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TnT

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CodeNet4Repair

like 0

s657764582

p03625

u711539583

Wrong Answer

We have N sticks with negligible thickness. The length of the i-th stick is A_i. Snuke wants to select four different sticks from these sticks and form a rectangle (including a square), using the sticks as its sides. Find the maximum possible area of the rectangle.

n = int(input()) d = {} for ai in input().split(): ai = int(ai) if ai not in d: d[ai] = 0 d[ai] += 1 n2 = [d[k] for k in d if d[k] > 1] n4 = [d[k] for k in d if d[k] > 3] n2.sort(reverse = True) n4.sort(reverse = True) ans = 0 if n4: ans = n4[0] ** 2 if len(n2) > 1: max(ans, n2[0] * n2[1]) print(ans)

s258912865

Accepted

n = int(input()) d = {} for ai in input().split(): ai = int(ai) if ai not in d: d[ai] = 0 d[ai] += 1 n2 = [k for k in d if d[k] > 1] n4 = [k for k in d if d[k] > 3] n2.sort(reverse = True) n4.sort(reverse = True) ans = 0 if n4: ans = n4[0] ** 2 if len(n2) > 1: ans = max(ans, n2[0] * n2[1]) print(ans)

s700065151

p04043

u163449343

Wrong Answer

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

n = input().split() print(["NO","YES"][n.count("7") ==2 and n.count("5") ==1])

s543216119

Accepted

n = input().split() print(["NO","YES"][n.count("7")==1 and n.count("5")==2])

s175073507

p03470

u702208001

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()) print(len(set([a for a in range(n)])))

s419744334

Accepted

n = int(input()) print(len(set([int(input()) for _ in range(n)])))

s049294801

p03192

u181215519

Wrong Answer

You are given an integer N that has exactly four digits in base ten. How many times does `2` occur in the base-ten representation of N?

N = list( input() ) istwo = [ x == 2 for x in N ] print( sum( istwo ) )

s936461241

Accepted

N = list( input() ) istwo = [ x == "2" for x in N ] print( sum( istwo ) )

s887360317

p02608

u673101577

Wrong Answer

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

# from pprint import pprint # import math # import collections from numba import jit N = int(input()) # a = list(map(int, input().split(' '))) @jit(cache=True) def calc(x, y, z): return x ** 2 + y ** 2 + z ** 2 + x * y + y * z + z * x p = [] for x in range(1, 100): for y in range(1, 100): for z in range(1, 100): p.append(calc(x, y, z)) # print(p) for i in range(1, N): if i == 6: print(1) elif i in p: print(3) else: print(0)

s245660212

Accepted

# from pprint import pprint # import math # import collections from numba import jit N = int(input()) # a = list(map(int, input().split(' '))) @jit(cache=True) def calc(x, y, z): return x ** 2 + y ** 2 + z ** 2 + x * y + y * z + z * x res = [0] * 1000009 for x in range(1, 100): for y in range(1, 100): for z in range(1, 100): n = calc(x, y, z) res[n - 1] += 1 for i in range(N): print(res[i])

s411637309

p03759

u690037900

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")

s584239983

Accepted

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

s786785565

p02927

u261260430

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 month in range(4, m): for day in range(22, d): sday = str(day) if int(sday[1]) < 2: continue if int(sday[0]) * int(sday[1]) == month: ans += 1 print(ans)

s266003916

Accepted

m,d = map(int, input().split()) ans = 0 for month in range(4, m+1): for day in range(22, d+1): sday = str(day) if int(sday[1]) < 2: continue if int(sday[0]) * int(sday[1]) == month: ans += 1 print(ans)

s841514303

p03826

u117629640

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.

print(1000000000+7)

s297552860

Accepted

# -*- coding: utf-8 -*- def main(): a, b, c, d = map(int, input().split()) print(a * b) if a * b > c * d else print(c * d) if __name__ == '__main__': main()

s545709554

p02612

u202641516

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.

x=int(input()) from math import ceil x=ceil(x/1000)*1000 print(x)

s876844234

Accepted

from math import ceil x=int(input()) print(ceil(x/1000)*1000-x)

s279826155

p03470

u363836311

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()) T=[0]*n for i in range(n): T[i]=int(input()) t=set(T) print(t)

s832329483

Accepted

n=int(input()) T=[0]*n for i in range(n): T[i]=int(input()) t=len(set(T)) print(t)

s303054489

p03854

u693953100

Wrong Answer

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

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

s002559752

Accepted

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

s992759778

p03759

u399721252

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 = [ int(v) for v in input().split() ] if c-b == b-a: ans = "yes" else: ans = "NO" print(ans)

s065737277

Accepted

a, b, c = [ int(v) for v in input().split() ] if (a-b) == (b-c): print("YES") else: print("NO")

s088528551

p03448

u838296343

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

s654796106

Accepted

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

s399430675

p02255

u182913399

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.

n = int(input()) a = list(map(int, input().split())) 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(' '.join(list(map(str, a))))

s398943414

Accepted

n = int(input()) a = list(map(int, input().split())) for i in range(1, n): print(' '.join(list(map(str, a)))) 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(' '.join(list(map(str, a))))

s117727184

p03163

u451206510

Wrong Answer

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

import numpy as np N, W = map(int, input().split()) dp = np.zeros(W + 1, int) for i in range(N): w, v = map(int, input().split()) np.maximum(dp[:-w] + v, dp[w:], out=dp[w:]) print(dp) print(dp[-1])

s066904028

Accepted

import numpy as np N, W = map(int, input().split()) dp = np.zeros(W + 1, int) for i in range(N): w, v = map(int, input().split()) np.maximum(dp[:-w] + v, dp[w:], out=dp[w:]) print(dp[-1])

s208372158

p03860

u728566015

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.

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

s459691933

Accepted

[print(i[0], end="") for i in input().split()] print()

s989499964

p02853

u014139588

Wrong Answer

We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned.

x,y = map(int,input().split()) ans = 0 if x == 1 and y == 1: print(1000000) elif x < 4 and y > 3: ans += 100000*(4-x) elif x > 3 and y < 4: ans += 100000*(4-y) elif x < 4 and y < 4: ans += 100000*(8-x-y) else: print(0) print(ans)

s992351628

Accepted

x,y = map(int,input().split()) if x == 1 and y == 1: print(1000000) elif x < 4 and y > 3: print(100000*(4-x)) elif x > 3 and y < 4: print(100000*(4-y)) elif x < 4 and y < 4: print(100000*(8-x-y)) else: print(0)

s740640824

p03643

u371467115

Wrong Answer

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

print(2**(len(bin(int(input())))-3))

s381661646

Accepted

print("ABC"+input())

s815281736

p00007

u506705885

Wrong Answer

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

week=int(input()) money=100000 for i in range(week): money=money*1.05 if money%1000!=0: money=1000*(money//1000+1) print(money)

s280001900

Accepted

week=int(input()) money=100000 for i in range(week): money=money*1.05 if money%1000!=0: money=1000*(money//1000+1) print(int(money))

s217651006

p02833

u798557584

Wrong Answer

For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).

