: return ab[1],ab[0] return ab[0],ab[1] def readQuestion(): ws = sys.stdin.readline().strip().split() A = int(ws[0]) B = int(ws[1]) C = int(ws[2]) return (A, B, C) def main(): print(solve(*readQuestion())) # Uncomment before submission # main()

s063397723

Accepted

import sys def solve(A,B,K): ab = [A,B] for i in range(K): if ab[0]%2!=0: ab[0]-=1 ab[1]+=ab[0]//2 ab[0]-=ab[0]//2 ab.reverse() if K%2!=0: print(ab[1],ab[0]) return print(ab[0],ab[1]) def readQuestion(): ws = sys.stdin.readline().strip().split() A = int(ws[0]) B = int(ws[1]) C = int(ws[2]) return (A, B, C) def main(): solve(*readQuestion()) # Uncomment before submission main()

s318552676

p03737

u779728630

Wrong Answer

You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.

a, b, c = input().split() print( a[0] + b[0] + c[0] )

s543797883

Accepted

a, b, c = input().split() print( (a[0] + b[0] + c[0]).upper() )

s968976056

p03860

u636775911

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.

#coding:utf-8 s=input() s="A"+s[0]+"C" print(s)

s569921749

Accepted

#coding:utf-8 s=input().split() s="A"+s[1][0]+"C" print(s)

s298183163

p03162

u680851063

Wrong Answer

Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains.

n = int(input()) l = [list(map(int, input().split())) for _ in range(n)] dp = [[0,0,0] for _ in range(n)] dp[0][0], dp[0][1], dp[0][2] = l[0][0], l[0][1], l[0][2] for i in range(1, n): for j in range(3): for k in range(3): if j != k: dp[i][k] = max(dp[i][k], dp[i-1][j] + l[i][k]) print(dp)

s965042533

Accepted

n = int(input()) l = [list(map(int, input().split())) for _ in range(n)] dp = [[0,0,0] for _ in range(n)] dp[0] = l[0] for i in range(1, n): for j in range(3): for k in range(3): if j != k: dp[i][j] = max(dp[i][j], dp[i-1][k] + l[i][j]) print(max(dp[-1]))

s890409728

p02748

u821265215

Wrong Answer

You are visiting a large electronics store to buy a refrigerator and a microwave. The store sells A kinds of refrigerators and B kinds of microwaves. The i-th refrigerator ( 1 \le i \le A ) is sold at a_i yen (the currency of Japan), and the j-th microwave ( 1 \le j \le B ) is sold at b_j yen. You have M discount tickets. With the i-th ticket ( 1 \le i \le M ), you can get a discount of c_i yen from the total price when buying the x_i-th refrigerator and the y_i-th microwave together. Only one ticket can be used at a time. You are planning to buy one refrigerator and one microwave. Find the minimum amount of money required.

s = input() s_len = len(s) if s_len % 2: print("No") else: prev = 0 count = 0 for i in range(1, (s_len//2)+1): if s[prev:i*2] != "hi": print("No") break else: count += 1 prev = i * 2 if count == s_len // 2: print("Yes")

s709139465

Accepted

a, b, m = input().split() a, b, m = int(a), int(b), int(m) a_values = input().split() b_values = input().split() m_values = {} for i in range(m): m_values[i] = input().split() min_num = int(a_values[int(m_values[0][0]) - 1]) + int(b_values[int(m_values[0][1]) - 1]) - int(m_values[0][2]) for i in range(1, m): now = int(a_values[int(m_values[i][0]) - 1]) + int(b_values[int(m_values[i][1]) - 1]) - int(m_values[i][2]) if min_num > now: min_num = now now = int(min(a_values)) + int(min(b_values)) if min_num > now: print(now) else: print(min_num)

s606113618

p02578

u440608859

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.

for _ in range(1): n=int(input()) a=list(map(int,input().split())) k=a[0] ans=0 for i in range(n): if i==0: m=a[i] a[i]=0 else: m=max(m,a[i]) a[i]=m-a[i] u=0 for i in range(n): ans+=max(0,a[i]-u) u=a[i] print(ans)

s370094785

Accepted

n=int(input()) a=list(map(int,input().split())) m=0 ans=0 for i in range(n): if i==0: m=max(a[i],m) else: if a[i]<m: ans+=m-a[i] else: m=a[i] print(ans)

s952976249

p03478

u168660441

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).

a = input().split() ans = 0 for i in range(int(a[0])+1): i=str(i) i.split() sum_num=0 for j in i: sum_num=int(j) if j>=a[1] and j<=a[2]: ans += 1 print(ans)

s723279800

Accepted

a = input().split() ans = 0 for i in range(int(a[0])+1): i=str(i) i.split() sum_num=0 for j in i: sum_num+=int(j) if sum_num>=int(a[1]) and sum_num<=int(a[2]): ans += int(i) print(ans)

s657346479

p02407

u264632995

Wrong Answer

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

input() a = input().split() print(a) a.reverse() print(" ".join(a))

s454484594

Accepted

input() a = input().split() a.reverse() print(" ".join(a))

s398158866

p02927

u115110170

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(2,m+1): for i in range(22,d+1): if m == ((d//10) * (d%10)): ans += 1 print(ans)

s403740043

Accepted

m,d = map(int,input().split()) ans = 0 for i in range(2,m+1): for j in range(22,d+1): if j%10 <= 1: continue if i == ((j//10) * (j%10)): ans += 1 print(ans)

s018330502

p03494

u494811189

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.

n=int(input()) a = list(map(int, input().split())) count =0 for k in range(n): if a[k]%2==0: count += 1 else: print(count)

s241158217

Accepted

n=int(input()) a = list(map(int, input().split())) count =0 flag = "true" while flag=="true": for k in range(n): if a[k]%2==0: a[k] //= 2 else: flag = "false" break count += 1 print(count-1)

s596875020

p00019

u567380442

Wrong Answer

Write a program which reads an integer n and prints the factorial of n. You can assume that n ≤ 20\.