# abc148_e.py from sys import stdin def f(n): sum = 1 while n >= 2: sum = sum * n n = n - 2 return sum N = int(stdin.readline().rstrip()) sum = f(N) print(sum) l = [int(x) for x in list(str(sum))] l.reverse(); cnt = 0 for i in l: if i == 0: cnt = cnt + 1 else: break print(cnt)

s528066951

Accepted

n = int(input()) if n % 2 == 1: print(0) else: i = 10 ans = 0 while i <= n: ans += n // i i *= 5 print(ans)

s259524918

p03795

u202570162

Wrong Answer

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

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

s589484395

Accepted

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

s555757237

p02744

u171077406

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 = 3 alpha = ["a","b","c","d","e","f","g","h","i","j"] s = "".join(alpha[:N]) print("a"*N) print(s)

s402654655

Accepted

N = int(input()) def DFS(s, mx): if len(s) == N: print(s) return c = 'a' while c <= chr(ord(mx)+1): t = s + c DFS(t, chr(max(ord(c), ord(mx)))) c = chr(ord(c) + 1) DFS("", chr(ord('a')-1))

s661856591

p02831

u434609232

Wrong Answer

Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests.

a, b = map(int, input().split()) def gcd(x, y): while x%y != 0: x, y = y, x%y return y ans = a * b / gcd(a, b) print(ans)

s903216945

Accepted

a, b = map(int, input().split()) def gcd(x, y): while x%y != 0: x, y = y, x%y return y ans = a * b / gcd(a, b) print(int(ans))

s612483921

p02396

u800408401

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.

i=1 while True: x = int(input()) print("Case %d: %d" % (i,x)) i+=1 if x== 0 : break

s382907652

Accepted

i=0 while True: i+=1 x = int(input()) if x== 0 : break print("Case %d: %d" % (i,x))

s353490650

p02603

u612496929

Wrong Answer

To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally?

n = int(input()) a = list(map(int,input().split())) dp = [0] * n dp[0] = 1000 for i in range(1, n): dp[i] = max(dp[i-1], dp[i-1] // a[i-1] * a[i] + dp[i-1] % a[i-1]) print(dp[i-1])

s251277195

Accepted

n = int(input()) a = list(map(int,input().split())) dp = [0] * n dp[0] = 1000 for i in range(1, n): dp[i] = max(dp[i-1], dp[i-1] // a[i-1] * a[i] + dp[i-1] % a[i-1]) print(dp[-1])

s220876599

p02601

u654240084

Wrong Answer

M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful.

a, b, c = map(int, input().split()) k = int(input()) if a * 2 < b < c or a < b * 2 < c or a < b < c * 2: print("Yes") else: print("No")

s637261330

Accepted

a, b, c = map(int, input().split()) k = int(input()) while a >= b: k -= 1 b *= 2 while b >= c: k -= 1 c *= 2 print("Yes" if k >= 0 else "No")

s781988018

p03456

u175034939

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.

a,b = input().split() ab = a+b for i in range(320): if int(ab) == i**2: print('yes') break else: print('No')

s565811566

Accepted

a,b = input().split() ab = a+b for i in range(4,320): if int(ab) == i**2: print('Yes') exit() else: print('No')

s349646615

p03378

u756988562

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.

import numpy as np import sys N,M,X = map(int,input().split(" ")) A = list(map(int,input().split(" "))) A = np.array(A) # print(A) test = (N+1)/2 count1 = np.where(A >= test) count2 = np.where(A <= test) # print(count1) # print(len(count2)) # print("test") if count1[0][0] == 0 and len(count1[0]) == 1: print(0) sys.exit() elif count2[0][0] == 0 and len(count2[0]) == 1: print(0) sys.exit()

s419014443

Accepted

import numpy as np import sys N,M,X = map(int,input().split(" ")) A = list(map(int,input().split(" "))) temp = [] ans1 = 0 ans2 = 0 for i in range(N+1): if (i == 0) or (i==N): temp.append(0) elif i in A: temp.append(1) # elif (i == 0) or (i==N): # temp.append(0) else: temp.append(0) for i in range(X,N+1): ans1 += temp[i] for i in range(X): ans2 += temp[i] print(min(ans1,ans2))

s971947825

p02694

u262481526

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()) year = 0 money = 100 while money > X: year += 1 money = int(money*1.01) print(int(year))

s305329008

Accepted

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

s372046069

p02406

u839008951

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; }

a = int(input()) i = 1 while True: x = i if x % 3 == 0: print(" {i}".format(i=i)) else: while True: if x % 10 == 3: print(" {i}".format(i=i)) x = x // 10 if x == 0: break i += 1 if i > a: break print('')

s900840945

Accepted

a = int(input()) res = '' i = 1 while True: x = i if x % 3 == 0: res += " " + str(i) else: while True: if x % 10 == 3: res += " " + str(i) break x = x // 10 if x == 0: break i += 1 if i > a: break print(res)

s275644221

p03150

u110619679

Wrong Answer

A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.

""" B - KEYENCE String """ def simple_input(): temp_input = input().rstrip().split(' ') if len(temp_input) > 1: return temp_input return temp_input[0] if __name__ == '__main__': s = input() keyence = "keyence" result = "NO" matchcount = 0 for i in range(7): si = s[i] ki = keyence[i] if(si == ki): matchcount = matchcount + 1 else: break for i in range(7): i2 = i * -1 - 1 si = s[i2] ki = keyence[i2] if(si == ki): matchcount = matchcount + 1 else: break if(matchcount >= 7): result = "YES" print(matchcount) print(result)

s321096547

Accepted

""" B - KEYENCE String """ def simple_input(): temp_input = input().rstrip().split(' ') if len(temp_input) > 1: return temp_input return temp_input[0] if __name__ == '__main__': s = input() keyence = "keyence" result = "NO" matchcount = 0 for i in range(7): si = s[i] ki = keyence[i] if(si == ki): matchcount = matchcount + 1 else: break for i in range(7): i2 = i * -1 - 1 si = s[i2] ki = keyence[i2] if(si == ki): matchcount = matchcount + 1 else: break if(matchcount >= 7): result = "YES" print(result)

s173276502

p03964

u698176039

Wrong Answer

AtCoDeer the deer is seeing a quick report of election results on TV. Two candidates are standing for the election: Takahashi and Aoki. The report shows the ratio of the current numbers of votes the two candidates have obtained, but not the actual numbers of votes. AtCoDeer has checked the report N times, and when he checked it for the i-th (1≦i≦N) time, the ratio was T_i:A_i. It is known that each candidate had at least one vote when he checked the report for the first time. Find the minimum possible total number of votes obtained by the two candidates when he checked the report for the N-th time. It can be assumed that the number of votes obtained by each candidate never decreases.