ret = 1 for i in range(2, int(input())): ret *= i print(ret)

s287758963

Accepted

ret = 1 for i in range(2, int(input()) + 1): ret *= i print(ret)

s283397806

p03380

u480138356

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 sys input = sys.stdin.readline def main(): N = int(input()) a = list(map(int, input().split())) ans1 = max(a) half = ans1 // 2 min_ = abs(a[0] - half) ans2 = a[0] for i in range(1, N): if a[i] != ans1: tmp = abs(a[i] - half) if tmp <= min_: ans2 = a[i] min_ = tmp print(ans1, ans2) if __name__ == "__main__": main()

s373210376

Accepted

import sys input = sys.stdin.readline def main(): N = int(input()) a = list(map(int, input().split())) ans1 = max(a) half = ans1 / 2 min_ = abs(a[0] - half) ans2 = a[0] for i in range(1, N): if a[i] != ans1: tmp = abs(a[i] - half) if tmp <= min_: ans2 = a[i] min_ = tmp print(ans1, ans2) if __name__ == "__main__": main()

s202930881

p02390

u175111751

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()) print(s // 3600, s // 60 % 60, s % 60)

s568552941

Accepted

s = int(input()) print('{0}:{1}:{2}'.format(s // 3600, s // 60 % 60, s % 60))

s991152455

p03370

u288786530

Wrong Answer

Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.

n, x = map(int, input().split()) m = [int(input()) for _ in range(n)] for i in m: x -= i s = 0 while x >= 0: x -= min(m) s += 1 ans = len(m) + s print(ans)

s776471397

Accepted

n, x = map(int, input().split()) m = [int(input()) for _ in range(n)] for i in m: x -= i s = 0 while x >= 0: x -= min(m) s += 1 ans = len(m) + s - 1 print(ans)

s012134583

p03729

u494748969

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`.

a, b, c = list(map(str, input().split())) if a[-1] == b[0] and b[-1] == c[0]: print("Yes") else: print("No")

s624451773

Accepted

a, b, c = list(map(str, input().split())) if a[-1] == b[0] and b[-1] == c[0]: print("YES") else: print("NO")

s005153832

p03524

u252828980

Wrong Answer

Snuke has a string S consisting of three kinds of letters: `a`, `b` and `c`. He has a phobia for palindromes, and wants to permute the characters in S so that S will not contain a palindrome of length 2 or more as a substring. Determine whether this is possible.

s = input() A = s.count("a") B = s.count("b") C = s.count("c") if max(A,B,C) - min(A,B,C) <= 1: print("Yes") else: print("No")

s966915302

Accepted

s = input() A = s.count("a") B = s.count("b") C = s.count("c") if max(A,B,C) - min(A,B,C) <= 1: print("YES") else: print("NO")

s225425127

p03485

u127856129

Wrong Answer

You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.

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

s868179061

Accepted

a,b=map(int,input().split()) print((a+b+1)//2)

s721220324

p03836

u253011685

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()) x=tx-sx y=ty-sy path="R"*x+"U"*(y+1)+"L"*x+"D"*(y+2)+"R"*(x+1)+"U"*(y+1)+"L"*(x+2)+"D"*y+"R"; print(path)

s053112693

Accepted

sx,sy,tx,ty=map(int,input().split()) x=tx-sx y=ty-sy path="U"*y+"R"*x+"D"*y+"L"*(x+1)+"U"*(y+1)+"R"*(x+1)+"D"+"R"+"D"*(y+1)+"L"*(x+1)+"U" print(path)

s186530793

p03476

u107639613

Time Limit Exceeded

We say that a odd number N is _similar to 2017_ when both N and (N+1)/2 are prime. You are given Q queries. In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i.

import sys from itertools import accumulate def input(): return sys.stdin.readline().strip() def get_sieve_of_eratosthenes(n): if not isinstance(n, int): raise TypeError('n is int type.') if n < 2: return [] prime = [2] limit = int(n**0.5) data = [i + 1 for i in range(2, n, 2)] while True: p = data[0] if limit < p: return prime + data prime.append(p) data = [e for e in data if e % p != 0] def main(): primes = get_sieve_of_eratosthenes(100000) like = [0] * 10**5 for i in range(3, 10**5): if i in primes and (i + 1) // 2 in primes: like[i] = 1 like = list(accumulate(like)) q = int(input()) for _ in range(q): l, r = map(int, input().split()) if l < 3: l = 3 if r < 3: print(0) continue print(like[r] - like[l - 1]) if __name__ == "__main__": main()

s500913963

Accepted

import sys from itertools import accumulate def input(): return sys.stdin.readline().strip() def get_sieve_of_eratosthenes(n): if not isinstance(n, int): raise TypeError('n is int type.') if n < 2: return [] prime = [1] * (n + 1) prime[0] = prime[1] = 0 for i in range(2, int(n**0.5) + 1): if not prime[i]: continue for j in range(i * 2, n + 1, i): prime[j] = 0 return prime def main(): primes = get_sieve_of_eratosthenes(100000) like = [0] * 10**5 for i in range(3, 10**5): if primes[i] and primes[(i + 1) // 2]: like[i] = 1 like = list(accumulate(like)) q = int(input()) for _ in range(q): l, r = map(int, input().split()) if l < 3: l = 3 if r < 3: print(0) continue print(like[r] - like[l - 1]) if __name__ == "__main__": main()

s481977084

p03433

u979078704

Wrong Answer

E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.

N = int(input()) A = int(input()) if N % 500 <= A: print("YES") else: print("NO")

s384446163

Accepted

N = int(input()) A = int(input()) if N % 500 <= A: print("Yes") else: print("No")

s007745054

p03545

u733608212

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.

s = input() li = [i for i in s] global ans def cal(left, count, remaining): global ans if len(remaining) > 0: cal(left + "+" + remaining[0], count + int(remaining[0]), remaining[1:]) cal(left + "-" + remaining[0], count - int(remaining[0]), remaining[1:]) elif count == 7: ans = left cal(li[0], int(li[0]), li[1:]) print(ans)

s905581509

Accepted

s = input() li = [i for i in s] global ans def cal(left, count, remaining): global ans if len(remaining) > 0: cal(left + "+" + remaining[0], count + int(remaining[0]), remaining[1:]) cal(left + "-" + remaining[0], count - int(remaining[0]), remaining[1:]) elif count == 7: ans = left cal(li[0], int(li[0]), li[1:]) print(ans+"=7")

s572018318

p03557

u557494880

Wrong Answer

The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower. He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i. To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar. How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.

N = int(input()) A = list(map(int,input().split())) B = list(map(int,input().split())) C = list(map(int,input().split())) A.sort() B.sort() C.sort() import bisect ans = 0 for i in range(N): b = B[i] x = bisect.bisect_left(A,b) y = bisect.bisect_right(C,b) if y == 0: break ans += x*y print(ans)

s984388278

Accepted

N = int(input()) A = list(map(int,input().split())) B = list(map(int,input().split())) C = list(map(int,input().split())) A.sort() B.sort() C.sort() import bisect ans = 0 for i in range(N): b = B[i] x = bisect.bisect_left(A,b) y = N - bisect.bisect_right(C,b) ans += x*y print(ans)

s000745232

p03351

u668503853

Wrong Answer

Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.

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

s849942812

Accepted

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

s314191165

p02659

u824237520

Wrong Answer

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

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

s526938878

Accepted

a, b = map(int, input().replace('.', '').split()) print(a * b // 100)

s165354447

p02690

u713165870

Wrong Answer

Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.

if __name__ == "__main__": x = int(input()) is_get = False for a in range(1000000000): if is_get: break for b in range(-1*a, a, 1): if x == a**5 - b**5: is_get = True break print(f"{a} {b}")

s721543723

Accepted

def get_answer(): x = int(input()) is_get = False for a in range(1000000000): if is_get: break for b in range(-1*a, a, 1): if x == a**5 - b**5: is_get = True break print(f"{a-1} {b}") if __name__ == "__main__": get_answer()

s441131577

p03626

u846372029

Wrong Answer

We have a board with a 2 \times N grid. Snuke covered the board with N dominoes without overlaps. Here, a domino can cover a 1 \times 2 or 2 \times 1 square. Then, Snuke decided to paint these dominoes using three colors: red, cyan and green. Two dominoes that are adjacent by side should be painted by different colors. Here, it is not always necessary to use all three colors. Find the number of such ways to paint the dominoes, modulo 1000000007. The arrangement of the dominoes is given to you as two strings S_1 and S_2 in the following manner: * Each domino is represented by a different English letter (lowercase or uppercase). * The j-th character in S_i represents the domino that occupies the square at the i-th row from the top and j-th column from the left.

# Coloring Dominoes N = int(input()) s1 = input() s2 = input() c = 3 for i in range(N-1): if (s1[i]==s2[i]): if (s1[i+1]==s2[i+1]): c*=2 else: if (s1[i]!=s1[i+1]): if (s1[i+1]!=s2[i+1]): c*=3 print(c%1000000007)

s645327863

Accepted

# Coloring Dominoes N = int(input()) s1 = input() s2 = input() if (s1[0]==s2[0]): c = 3 else: c = 6 for i in range(N-1): if (s1[i]==s2[i]): c*=2 else: if (s1[i]!=s1[i+1]): if (s1[i+1]!=s2[i+1]): c*=3 print(c%1000000007)

s482659321

p03351

u802977614

Wrong Answer

Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.

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

s017031666

Accepted

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

s588811097

p03623

u600261652

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|.

x, a, b = map(int, input().split()) print(min(abs(x-a), abs(x-b)))

s530412897

Accepted

x, a, b = map(int, input().split()) print("A" if abs(x-a) < abs(x-b) else "B")

s336621887

p02612

u517630860

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.

print(int(input())%1000)

s658301205

Accepted

N = int(input()) if N == 0 or N%1000 == 0: print('0') else: print(1000 - N%1000)

s508782745

p03448

u505420467

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

s682455555

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 i*500+j*100+k*50==x: ans+=1 print(ans)

s494034862

p04029

u792512290

Wrong Answer

There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?

n = int(input()) print((1 + n) * n / 2)

s091605875

Accepted

n = int(input()) print((1 + n) * n // 2)

s154412315

p03351

u485305016

Wrong Answer

Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.

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

s710919126

Accepted

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

s841879790

p03048

u075628913

Wrong Answer

Snuke has come to a store that sells boxes containing balls. The store sells the following three kinds of boxes: * Red boxes, each containing R red balls * Green boxes, each containing G green balls * Blue boxes, each containing B blue balls Snuke wants to get a total of exactly N balls by buying r red boxes, g green boxes and b blue boxes. How many triples of non-negative integers (r,g,b) achieve this?

r, g, b, n = map(int, input().split()) count = 0 for ri in range(0, n+1, r): for gi in range(0, n+1, g): for bi in range(0, n+1, b): if ri + gi + bi == n: count += 1 elif ri + gi + bi < n: break print(count)

s085946437

Accepted

R, G, B, N = map(int, input().split()) count = 0 for r in range(0, N+1, R): for g in range(0, N+1-r, G): val = (N - r - g) if val % B == 0 and val >= 0: count += 1 print(count)

s989086758

p03129

u858428199

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()) a = n // 2 a += n % 2 print(a) if a >= k: print("YES") else: print("NO")

s145604838

Accepted

n,k=map(int, input().split()) a = n // 2 a += n % 2 if a >= k: print("YES") else: print("NO")

s987429303

p03387

u757117214

Wrong Answer

You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order: * Choose two among A, B and C, then increase both by 1. * Choose one among A, B and C, then increase it by 2. It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.

lst = list(map(int,input().split())) lst2 = list(map(lambda x: max(lst) - x, lst)) if lst2[0] % 2 == 0 and lst2[1] % 2 == 0: print(sum(lst2) // 2) elif lst2[0] % 2 != 0 and lst2[1] % 2 != 0: _sum = (lst2[0] - 1) + (lst2[1] - 1) print(_sum // 2 + 1) else: print((sum(lst2) + 1) // 2 + 1)

s083853861

Accepted

lst = list(map(int,input().split())) lst2 = list(map(lambda x: max(lst) - x, lst)) lst2.remove(0) if lst2[0] % 2 == lst2[1] % 2: print(sum(lst2) // 2) else: print((sum(lst2) + 3) // 2)

s850682492

p03547

u368796742

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("<") elif x < y: print(">") else: print("=")

s740386331

Accepted

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

s913620903

p03971

u068142202

Wrong Answer

There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.

n, a, b = map(int, input().split()) s = list(input()) for i in s: if i == "a" and a != 0: a -= 1 print("Yes") elif i == b and b != 0: b -=1 print("Yes") else: print("No")

s305757329

Accepted

n, a, b = map(int, input().split()) s = list(input()) for i in s: if i == "a" and (a + b) != 0: a -= 1 print("Yes") elif i == "b" and (a + b) != 0 and b != 0: b -=1 print("Yes") else: print("No")

s320778404

p03543

u882831132

Wrong Answer

We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?