N = int(input()) TA = [list(map(int,input().split())) for _ in range(N)] lcms = TA[0] for t,a in TA[1:]: print(t,a,lcms) tmp = [t,a] while tmp[0] < lcms[0]: tmp[0] += t tmp[1] += a while tmp[1] < lcms[1]: tmp[0] += t tmp[1] += a lcms = tmp print(sum(lcms))

s715506429

Accepted

N = int(input()) TA = [list(map(int,input().split())) for _ in range(N)] lcms = TA[0] for t,a in TA[1:]: tmp = [t,a] m1 = (lcms[0]+tmp[0]-1)//tmp[0] m2 = (lcms[1]+tmp[1]-1)//tmp[1] m = max(m1,m2) if m != 0: tmp[0] *= m tmp[1] *= m lcms = tmp print(sum(lcms))

s963387348

p03719

u763963344

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 c>=a and c<=b: print("YES") else: print("NO")

s657940688

Accepted

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

s119827763

p02396

u726978687

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.

def solve(test): for i in range(len(test)): print("Case {0}: {1}".format(i,test[i])) def answer(): test =[] while True : n = input() if n=='0': break else: test.append(n) solve(test) if __name__ =='__main__': answer()

s876114125

Accepted

def solve(test): for i in range(len(test)): print("Case {0}: {1}".format(i+1,test[i])) def answer(): test =[] while True : n = input() if n=='0': break else: test.append(n) solve(test) if __name__ =='__main__': answer()

s069068452

p02646

u989074104

Wrong Answer

Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.

A,V = map(int,input().split()) B,W = map(int,input().split()) T=int(input()) if V<=W: print("NO") elif abs(A-B)/(V-W)<=T: print("Yes") else: print("NO")

s900334360

Accepted

A,V = map(int,input().split()) B,W = map(int,input().split()) T=int(input()) if V<=W: print("NO") elif abs(A-B)/abs(V-W)<=T: print("YES") else: print("NO")

s126241936

p02850

u029000441

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.

import sys sys.setrecursionlimit(500000) N = int(input()) E = [[] for _ in range(N+1)] for i in range(N-1): a, b = map(int, input().split()) E[a].append((b, i)) E[b].append((a, i)) K = max(len(e) for e in E) print(K) Ans = [-1] * (N-1) def dfs(v=1, p=0, p_col=-1): col = 1 for u, idx in E[v]: if u!=p: if col == p_col: col += 1 Ans[idx] = col print(p_col) dfs(u, v, col) col += 1 dfs() print("\n".join(map(str, Ans)))

s665536298

Accepted

import sys sys.setrecursionlimit(500000) N = int(input()) E = [[] for _ in range(N+1)] for i in range(N-1): a, b = map(int, input().split()) E[a].append((b, i)) E[b].append((a, i)) K = max(len(e) for e in E) print(K) Ans = [-1] * (N-1) def dfs(v=1, p=0, p_col=-1): col = 1 for u, idx in E[v]: if u!=p: if col == p_col: col += 1 Ans[idx] = col dfs(u, v, col) col += 1 dfs() print("\n".join(map(str, Ans)))

s339747374

p03624

u220345792

Wrong Answer

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

alpha = "abcdefghijklmnopqrstuvwxyz" S = input() flag = True for i in alpha: if i in S: pass else: print(i) flag = False if flag: print("None")

s114986841

Accepted

alpha = "abcdefghijklmnopqrstuvwxyz" S = input() flag = True for i in alpha: if i in S: pass else: print(i) flag = False break if flag: print("None")

s845892735

p04043

u814986259

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.

s=list(input()) s.sort(key=lambda x:len(x)) if len(s[0])==7 and len(s[1])==5 and len(s[2])==5: print("YES") else: print("NO")

s641521127

Accepted

s=list(map(int,input().split())) s.sort() if s[0]==5 and s[1]==5 and s[2]==7: print("YES") else: print("NO")

s499879755

p03731

u791838908

Wrong Answer

In a public bath, there is a shower which emits water for T seconds when the switch is pushed. If the switch is pushed when the shower is already emitting water, from that moment it will be emitting water for T seconds. Note that it does not mean that the shower emits water for T additional seconds. N people will push the switch while passing by the shower. The i-th person will push the switch t_i seconds after the first person pushes it. How long will the shower emit water in total?

n, T = map(int, input().split()) t = list(map(int, input().split())) total = T for i in t: total += T - i print(total)

s949375160

Accepted

n, T = map(int, input().split()) t = list(map(int, input().split())) total = T lastnum = 0 for i in t: time = i - lastnum if T < time: total += T else: total += time lastnum = i print(total)

s509639087

p03945

u189326411

Wrong Answer

Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa. In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white. Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.

s = input() while s!="": if s[-5:]=="dream" or s[-5:]=="erase": s = s[:-5] elif s[-6:]=="eraser": s = s[:-6] elif s[-7:]=="dreamer": s = s[:-7] else: print("NO") exit() print("YES")

s937287337

Accepted

s = input() t = "" count = -1 for i in s: if i!=t: t = i count += 1 print(count)

s523196915

p03433

u167908302

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.

#coding:utf-8 n = int(input()) a = int(input()) need = n % 500 if a <= need: print("yes") else: print("no")

s466493531

Accepted

#coding:utf-8 n = int(input()) a = int(input()) need = n % 500 if a >= need: print("Yes") else: print("No")

s329987244

p03730

u357949405

Time Limit Exceeded

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

A, B, C = map(int, input().split()) count = 1 while B > count: if A * count % B == C: print("YES") exit(0) print("NO")

s052479562

Accepted

A, B, C = map(int, input().split()) count = 1 while B > count: if A * count % B == C: print("YES") exit(0) count += 1 print("NO")

s604504878

p03386

u880277518

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=list(map(int,input().split())) l=[] v=[] for i in range(a,b+1): l.append(i) for i in range(k): if i>=len(l): break v.append(l[i]) l.reverse() for i in range(k-1,-1,-1): if i>=len(l): continue v.append(l[i]) s=set(v) for i in s: print(i)

s772066163

Accepted

A,B,K=list(map(int,input().split())) L=[] k=0 for i in range(A,B+1): L.append(i) k+=1 if(k>=K): break k=0 for i in range(B,A-1,-1): L.append(i) k+=1 if(k>=K): break S=set(L) for s in sorted(list(S)): print(s)

s640019345

p03828

u218843509

Wrong Answer

You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7.