N=input() ans=1 for i in range(len(N)-1): if N[i] == N[i+1]: ans+=1 else: ans=1 if ans>=3: print("Yes") else: print("No")

s552291835

Accepted

N=input() ans=1 for i in range(len(N)-1): if N[i] == N[i+1]: ans+=1 else: ans=1 if ans>=3: print("Yes") exit() print("No")

s995318357

p04044

u379424722

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()) L = sorted(input().split()) print(''.join(L))

s502966470

Accepted

N,L = map(int,input().split()) M = [] for i in range(N): M.append(input()) M = sorted(M) print(''.join(M))

s502033077

p03485

u992910889

Wrong Answer

You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.

A,B=map(int,input().split()) print((A+B)//2+1)

s558556101

Accepted

import math A,B=map(int,input().split()) print(math.ceil((A+B)/2))

s750072954

p03486

u138486156

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 = list(input()) t = list(input()) a = b = [] for i in range(len(s)): a.append(ord(s[i])) for i in range(len(t)): b.append(ord(t[i])) if min(a) < max(b): print("Yes") elif len(s) < len(t): for i in range(len(s)): if s[i] != t[i]: print("No") exit() print("Yes") exit() print("No")

s079132385

Accepted

s = list(input()) t = list(input()) n = len(s) m = len(t) s.sort() t.sort() t = t[-1::-1] a = b = "" for i in range(n): a += s[i] for i in range(m): b += t[i] if a < b: print("Yes") else: print("No")

s039843900

p03610

u013756322

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() ans = '' for c in s: if (len(s) % 2 == 1): ans += c print(ans)

s270364120

Accepted

print(input()[0::2])

s672966712

p03997

u638648846

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

s898155305

Accepted

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

s642085841

p03854

u225388820

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() i=0 ans=True while i<=len(s)-1: if s[i:i+7]=='dreamer': i+=7 elif s[i:i+5]=='dream': i+=5 elif s[i:i+6]=='eraser': i+=6 elif s[i:i+5]=='erase': i+=5 else: ans=False break if ans: print('Yes') else: print('No')

s992149759

Accepted

s=input() i=len(s) ans=True while 3<=i: if s[i-7:i]=='dreamer': i-=7 elif s[i-5:i]=='dream': i-=5 elif s[i-6:i]=='eraser': i-=6 elif s[i-5:i]=='erase': i-=5 else: ans=False break if ans: print('YES') else: print('NO')

s348398519

p03636

u889989259

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 = list(input()) print(s[0], len(s)-2, s[len(s)-1])

s876444100

Accepted

s = list(input()) print(s[0], len(s)-2, s[len(s)-1], sep='')

s594267326

p02742

u455533363

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:

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

s748539060

Accepted

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

s134600888

p03447

u475675023

Wrong Answer

You went shopping to buy cakes and donuts with X yen (the currency of Japan). First, you bought one cake for A yen at a cake shop. Then, you bought as many donuts as possible for B yen each, at a donut shop. How much do you have left after shopping?

print((int(input())-int(input()))//int(input()))

s695858960

Accepted

x=int(input()) a=int(input()) b=int(input()) print(x-a-((x-a)//b)*b)

s861308338

p03997

u441320782

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

s178306412

Accepted

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

s669135897

p02612

u849433300

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.

N = int(input()) print(N//1000)

s598948518

Accepted

N = int(input()) if N%1000 == 0: print(0) else: print(1000- N%1000)

s545579834

p02606

u730807152

Wrong Answer

How many multiples of d are there among the integers between L and R (inclusive)?

l,r,d=map(int,input().split()) ans=0 for i in range(l,r): if i%d == 0: ans += 1 print(ans)

s596710209

Accepted

l,r,d=map(int,input().split()) ans=0 for i in range(l,r+1): if i%d == 0: ans += 1 print(ans)

s413848362

p03860

u581187895

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 = list(input().split()) print(s[0][0],s[1][0],s[2][0], end="")

s810669787

Accepted

a, b, c = input().split() print(a[0]+b[0]+c[0])

s331802121

p03501

u477343425

Wrong Answer

You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours.

n,a,b = map(int, input().split()) ans = min(n*a, b)

s678116663

Accepted

n,a,b = map(int,input().split()) ans = min(n*a, b) print(ans)

s327981358

p03456

u110580875

Wrong Answer

AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.

import math a,b=(map(int,input().split())) num=str(a)+str(b) ans=int(math.sqrt(int(num))) print(ans if ans*ans==int(num) else"No")

s662540943

Accepted

import math a,b=(map(int,input().split())) num=str(a)+str(b) ans=int(math.sqrt(int(num))) print("Yes" if ans*ans==int(num) else"No")

s610318565

p03543

u626228246

Wrong Answer

We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?

N = list(input()) print(N[0] == N[1] == N[2] or N[1] == N[2] == N[3])

s611401222

Accepted

N = list(input()) print("Yes" if N[0] == N[1] == N[2] or N[1] == N[2] == N[3] else "No")

s579839866

p03433

u574464625

Wrong Answer

E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.

N=int(input()) A=int(input()) if N%500 < A : print("YES") else : print("NO")

s900645100

Accepted

#infinity Coins n=int(input()) a=int(input()) if n%500 <= a : print("Yes") else : print("No")

s899887242

p03448

u806257533

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.

big = int(input()) mid = int(input()) small = int(input()) input_money = int(input()) cnt = 0 for i in range(big+1): for j in range(mid+1): for k in range(small+1): if (input_money-500*i)>=0: money = (input_money-500*i) if money==0: cnt += 1 break if (money-100*j)>=0: money = (money-100*i) if money==0: cnt += 1 break if (money-50*j)>=0: money = (money-50*i) if money==0: cnt += 1 break else: continue break print(cnt)

s826928083

Accepted

big = int(input()) mid = int(input()) small = int(input()) input_money = int(input()) cnt = 0 for i in range(big+1): for j in range(mid+1): for k in range(small+1): if input_money==(500*i + 100*j + 50*k): cnt += 1 print(cnt)

s425500025

p03997

u708211626

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,b,c=[int(input()) for i in range(3)] print((a+b)*c/2)

s921304190

Accepted

a,b,c=[int(input()) for i in range(3)] print(int(a+b)*c//2)

s786201249

p03549

u329407311

Wrong Answer

Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`. When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases. Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds. Now, he goes through the following process: * Submit the code. * Wait until the code finishes execution on all the cases. * If the code fails to correctly solve some of the M cases, submit it again. * Repeat until the code correctly solve all the cases in one submission. Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).

N,M = map(int,input().split()) T = M * 1900 + (N-M) * 100 print( int(( (1/2)**M / (1-(1-(1/2)**M)) ) *T))

s933872673

Accepted

import math N,M = map(int,input().split()) T = M * 1900 + (N-M) * 100 p = (1/2)**M p2 = (1/2)**M ans = 0 for i in range(1,500000): ans += T * p2 * i p2 = p2 * (1-p) print(round(ans))

s541812719

p03854

u044442911

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() while True: if S[:5] == "dream": S = S[5:] elif S[:6] == "eraser": S = S[6:] elif S[:5] == "erase": S = S[5:] elif S[:6] == "dreamer": S = S[7:] else: print("No") break if len(S) == 0: print("Yes") break if __name__ == "__main__": main()

s782885634

Accepted

from re import match # f = open("./16/input_c2_2.txt") # return f.readline().rstrip() def main(): S = input() if match('(dreamer|eraser|erase|dream)+$', S): print("YES") else: print("NO") # while True: # result_1 = search('dreamer$', S) # result_4 = search('dream$', S) # if result_1: # S = S[:result_1.start()] # elif result_2: # S = S[:result_2.start()] # elif result_3: # S = S[:result_3.start()] # elif result_4: # S = S[:result_4.start()] # else: # print("NO") # break # if len(S) == 0: # print("YES") # break if __name__ == "__main__": main()

s879254667

p03369

u623814058

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.

print(input().count('o') * 200 + 700)

s564627426

Accepted

print(input().count('o') * 100 + 700)

s833276146

p03095

u143492911

Wrong Answer

You are given a string S of length N. Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings. Here, a subsequence of a string is a concatenation of **one or more** characters from the string without changing the order.

n=int(input()) s=input() cnt=[0]*26 for i in s: cnt[ord(i)-97]+=1 print(cnt) ans=1 for i in range(26): ans*=cnt[i]+1 print(ans-1)

s072610466

Accepted

n=int(input()) s=input() m=10**9+7 cnt=[0]*26 for i in s: cnt[ord(i)-97]+=1 ans=1 for i in range(26): ans*=cnt[i]+1 ans%=m print(ans-1)

s891891922

p03919

u619819312

Wrong Answer

There is a grid with H rows and W columns. The square at the i-th row and j-th column contains a string S_{i,j} of length 5. The rows are labeled with the numbers from 1 through H, and the columns are labeled with the uppercase English letters from `A` through the W-th letter of the alphabet. Exactly one of the squares in the grid contains the string `snuke`. Find this square and report its location. For example, the square at the 6-th row and 8-th column should be reported as `H6`.

h,w=map(int,input().split()) q=[list(map(str,input().split()))for i in range(h)] p=["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"] for i in range(h): for j in range(w): if q[i][j]=="snuke": print(p[i]+str(j+1))

s435947784

Accepted

h,w=map(int,input().split()) for i in range(h): q=list(input().split()) for j in range(w): if q[j]=="snuke": print(chr(ord("A")+j)+str(i+1))

s299581005

p03693

u039360403

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?

a=input().split() print('Yes' if int(a[0]+a[1]+a[2])%4==0 else 'No')

s716335034

Accepted

a=input().split() print('YES' if int(a[0]+a[1]+a[2])%4==0 else 'NO')

s583568162

p02401

u655518263

Wrong Answer

Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.

l = [str(i) for i in input().split()] a = int(l[0]) b = int(l[2]) op = l[1] if op == "+": print(a+b) elif op == "-": print(a-b) elif op == "*": print(a*b) elif op == "/": print(int(a/b)) elif l == ["0","?","0"]: pass

s815612538

Accepted

while True: l = [str(i) for i in input().split()] a = int(l[0]) b = int(l[2]) op = l[1] if op == "+": print(a+b) elif op == "-": print(a-b) elif op == "*": print(a*b) elif op == "/": print(a//b) elif op == "?": break

s684258947

p03473

u776871252

Wrong Answer

How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?

h = int(input()) print(24 - h)

s072912142

Accepted

h = int(input()) print(24 + (24 - h))

s351722733

p03545

u528482684

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.