n = int(input()) prime = [] for i in range(2, n): j = 2 while j ** 2 <= i: if i % j == 0: break j += 1 if j ** 2 > i: prime.append(i) ans = 1 for p in prime: soinsu = 1 for i in range(2, n + 1): while True: if i % p == 0: soinsu += 1 i = int(i / p) else: break ans = (ans * soinsu) % (10 ** 9 + 7) print(ans)

s026138647

Accepted

n = int(input()) prime = [] for i in range(2, n + 1): j = 2 while j ** 2 <= i: if i % j == 0: break j += 1 if j ** 2 > i: prime.append(i) ans = 1 for p in prime: soinsu = 1 for i in range(2, n + 1): while True: if i % p == 0: soinsu += 1 i = int(i / p) else: break ans = (ans * soinsu) % (10 ** 9 + 7) print(ans)

s371784262

p03434

u359787029

Wrong Answer

We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.

def ret_max(n, list): i = 0 tmp = 0 max = 0 for i in range(n): if(max < list[i]): tmp = i max = list[i] return tmp N = int(input()) lis = [int(i) for i in input().split()] j = 0 k = 0 alice = 0 bob = 0 max = 0 lis_elem = len(lis) for j in range(N): lis_elem = len(lis) max = lis[ret_max(lis_elem, lis)] print("max:",max) del lis[ret_max(lis_elem, lis)] if(j%2 == 0): alice = alice + max elif(j%2 == 1): bob = bob + max print(alice - bob)

s250182414

Accepted

def ret_max_index(n, list): i = 0 tmp = 0 max = 0 for i in range(n): if(max < list[i]): tmp = i max = list[i] return tmp N = int(input()) lis = [int(i) for i in input().split()] j = 0 k = 0 alice = 0 bob = 0 max = 0 lis_elem = len(lis) for j in range(N): lis_elem = len(lis) max = lis[ret_max_index(lis_elem, lis)] #print("max:",max) del lis[ret_max_index(lis_elem, lis)] if(j%2 == 0): alice = alice + max elif(j%2 == 1): bob = bob + max print(alice - bob)

s408881252

p03998

u442877951

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.

from collections import deque S = [list(input()) for _ in range(3)] SA = deque(S[0]) SB = deque(S[1]) SC = deque(S[2]) i = 'a' ans = 'now playing' while ans == 'now playing': if i == 'a': if SA == deque([]): ans = 'A' else: i = SA.popleft() elif i == 'b': if SB == deque([]): ans = 'B' else: i = SB.popleft() else: if SC == deque([]): ans = 'C' else: i = SC.popleft() print(SA,SB,SC,ans)

s569196503

Accepted

from collections import deque S = [list(input()) for _ in range(3)] SA = deque(S[0]) SB = deque(S[1]) SC = deque(S[2]) i = 'a' ans = 'now playing' while ans == 'now playing': if i == 'a': if SA == deque([]): ans = 'A' else: i = SA.popleft() elif i == 'b': if SB == deque([]): ans = 'B' else: i = SB.popleft() else: if SC == deque([]): ans = 'C' else: i = SC.popleft() print(ans)

s160354631

p03679

u140251125

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.

# input X, A, B = map(int, input().split()) if A - B < 0: print('dangerous') elif A - B == 0: print('safe') else: print('delicious')

s817255849

Accepted

# input X, A, B = map(int, input().split()) if A >= B: print('delicious') elif B - A <= X: print('safe') else: print('dangerous')

s146767986

p02613

u972474792

Wrong Answer

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

N=int(input()) I=[] for i in range(N): I.append(input()) a=0 b=0 c=0 d=0 for i in range(N): if I[i]=='AC': a+=1 if I[i]=='WA': b+=1 if I[i]=='TLE': c+=1 if I[i]=='RE': d+=1 print('AC × '+str(a)) print('WA × '+str(b)) print('TLE × '+str(c)) print('RE × '+str(d))

s084343653

Accepted

N=int(input()) I=[] for i in range(N): I.append(input()) a=0 b=0 c=0 d=0 for i in range(N): if I[i]=='AC': a+=1 if I[i]=='WA': b+=1 if I[i]=='TLE': c+=1 if I[i]=='RE': d+=1 print('AC x '+str(a)) print('WA x '+str(b)) print('TLE x '+str(c)) print('RE x '+str(d))

s243850162

p03854

u708019102

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() a = s.count("dream") b = s.count("dreamer") c = s.count("erase") d = s.count("eraser") e = s.count("dreamera") f = s.count("erasera") b = b-e d = d-f a = a-b c = c-e if (a+c)*5 + b*7 + d*6 == len(s): print("Yes") else: print("No")

s993284055

Accepted

l = list(input()) n = len(l) cursor = n-1 while True: if cursor >= 5 and l[cursor] == "r" and l[cursor-1] == "e" and l[cursor-2] == "s" and l[cursor-3] == "a" and l[cursor-4] == "r" and l[cursor-5] == "e": cursor = cursor - 6 continue if cursor >= 4 and l[cursor] == "e" and l[cursor-1] == "s" and l[cursor-2] == "a" and l[cursor-3] == "r" and l[cursor-4] == "e": cursor = cursor -5 continue if cursor >= 6 and l[cursor] == "r" and l[cursor-1] == "e" and l[cursor-2] == "m" and l[cursor-3] == "a" and l[cursor-4] == "e" and l[cursor-5] == "r" and l[cursor-6] == "d": cursor -= 7 continue if cursor >= 4 and l[cursor] == "m" and l[cursor-1] == "a" and l[cursor-2] == "e" and l[cursor-3] == "r" and l[cursor-4] == "d": cursor -= 5 continue if cursor == -1: print("YES") break else: print("NO") break

s144279698

p00718

u078042885

Wrong Answer

Prof. Hachioji has devised a new numeral system of integral numbers with four lowercase letters "m", "c", "x", "i" and with eight digits "2", "3", "4", "5", "6", "7", "8", "9". He doesn't use digit "0" nor digit "1" in this system. The letters "m", "c", "x" and "i" correspond to 1000, 100, 10 and 1, respectively, and the digits "2", ...,"9" correspond to 2, ..., 9, respectively. This system has nothing to do with the Roman numeral system. For example, character strings > "5m2c3x4i", "m2c4i" and "5m2c3x" correspond to the integral numbers 5234 (=5*1000+2*100+3*10+4*1), 1204 (=1000+2*100+4*1), and 5230 (=5*1000+2*100+3*10), respectively. The parts of strings in the above example, "5m", "2c", "3x" and "4i" represent 5000 (=5*1000), 200 (=2*100), 30 (=3*10) and 4 (=4*1), respectively. Each of the letters "m", "c", "x" and "i" may be prefixed by one of the digits "2", "3", ..., "9". In that case, the prefix digit and the letter are regarded as a pair. A pair that consists of a prefix digit and a letter corresponds to an integer that is equal to the original value of the letter multiplied by the value of the prefix digit. For each letter "m", "c", "x" and "i", the number of its occurrence in a string is at most one. When it has a prefix digit, it should appear together with the prefix digit. The letters "m", "c", "x" and "i" must appear in this order, from left to right. Moreover, when a digit exists in a string, it should appear as the prefix digit of the following letter. Each letter may be omitted in a string, but the whole string must not be empty. A string made in this manner is called an _MCXI-string_. An MCXI-string corresponds to a positive integer that is the sum of the values of the letters and those of the pairs contained in it as mentioned above. The positive integer corresponding to an MCXI-string is called its MCXI-value. Moreover, given an integer from 1 to 9999, there is a unique MCXI-string whose MCXI-value is equal to the given integer. For example, the MCXI-value of an MCXI-string "m2c4i" is 1204 that is equal to `1000 + 2*100 + 4*1`. There are no MCXI-strings but "m2c4i" that correspond to 1204. Note that strings "1m2c4i", "mcc4i", "m2c0x4i", and "2cm4i" are not valid MCXI-strings. The reasons are use of "1", multiple occurrences of "c", use of "0", and the wrong order of "c" and "m", respectively. Your job is to write a program for Prof. Hachioji that reads two MCXI-strings, computes the sum of their MCXI-values, and prints the MCXI-string corresponding to the result.

a={'m':1000,'c':100,'x':10,'i':1} for _ in range(int(input())): b,s,t=input(),0,1 for x in b: if x==' ':continue if x in a:s+=a[x]*t;t=1 else:t=int(x) ans='' for k in ['m','c','x','i']: c,s=divmod(s,a[k]) if c:ans+=['',str(c)][c!=0]+k print(ans)

s527405047

Accepted

a={'m':1000,'c':100,'x':10,'i':1} for _ in range(int(input())): b,s,t=input().replace(' ',''),0,1 for x in b: if x in a:s+=a[x]*t;t=1 else:t=int(x) d='' for k in ['m','c','x','i']: c,s=divmod(s,a[k]) if c:d+=['',str(c)][c!=1]+k print(d)

s269831237

p02646

u842797390

Wrong Answer

Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.

A, V = [int(x) for x in input().split()] B, W = [int(x) for x in input().split()] T = int(input()) if W >= V: print('No') elif abs(A - B) <= (V - W) * T: print('Yes') else: print('No')

s468069371

Accepted

A, V = [int(x) for x in input().split()] B, W = [int(x) for x in input().split()] T = int(input()) if W >= V: print('NO') elif abs(A - B) <= (V - W) * T: print('YES') else: print('NO')

s058351294

p02277

u253463900

Wrong Answer

Let's arrange a deck of cards. Your task is to sort totally n cards. A card consists of a part of a suit (S, H, C or D) and an number. Write a program which sorts such cards based on the following pseudocode: Partition(A, p, r) 1 x = A[r] 2 i = p-1 3 for j = p to r-1 4 do if A[j] <= x 5 then i = i+1 6 exchange A[i] and A[j] 7 exchange A[i+1] and A[r] 8 return i+1 Quicksort(A, p, r) 1 if p < r 2 then q = Partition(A, p, r) 3 run Quicksort(A, p, q-1) 4 run Quicksort(A, q+1, r) Here, A is an array which represents a deck of cards and comparison operations are performed based on the numbers. Your program should also 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).

def Partition(A,p,r): x = A[r] i = p-1 for j in range(p,r): if (A[j] <= x): i += 1 A[i],A[j] = A[j],A[i] A[i+1],A[r] = A[r],A[i+1] return i+1 def qsort(A,p,r): if p < r: q = Partition(A,p,r) qsort(A,p,q-1) qsort(A,q+1,r) class card(object): def __init__(self, str): data = str.split() self.symbol,self.num = data[0],int(data[1]) def __le__(self,obj): if(self.num < obj.num): return True else: return False def __ge__(self,obj): if(self.num > obj.num): return True else: return False def display_card(self): print("{0} {1}".format(self.symbol,self.num)) c = [] n = int(input()) while(n>0): c.append(card(input())) n -= 1 c_org = c.copy() qsort(c,0,len(c)-1) notstable = False for i in range(1,len(c)): if(c[i-1].num == c[i].num): firstCard = c[i-1] secondCard = c[i] is_found_1st = False is_found_2nd = False for org in c_org: if(org == firstCard): if is_found_2nd: notstable = True if(org == secondCard): is_found_2nd = True if(notstable == 1): print('Not stable') else: print('Stable') for x in c: x.display_card()

s021530687

Accepted

def Partition(A,p,r): x = A[r] i = p-1 for j in range(p,r): if (A[j] <= x): i += 1 A[i],A[j] = A[j],A[i] A[i+1],A[r] = A[r],A[i+1] return i+1 def qsort(A,p,r): if p < r: q = Partition(A,p,r) qsort(A,p,q-1) qsort(A,q+1,r) def marge(A,left,mid,right): n1 = mid -left n2 = right -mid L = [] R = [] for i in range(left,n1+left): L.append(A[i]) for i in range(mid,n2+mid): R.append(A[i]) L.append(card('S 10000000000')) R.append(card('S 10000000000')) r_id,l_id = 0,0 for k in range(left,right): if(L[l_id] <= R[r_id]): A[k] = L[l_id] l_id += 1 else: A[k] = R[r_id] r_id += 1 def margeSort(A,left,right): if left+1 < right: mid = int((left+right)/2) margeSort(A,left,mid) margeSort(A,mid,right) marge(A,left,mid,right) class card(object): def __init__(self, str): data = str.split() self.symbol,self.num = data[0],int(data[1]) def __le__(self,obj): if(self.num <= obj.num): return True else: return False def __ge__(self,obj): if(self.num >= obj.num): return True else: return False def display_card(self): print("{0} {1}".format(self.symbol,self.num)) c = [] n = int(input()) while(n>0): c.append(card(input())) n -= 1 c_marge = c.copy() margeSort(c_marge,0,len(c)) qsort(c,0,len(c)-1) ''' notStable = False for i in range(1,len(c)): if(c[i-1].num == c[i].num): firstCard = c[i-1] secondCard = c[i] is_found_1st = False is_found_2nd = False for org in c_org: if(org == firstCard): is_found_1st = True if(org == secondCard): is_found_2nd = True if(is_found_1st == True): break else: notStable = True break if(notstable == 1): print('Not stable') else: print('Stable') for x in c: x.display_card() ''' is_stable = True for i in range(0,len(c)): if(c[i] != c_marge[i]): is_stable = False break if(is_stable == True): print('Stable') else: print('Not stable') for i in c: i.display_card()

s024318321

p03380

u047197186

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 math def combinations_count(n, r): return math.factorial(n) // (math.factorial(n - r) * math.factorial(r)) n = int(input()) a_lst = list(map(int, input().split())) max_ = 0 for i, a1 in enumerate(a_lst): for a2 in a_lst[i+1:]: if a1 > a2: cand = combinations_count(a1, a2) else: cand = combinations_count(a2, a1) if max_ < cand: max_ = cand print(max_)

s395812139

Accepted

n = int(input()) a = list(map(int, input().split())) a.sort() large = max(a) a.remove(large) min_list = [0, large/2] for i in a: if abs(i - large/2) < min_list[1]: min_list = [i, abs(i - large/2)] print(large, min_list[0])

s517673969

p00002

u298999032

Wrong Answer

Write a program which computes the digit number of sum of two integers a and b.

a,b=map(int,input().split()) print(len(str(a+b)))

s009855334

Accepted

while 1: try: a,b=map(int,input().split()) print(len(str(a+b))) except: break

s891569392

p03593

u102126195

Wrong Answer

We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter. Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition: * Every row and column in A' can be read as a palindrome. Determine whether he can create a matrix satisfying the condition.