S = input() op = ["+","-"] for a in op: for b in op: for c in op: shiki = S[0]+a+S[1]+b+S[2]+c+S[3]+"==7" if eval(shiki): print(shiki) exit()

s953984548

Accepted

S = input() op = ["+","-"] for a in op: for b in op: for c in op: shiki = S[0]+a+S[1]+b+S[2]+c+S[3] if eval(shiki)==7: print(shiki+"=7") exit()

s540396811

p02261

u390995924

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).

import copy N = int(input()) bscs = [(int(c[-1]), c) for c in input().split(" ")] sscs = copy.copy(bscs) for i in range(N): j = N - 1 while j > i: if bscs[j][0] < bscs[j - 1][0]: tmp = bscs[j] bscs[j] = bscs[j - 1] bscs[j - 1] = tmp j -= 1 print(bscs) print("Stable") for i in range(N): minj = i for j in range(i, N): if sscs[j][0] < sscs[minj][0]: minj = j tmp = sscs[i] sscs[i] = sscs[minj] sscs[minj] = tmp print(sscs) print("Stable" if bscs == sscs else "Not Stable")

s411859932

Accepted

import copy N = int(input()) bscs = [(int(c[-1]), c) for c in input().split(" ")] sscs = copy.copy(bscs) for i in range(N): j = N - 1 while j > i: if bscs[j][0] < bscs[j - 1][0]: tmp = bscs[j] bscs[j] = bscs[j - 1] bscs[j - 1] = tmp j -= 1 print(" ".join([c[1] for c in bscs])) print("Stable") for i in range(N): minj = i for j in range(i, N): if sscs[j][0] < sscs[minj][0]: minj = j tmp = sscs[i] sscs[i] = sscs[minj] sscs[minj] = tmp print(" ".join([c[1] for c in sscs])) print("Stable" if bscs == sscs else "Not stable")

s904632355

p03556

u363836311

Wrong Answer

Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.

n=int(input()) t=1 while t*t<n: t+=1 print(t)

s321796506

Accepted

n=int(input()) t=1 while t*t<=n: t+=1 print((t-1)**2)

s873211581

p03836

u591295155

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()) keiro = "" for i in range(ty-sy): keiro += "U" for i in range(tx-sx): keiro += "R" keiro0 = keiro keiro = "" for i in range(ty-sy): keiro += "D" for i in range(tx-sx): keiro += "L" keiro1 = keiro ans = keiro0+keiro1+"LU"+keiro0+"RDDR"+keiro1+"LU" print(ans)

s789213878

Accepted

sx, sy, tx, ty = map(int,input().split()) keiro = "" for i in range(ty-sy): keiro += "U" for i in range(tx-sx): keiro += "R" keiro0 = keiro keiro = "" for i in range(ty-sy): keiro += "D" for i in range(tx-sx): keiro += "L" keiro1 = keiro ans = keiro0+keiro1+"LU"+keiro0+"RDRD"+keiro1+"LU" print(ans)

s002850682

p03477

u670180528

Wrong Answer

A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.

a,b,c,d=map(int,input().split()) if a+b<c+d: print("Left") elif a+b>c+d: print("Right") else: print("Balanced")

s536660907

Accepted

a,b,c,d=map(int,input().split()) if a+b>c+d: print("Left") elif a+b<c+d: print("Right") else: print("Balanced")

s792577690

p02861

u375681664

Wrong Answer

There are N towns in a coordinate plane. Town i is located at coordinates (x_i, y_i). The distance between Town i and Town j is \sqrt{\left(x_i- x_j\right)^2+\left(y_i-y_j\right)^2}. There are N! possible paths to visit all of these towns once. Let the length of a path be the distance covered when we start at the first town in the path, visit the second, third, \dots, towns, and arrive at the last town (assume that we travel in a straight line from a town to another). Compute the average length of these N! paths.

import math N=int(input()) X=[0]*N Y=[0]*N S=0 T=math.factorial(N) for i in range(N): X[i],Y[i]=map(int,input().split()) for j in range(N-1): for k in range(j+1,N): print(math.sqrt((X[j]-X[k])**2+(Y[j]-Y[k])**2)) S+=math.sqrt((X[j]-X[k])**2+(Y[j]-Y[k])**2) print(S*2/N)

s139528578

Accepted

import math N=int(input()) X=[0.0]*N Y=[0.0]*N S=0.0 for i in range(N): X[i],Y[i]=map(int,input().split()) for j in range(N-1): for k in range(j+1,N): S+=math.sqrt((X[j]-X[k])**2+(Y[j]-Y[k])**2) print(round(S*2/N,10))

s181515056

p02731

u317423698

Wrong Answer

Given is a positive integer L. Find the maximum possible volume of a rectangular cuboid whose sum of the dimensions (not necessarily integers) is L.

import sys def resolve(in_): L = int(in_.read()) return L ** -3 def main(): answer = resolve(sys.stdin) print(answer) if __name__ == '__main__': main()

s450999110

Accepted

import sys def resolve(in_): L = int(in_.read()) return (L / 3.0) ** 3 def main(): answer = resolve(sys.stdin) print(f'{answer:.12f}') if __name__ == '__main__': main()

s888479112

p03544

u100858029

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)

L = [2 if i==0 else 1 for i in range(100)] for n in range(100): L[n] = L[n-1]+L[n-2] print(L[int(input())])

s422942021

Accepted

L = [2 if i==0 else 1 for i in range(100)] for n in range(2,100): L[n] = L[n-1]+L[n-2] print(L[int(input())])

s408388255

p03759

u702018889

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()) print("Yes" if b-a==c-b else "No")

s942541255

Accepted

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

s388053025

p02240

u662418022

Wrong Answer

Write a program which reads relations in a SNS (Social Network Service), and judges that given pairs of users are reachable each other through the network.

# -*- coding: utf-8 -*- from collections import deque if __name__ == '__main__': n, m = [int(s) for s in input().split(" ")] M = [[] for j in range(n)] for _ in range(m): u, v = [int(s) for s in input().split(" ")] M[u].append(v) M[v].append(u) color = [0] * n print(M) def dfs(u, c): color[u] = c for v in range(n): if v in M[u] and color[v] == 0: dfs(v, c) color[u] = c c = 1 for u in range(n): if color[u] == 0: dfs(u, c) c += 1 l = int(input()) for _ in range(l): p, q = [int(s) for s in input().split(" ")] if color[p] == color[q]: print("yes") else: print("no")

s100408127

Accepted

# -*- coding: utf-8 -*- from collections import deque import sys sys.setrecursionlimit(200000) if __name__ == '__main__': n, m = [int(s) for s in input().split(" ")] M = [set() for j in range(n)] for _ in range(m): u, v = [int(s) for s in input().split(" ")] M[u].add(v) M[v].add(u) color = [0] * n def dfs(u, c): color[u] = c for v in M[u]: if color[v] == 0: dfs(v, c) c = 1 for u in range(n): if color[u] == 0: dfs(u, c) c += 1 l = int(input()) for _ in range(l): p, q = [int(s) for s in input().split(" ")] if color[p] == color[q]: print("yes") else: print("no")

s087626917

p03679

u231122239

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.

x, a, b = map(int, input().split(' ')) if b > a: print('delicious') elif b - a <= x: print('safe') else: print('dangerous')

s232840135

Accepted

x, a, b = map(int, input().split(' ')) if b <= a: print('delicious') elif b - a <= x: print('safe') else: print('dangerous')

s505032645

p03644

u106778233

Wrong Answer

Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.

ans =0 n = int(input()) while n%2 ==0 : ans+= 1 n//=2 print(ans)

s473095529

Accepted

n = int(input()) for i in range(8): if 2**i> n: break print(2**(i-1))

s956459351

p03610

u140251125

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.