H, W = map(int, input().split()) A = [] for i in range(H): A.extend(list(input())) A.sort() print(A) a = [0, 0, 0] if H % 2 != 0 and W % 2 != 0: a[0] = 1 a[1] = int(H / 2) + int(W / 2) a[2] = int((H * W - a[0] - a[1]) / 4) elif H % 2 != 0: a[0] = 0 a[1] = int(W / 2) a[2] = int((H * W - a[0] - a[1]) / 4) elif W % 2 != 0: a[0] = 0 a[1] = int(H / 2) a[2] = int((H * W - a[0] - a[1]) / 4) else: a[0] = 0 a[1] = 0 a[2] = int(H * W / 4) c = 1 cnt = [0, 0, 0] for i in range(1, H * W): if A[i] == A[i - 1]: c += 1 else: if c == 1: cnt[0] += 1 elif c == 2: cnt[1] += 1 elif c == 3: cnt[0] += 1 cnt[1] += 1 elif c == 4: cnt[2] += 1 elif c > 4: cnt[2] += int(c / 4) c %= 4 if c == 3: cnt[0] += 1 cnt[1] += 1 if c == 2: cnt[1] += 1 c = 1 if c == 1: cnt[0] += 1 elif c == 2: cnt[1] += 1 elif c == 3: cnt[0] += 1 cnt[1] += 1 elif c == 4: cnt[2] += 1 elif c > 4: cnt[2] += int(c / 4) c %= 4 if c == 3: cnt[0] += 1 cnt[1] += 1 if c == 2: cnt[1] += 1 if a == cnt: print("Yes") else: print("No")

s441181719

Accepted

H, W = map(int, input().split()) A = [] for i in range(H): A.extend(list(input())) A.sort() a = [0, 0, 0] if H % 2 != 0 and W % 2 != 0: a[0] = 1 a[1] = int(H / 2) + int(W / 2) a[2] = int((H * W - a[0] - a[1] * 2) / 4) elif H % 2 != 0: a[0] = 0 a[1] = int(W / 2) a[2] = int((H * W - a[0] - a[1] * 2) / 4) elif W % 2 != 0: a[0] = 0 a[1] = int(H / 2) a[2] = int((H * W - a[0] - a[1] * 2) / 4) else: a[0] = 0 a[1] = 0 a[2] = int(H * W / 4) c = 1 cnt = [0, 0, 0] for i in range(1, H * W): if A[i] == A[i - 1]: c += 1 else: if c == 1: cnt[0] += 1 elif c == 2: cnt[1] += 1 elif c == 3: cnt[0] += 1 cnt[1] += 1 elif c == 4: cnt[2] += 1 elif c > 4: cnt[2] += int(c / 4) c %= 4 if c == 3: cnt[0] += 1 cnt[1] += 1 if c == 2: cnt[1] += 1 if c == 1: cnt[0] += 1 c = 1 if c == 1: cnt[0] += 1 elif c == 2: cnt[1] += 1 elif c == 3: cnt[0] += 1 cnt[1] += 1 elif c == 4: cnt[2] += 1 elif c > 4: cnt[2] += int(c / 4) c %= 4 if c == 3: cnt[0] += 1 cnt[1] += 1 if c == 2: cnt[1] += 1 if c == 1: cnt[0] += 1 ans = 1 #print(a, cnt, ans) if a[2] > cnt[2]: ans = -1 cnt[2] -= a[2] cnt[1] += cnt[2] * 2 #print(a, cnt, ans) if a[1] > cnt[1]: ans = -1 cnt[1] -= a[1] cnt[0] += cnt[1] * 2 #print(a, cnt, ans, 2) if cnt[0] != a[0]: ans = -1 if ans == 1: print("Yes") else: print("No")

s211341275

p03673

u023229441

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()) A=list(map(str,input().split())) ans="" for i in range(n): ans+=A[i] ans=ans[::-1] print(ans)

s605619872

Accepted

n=int(input()) A=list(map(str,input().split())) ans=A[1::2][::-1]+A[::2] if n%2==1: print(*[ans[::-1]][0]) else: print(*[ans][0])

s883440554

p03214

u735906430

Wrong Answer

Niwango-kun is an employee of Dwango Co., Ltd. One day, he is asked to generate a thumbnail from a video a user submitted. To generate a thumbnail, he needs to select a frame of the video according to the following procedure: * Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video. * Select t-th frame whose representation a_t is nearest to the average of all frame representations. * If there are multiple such frames, select the frame with the smallest index. Find the index t of the frame he should select to generate a thumbnail.

import numpy as np N = int(input()) a = np.array([int(x) for x in input().split(" ")]) indices = np.abs(np.subtract.outer(a, a.mean())).argmin(0) print(a[indices])

s726159581

Accepted

import numpy as np N = int(input()) a = np.array([int(x) for x in input().split(" ")]) indices = np.abs(np.subtract.outer(a, a.mean())).argmin(0) print(indices)

s850822725

p03485

u643840641

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)

s745099148

Accepted

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

s424542534

p03563

u385244248

Wrong Answer

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

R,G = map(int,open(0).read().split()) print(3*R-2*G)

s924402264

Accepted

R = int(input()) G = int(input()) print(2*G -R)

s716499894

p03827

u607741489

Wrong Answer

You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).

n = int(input()) s = list(input()) ans=0 tmp = 0 for i in range(n): if s[i] == 'D': tmp += 1 else: tmp -= 1 ans = max(ans,tmp) print(ans)

s403426290

Accepted

n = int(input()) s = list(input()) ans=0 tmp = 0 for i in range(n): if s[i] == 'I': tmp += 1 else: tmp -= 1 ans = max(ans,tmp) print(ans)

s937510155

p02659

u906501980

Wrong Answer

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

from decimal import Decimal as dec def main(): a, b = map(dec, input().split()) print(a*b) if __name__ == "__main__": main()

s995891177

Accepted

from decimal import Decimal as dec def main(): a, b = map(dec, input().split()) print(int(a*b)) if __name__ == "__main__": main()

s717418990

p03351

u859897687

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()) print('No'if max(abs(a-b),abs(b-c),abs(c-a))>d else'Yes')

s282411303

Accepted

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

s696689234

p02612

u092061507

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)

s401361957

Accepted

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

s165195215

p03228

u905582793

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.

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

s005671169

Accepted

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

s589150483

p03854

u496975476

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`.

#!/usr/bin/env python3 s = input() words = ['maerd', 'remaerd', 'resare', 'esare'] for word in words: if s[::-1].find(word) != -1: s = s.replace(word[::-1], '') print(s) if s == '': print('YES') else: print('NO')

s678824339

Accepted

#!/usr/bin/env python3 s = input()[::-1] collector = '' t = '' words = ['remaerd', 'maerd', 'resare', 'esare'] for c in s: collector += c if collector in words: t += collector collector = '' if s == t: print('YES') else: print('NO')

s822646530

p02389

u965112171

Wrong Answer

Write a program which calculates the area and perimeter of a given rectangle.

def rect(a,b): return a * b, b+2 s = input() x = s.split() a = int(x[0]) b = int(x[1]) area,perimeter = rect(a,b) print(area,perimeter)

s281584604

Accepted

s = input() x = s.split() a = int(x[0]) b = int(x[1]) print(a*b,a+b+a+b)

s362861084

p03611

u498575211

Wrong Answer

You are given an integer sequence of length N, a_1,a_2,...,a_N. For each 1≤i≤N, you have three choices: add 1 to a_i, subtract 1 from a_i or do nothing. After these operations, you select an integer X and count the number of i such that a_i=X. Maximize this count by making optimal choices.

N = int(input()) a = list(map(int, input().split())) Memory = {} for i in a: p = i + 1 q = i r = -1 if i >= 0: r = i-1 if p in Memory: Memory[p] += 1 else: Memory[p] = 1 if q in Memory: Memory[q] += 1 else: Memory[q] = 1 if r != -1: if r in Memory: Memory[r] += 1 else: Memory[r] = 1 #print(Memory) max = 0 maxVal = 0 for k,i in Memory.items(): if max < i: max = i maxVal = k print(maxVal)

s632012864

Accepted

N = int(input()) a = list(map(int, input().split())) Memory = {} for i in a: p = i + 1 q = i r = -1 if i >= 0: r = i-1 if p in Memory: Memory[p] += 1 else: Memory[p] = 1 if q in Memory: Memory[q] += 1 else: Memory[q] = 1 if r != -1: if r in Memory: Memory[r] += 1 else: Memory[r] = 1 #print(Memory) max = 0 maxVal = 0 for k,i in Memory.items(): if max < i: max = i maxVal = k print(max)

s875375339

p02396

u967268722

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.

s=1 while True: i=input() if i=='0': break print('case '+str(s)+':', i) s+=1

s078194921

Accepted

s=1 while True: i=input() if i=='0': break print('Case '+str(s)+':', i) s+=1

s685123652

p04044

u187737912

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.

val = input() splited_val = val.split() val_list=[] for x in range(int(splited_val[0])): val = input() val_list.append(val) print(sorted(val_list))

s814240185

Accepted

val = input() splited_val = val.split() val_list=[] for x in range(int(splited_val[0])): val = input() val_list.append(val) print("".join(sorted(val_list)))

s218662672

p03943

u982591663

Wrong Answer

Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.

A, B, C = map(int, input().split()) AB = A+B BC = B+C CA = C+A if AB==BC or BC==CA or CA==AB: print("Yes") else: print("No")

s245585357

Accepted

A, B, C = map(int, input().split()) AB = A+B BC = B+C CA = C+A if AB==C or BC==A or CA==B: print("Yes") else: print("No")

s745818849

p02314

u011621222

Wrong Answer

Find the minimum number of coins to make change for n cents using coins of denominations d1, d2,.., dm. The coins can be used any number of times.

n,m = map(int, input().split()) cs = list(map(int, input().split())) dp = [True]*(n+1) dp[0] = 0 while 1: if dp[n]: break for d in range(n+1): if (dp[d]): continue for c in cs: if (dp[d+c]): dp[d+c] = dp[d]+1 elif d+c < n+1: dp[d+c] = min(dp[d+c], dp[d] + 1) print(dp[n])

s163035965

Accepted

n,m = map(int, input().split()) mo = list(map(int,input().split())) dp = [float("inf") for i in range(n+2)] dp[0] = 0 for i in range(n+1): for j in mo: if i+j <= n: dp[i+j] = min(dp[i+j],dp[i]+1) print(dp[n])

s224673973

p03387

u057079894

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.

da = list(map(int, input().split())) da = sorted(da) ans = 0 ans += (da[2] - da[0]) // 2 da[0] += (da[2] - da[0]) // 2 * 2 ans += (da[2] - da[1]) // 2 da[1] += (da[2] - da[1]) // 2 * 2 da = sorted(da) if da[0] == da[2]: pass elif da[1] == da[2]: ans += 1 else: ans += 2 print(ans)

s453913087

Accepted

da = list(map(int, input().split())) da = sorted(da) ans = 0 ans += (da[2] - da[0]) // 2 da[0] += (da[2] - da[0]) // 2 * 2 ans += (da[2] - da[1]) // 2 da[1] += (da[2] - da[1]) // 2 * 2 da = sorted(da) if da[0] == da[2]: pass elif da[0] == da[1]: ans += 1 else: ans += 2 print(ans)

s528879864

p03352

u650700749

Wrong Answer

You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.

import math x = int(input()) print(math.pow(math.floor(math.sqrt(x)), 2))

s892225047

Accepted

import math x = int(input()) def func(num, n): root = num ** (1/n) return round(root) if num == round(root) ** n else root slv = lambda b, p: math.floor(func(b, p)) ** p print(max([slv(x, p) for p in range(2, 10)]))

s204212221

p03943

u635339675

Wrong Answer

Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.

x,y,z=map(int,input().split()) if x+y==z or y+z==z or y+z==x: print("Yes") else: print("No")

s101204489

Accepted

x,y,z=map(int,input().split()) if x+y==z or y+z==x or z+x==y: print("Yes") else: print("No")

s269160102

p04044

u248670337

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=sorted([input() for i in range(N)]) print(*S,end="")

s309396395

Accepted

N,L=map(int,input().split()) S=sorted([input() for i in range(N)]) print(*S,sep="")

s277361113

p03854

u843318925

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('erace', '').replace('dreamer', '').replace('dream', '') if s: print('no') else: print('yes')

s766632645

Accepted

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

s909261058

p00767

u506554532

Wrong Answer

Let us consider rectangles whose height, _h_ , and width, _w_ , are both integers. We call such rectangles _integral rectangles_. In this problem, we consider only wide integral rectangles, i.e., those with _w_ > _h_. We define the following ordering of wide integral rectangles. Given two wide integral rectangles, 1. The one shorter in its diagonal line is smaller, and 2. If the two have diagonal lines with the same length, the one shorter in its height is smaller. Given a wide integral rectangle, find the smallest wide integral rectangle bigger than the given one.