# input s = input() ans = s[0] for i in range(len(s) // 2): ans = ans + s[2 * i] print(ans)

s014286836

Accepted

import math # input s = input() ans = '' for i in range(math.ceil(len(s) / 2)): ans = ans + s[2 * i] print(ans)

s437466167

p03574

u591287669

Wrong Answer

You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.

h,w = map(int,input().split()) arr=[] arr.append(['.']*(w+2)) for _ in range(h): arr.append(list('.'+input()+'.')) arr.append(['.']*(w+2)) ans=[] for i in range(h): ans.append([]) for j in range(w): tmpans=0 if arr[i+1][j+1]=='#': tmpans='#' else: for hh in range(3): for ww in range(3): if not(hh==1 and ww==1) and arr[i+hh][j+ww]=='#': tmpans+=1 tmpans=str(tmpans) ans[i].append(tmpans) print(arr) print(ans) for line in ans: print(''.join(line))

s379351906

Accepted

h,w = map(int,input().split()) arr=[] arr.append(['.']*(w+2)) for _ in range(h): arr.append(list('.'+input()+'.')) arr.append(['.']*(w+2)) ans=[] for i in range(h): ans.append([]) for j in range(w): tmpans=0 if arr[i+1][j+1]=='#': tmpans='#' else: for hh in range(3): for ww in range(3): if not(hh==1 and ww==1) and arr[i+hh][j+ww]=='#': tmpans+=1 tmpans=str(tmpans) ans[i].append(tmpans) for line in ans: print(''.join(line))

s918316768

p03796

u501451051

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()) double = 1 for i in range(1, N): double = double*i print(double%(10**9+7))

s651800841

Accepted

import math a = 10**9 + 7 print(math.factorial(int(input())) % a)

s971146305

p03485

u514118270

Wrong Answer

You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.

import sys import math import numpy as np import itertools import bisect from copy import copy from collections import deque,Counter from decimal import Decimal def s(): return input() def i(): return int(input()) def S(): return input().split() def I(): return map(int,input().split()) def L(): return list(input().split()) def l(): return list(map(int,input().split())) def lcm(a,b): return a*b//math.gcd(a,b) sys.setrecursionlimit(10 ** 9) INF = 10**9 mod = 10**9+7 a,b = I() print(a*b//2)

s686970997

Accepted

import sys import math import numpy as np import itertools import bisect from copy import copy from collections import deque,Counter from decimal import Decimal def s(): return input() def i(): return int(input()) def S(): return input().split() def I(): return map(int,input().split()) def L(): return list(input().split()) def l(): return list(map(int,input().split())) def lcm(a,b): return a*b//math.gcd(a,b) sys.setrecursionlimit(10 ** 9) INF = 10**9 mod = 10**9+7 a,b = I() print((a+b+1)//2)

s791963627

p02612

u698919163

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.

N = int(input()) for i in range(11): if N < 1000: break N = N - 1000 print(N)

s475335246

Accepted

N = int(input()) tmp = 0 for i in range(11): tmp += 1000 if tmp >= N: print(tmp-N) break

s937616305

p03303

u150641538

Wrong Answer

You are given a string S consisting of lowercase English letters. We will write down this string, starting a new line after every w letters. Print the string obtained by concatenating the letters at the beginnings of these lines from top to bottom.

s = input() w = int(input()) length = len(s) output = s[0] i = w while(i<length): output += s[w] i += w print(output)

s832769588

Accepted

s = input() w = int(input()) length = len(s) output = s[0] i = w while(i<length): output += s[i] i += w print(output)

s176923400

p03160

u443536265

Wrong Answer

There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.

N = int(input()) li = list(map(int,input().split())) if len(li) == 1: print(li[0]) exit() dp = [0,abs(li[1]-li[0])] print(dp) for i in range(2,N): dp.append(min(dp[i-1]+abs(li[i]-li[i-1]), dp[i-2]+abs(li[i]-li[i-2]))) print(dp[N-1])

s151837840

Accepted

N = int(input()) li = list(map(int,input().split())) if len(li) == 1: print(li[0]) exit() dp = [0,abs(li[1]-li[0])] for i in range(2,N): dp.append(min(dp[i-1]+abs(li[i]-li[i-1]), dp[i-2]+abs(li[i]-li[i-2]))) print(dp[N-1])

s154127448

p03162

u606878291

Wrong Answer

Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains.

def main(n, choices): results = [[0, 0, 0] for _ in range(n)] results[0] = choices[0] for i in range(1, n): results[i][0] = choices[i][0] + max(results[i - 1][1], results[i - 1][2]) results[i][1] = choices[i][1] + max(results[i - 1][0], results[i - 1][2]) results[i][2] = choices[i][2] + max(results[i - 1][0], results[i - 1][1]) return max(results[-1]) if __name__ == '__main__': N = int(input()) C = [] for _ in range(N): C.append(list(map(int, input().split()))) print(N, C)

s820776023

Accepted

import numpy as np N = int(input()) values = [tuple(map(int, input().split(' '))) for _ in range(N)] dp = [values[0]] for _ in range(N - 1): dp.append([0, 0, 0]) for i in range(1, N): prev = dp[i - 1] dp[i][0] = values[i][0] + max(prev[1], prev[2]) dp[i][1] = values[i][1] + max(prev[0], prev[2]) dp[i][2] = values[i][2] + max(prev[0], prev[1]) print(np.max(dp[N - 1]))

s464411966

p03589

u094191970

Wrong Answer

You are given an integer N. Find a triple of positive integers h, n and w such that 4/N = 1/h + 1/n + 1/w. If there are multiple solutions, any of them will be accepted.

n=int(input()) for h in range(1,3500): for n in range(1,3500): if h+n>=3500: continue w=3500-h-n if 4/n == 1/h+1/n+1/w: print(h,n,w) exit()

s104116708

Accepted

n=int(input()) for a in range(n//4+1,3501): for b in range((n*a)//(4*a-n),3501): if 4*a*b-n*b-a*n==0: continue c=(n*a*b)/(4*a*b-n*b-a*n) if c==int(c) and c>=1: print(a,b,int(c)) exit()

s801786380

p02388

u997714038

Wrong Answer

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

x = float(input()) print(x ** 3)

s822200586

Accepted

x = int(input()) print(x ** 3)

s500677196

p03470

u490464064

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()) N_list=[input() for i in range(N)] result_list=[] for i in range(len(N_list)): if i==(a for a in N_list): result_list.append(i) print(len(result_list))

s368948269

Accepted

N=int(input()) c=[] for i in range(N): c.append(int(input())) print(len(set(c)))

s859740695

p03007

u226108478

Wrong Answer

There are N integers, A_1, A_2, ..., A_N, written on a blackboard. We will repeat the following operation N-1 times so that we have only one integer on the blackboard. * Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y. Find the maximum possible value of the final integer on the blackboard and a sequence of operations that maximizes the final integer.

# -*- coding: utf-8 -*- def main(): from collections import deque n = int(input()) a = sorted(list(map(int, input().split()))) d = deque() for ai in a: d.append(ai) ans = list() count = n while count > 1: left = d.popleft() right = d.pop() ans.append((left, right)) tmp = left - right d.append(tmp) count -= 1 print(ans[-1][0] - ans[-1][1]) for i in range(len(ans)): print(ans[i][0], ans[i][1]) if __name__ == '__main__': main()

s244244926

Accepted

# -*- coding: utf-8 -*- def main(): n = int(input()) a = sorted(list(map(int, input().split()))) plus = [a[-1]] minus = [a[0]] # KeyInsight # See: # https://www.youtube.com/watch?v=TbobvdA6AlE for ai in a[1:-1]: if ai >= 0: plus.append(ai) else: minus.append(ai) print(sum(plus) - sum(minus)) for p in plus[1:]: print(minus[-1], p) minus[-1] -= p for m in minus: print(plus[0], m) plus[0] -= m if __name__ == '__main__': main()

s199827600

p03994

u623687794

Wrong Answer

Mr. Takahashi has a string s consisting of lowercase English letters. He repeats the following operation on s exactly K times. * Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of `z` is `a`. For example, if you perform an operation for the second letter on `aaz`, `aaz` becomes `abz`. If you then perform an operation for the third letter on `abz`, `abz` becomes `aba`. Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s. Find the such string.

s=list(input()) print(s) K=int(input()) i=0 while i<len(s): if i==len(s)-1: s[i]=chr(97+(ord(s[i])-97+K)%26) break if K<abs(123-ord(s[i]))%26: i+=1 elif K>=abs(123-ord(s[i]))%26: sub=abs(123-ord(s[i]))%26 s[i]="a" K-=sub i+=1 print("".join(s))

s152312608

Accepted

s=list(input()) K=int(input()) i=0 while i<len(s): if i==len(s)-1: s[i]=chr(97+(ord(s[i])-97+K)%26) break if K<abs(123-ord(s[i]))%26: i+=1 elif K>=abs(123-ord(s[i]))%26: sub=abs(123-ord(s[i]))%26 s[i]="a" K-=sub i+=1 print("".join(s))

s145118115

p02401

u088372268

Wrong Answer

Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.

while True: data = list(input().split(" ")) a = int(data[0]) b = int(data[2]) if data[1] == "?": break elif data[1] == "+": print(a+b) elif data[1] == "-": print(a-b) elif data[1] == "*": print(a*b) else: print(a/b)

s086552583

Accepted

while True: data = list(input().split(" ")) a = int(data[0]) b = int(data[2]) if data[1] == "?": break elif data[1] == "+": print(a+b) elif data[1] == "-": print(a-b) elif data[1] == "*": print(a*b) else: print(a//b)

s210261719

p04012

u862757671

Wrong Answer

Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.

w1 = [s for s in input()] w2 = [w1.count(a) for a in list(set(w1))] w3 = [w for w in w2 if w % 2 == 0] print(('YES' if len(w3) == len(w2) else 'NO'))

s569335610

Accepted

w1 = [s for s in input()] w2 = [w1.count(a) for a in list(set(w1))] w3 = [w for w in w2 if w % 2 == 0] print(('Yes' if len(w3) == len(w2) else 'No'))

s946389413

p03377

u278670845

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()) if a>x: print("NO") elif a+b>x: print("NO") else: print("YES")

s448179553

Accepted

a,b,x = map(int,input().split()) if a>x: print("NO") elif a+b<x: print("NO") else: print("YES")

s149118932

p02397

u244493040

Wrong Answer

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

print(*sorted(input().split()))

s615131351

Accepted

while 1: x,y = sorted(map(int,input().split())) if not (x or y): break print(x,y)

s381526102

p03378

u068142202

Wrong Answer

There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right. Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N. Find the minimum cost incurred before reaching the goal.

n, m, x = map(int, input().split()) A = list(map(int, input().split())) cost = 0 goal = min(x, (n - x)) for i in range(m): if A[i] >= x and x < goal: cost += 1 elif A[i] <= x and x > goal: cost += 1 print(cost)

s695303940

Accepted

n, m, x = map(int, input().split()) A = list(map(int, input().split())) cost = 0 for i in range(x, n+1): if A.count(i) > 0: cost += 1 cost_nomal = cost cost = 0 for i in range(1, x+1): if A.count(i) > 0: cost += 1 cost_reverse = cost print(min(cost_nomal,cost_reverse))