while True: H,W = map(int,input().split()) if H == 0: break d0 = H*H + W*W ans_d = ans_h = ans_w = 10**9 for w in range(2,151): for h in range(1,w): d = h*h + w*w if d <= d0: continue if d <= ans_d: ans_h = h ans_w = w if d < ans_d: ans_d = d break print('{0} {1}'.format(ans_h, ans_w))

s097529959

Accepted

while True: H,W = map(int,input().split()) if H == 0: break d0 = H*H + W*W ans = (10**9,H,W) for w in range(2,151): for h in range(1,w): d = h*h + w*w if (d,h,w) < (d0,H,W) or (h,w) == (H,W): continue ans = min(ans, (d,h,w)) break print('{0} {1}'.format(ans[1], ans[2]))

s912715253

p02618

u384793271

Wrong Answer

AtCoder currently hosts three types of contests: ABC, ARC, and AGC. As the number of users has grown, in order to meet the needs of more users, AtCoder has decided to increase the number of contests to 26 types, from AAC to AZC. For convenience, we number these 26 types as type 1 through type 26. AtCoder wants to schedule contests for D days so that user satisfaction is as high as possible. For every day, AtCoder will hold exactly one contest, and each contest will end on that day. The satisfaction is calculated as follows. * The satisfaction at the beginning of day 1 is 0. Satisfaction can be negative. * Holding contests increases satisfaction. The amount of increase will vary depending on a variety of factors. Specifically, we know in advance that holding a contest of type i on day d will increase the satisfaction by s_{d,i}. * If a particular type of contest is not held for a while, the satisfaction decreases. Each contest type i has an integer c_i, and at the end of each day d=1,2,...,D, the satisfaction decreases as follows. Let \mathrm{last}(d,i) be the last day before day d (including d) on which a contest of type i was held. If contests of type i have never been held yet, we define \mathrm{last}(d,i)=0. At the end of day d, the satisfaction decreases by \sum _{i=1}^{26}c_i \times (d-\mathrm{last}(d,i)). Please schedule contests on behalf of AtCoder. If the satisfaction at the end of day D is S, you will get a score of \max(10^6 + S, 0). There are 50 test cases, and the score of a submission is the total scores for each test case. You can make submissions multiple times, and the highest score among your submissions will be your score.

D = int(input()) C = [0] + list(map(int, input().split())) S = [0] + [[0] + list(map(int, input().split())) for _ in range(D)] last = [0] + [0 for _ in range(26)] for d in range(1, D+1): vmax = -float('inf') for i in range(1, 27): v = 0 last_ = [s for s in last] v += S[d][i] last_[i] = d for j in range(1, 27): v -= C[j] * (d - last_[j]) if v > vmax: vmax = v t = i print(i)

s980264075

Accepted

D = int(input()) C = [0] + list(map(int, input().split())) S = [0] + [[0] + list(map(int, input().split())) for _ in range(D)] last = [0] + [0 for _ in range(26)] for d in range(1, D+1): vmax = -float('inf') for i in range(1, 27): v = 0 last_ = [s for s in last] v += S[d][i] last_[i] = d for j in range(1, 27): v -= C[j] * (d - last_[j]) if v > vmax: vmax = v t = i print(t)

s469415215

p03068

u095756391

Wrong Answer

You are given a string S of length N consisting of lowercase English letters, and an integer K. Print the string obtained by replacing every character in S that differs from the K-th character of S, with `*`.

N = int(input()) S = input() K = int(input()) for i in range(N): if S[K-1] != S[i]: S[i] == '*' print(S)

s545814590

Accepted

N = int(input()) S = input() K = int(input()) ans = '' for i in range(N): if S[K-1] != S[i]: ans += '*' else: ans += S[i] print(ans)

s960328583

p03447

u329049771

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?

x, a, b = [int(input()) for _ in range(3)] x -= a while x > 0: x -= b

s254435484

Accepted

x, a, b = [int(input()) for _ in range(3)] x -= a while x >= b: x -= b print(x)

s287074234

p02853

u663861584

Wrong Answer

We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned.

x,y = map(int, input().split()) point = 0 if x == 1 and y == 1: point = 40000 if x == 1: point += 30000 elif x == 2: point += 20000 elif x == 3: point += 10000 else: pass if y == 1: point += 30000 elif y == 2: point += 20000 elif y == 3: point += 10000 else: pass print(point)

s892130608

Accepted

x,y = map(int, input().split()) point = 0 if x == 1 and y == 1: point = 400000 if x == 1: point += 300000 elif x == 2: point += 200000 elif x == 3: point += 100000 else: pass if y == 1: point += 300000 elif y == 2: point += 200000 elif y == 3: point += 100000 else: pass print(point)

s557077205

p03543

u583276018

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**?

if(int(input()) % 1111 == 0): print("Yes") else: print("No")

s080112723

Accepted

t = input() if((t[0:1] == t[1:2] == t[2:3]) or (t[1:2] == t[2:3] == t[3:4])): print("Yes") else: print("No")

s748841493

p03139

u428341537

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()) q=min(a, b) w=a+b-n print(q,w)

s637394497

Accepted

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

s262573638

p02422

u362104929

Wrong Answer

Write a program which performs a sequence of commands to a given string $str$. The command is one of: * print a b: print from the a-th character to the b-th character of $str$ * reverse a b: reverse from the a-th character to the b-th character of $str$ * replace a b p: replace from the a-th character to the b-th character of $str$ with p Note that the indices of $str$ start with 0.

def main(): class MyClass: def __init__(self): self.strings = "" def rpl(self, int1, int2, str1): self.strings = self.strings.replace(self.strings[int1:int2+1], str1) print(self.strings) def rvr(self, int1, int2): if int1 >= 1: revstr = self.strings[int2:int1-1:-1] else: revstr = self.strings[int2::-1] self.strings = self.strings[:int1] + revstr + self.strings[int2+1:] print(self.strings) def pri(self, int1, int2): print(self.strings[int1:int2+1]) cls = MyClass() cls.strings = input() n = int(input()) for _ in range(n): li = input().split() if li[0] == "replace": cls.rpl(int(li[1]), int(li[2]), li[3]) elif li[0] == "reverse": cls.rvr(int(li[1]), int(li[2])) elif li[0] == "print": cls.pri(int(li[1]), int(li[2])) if __name__ == "__main__": main()

s173160521

Accepted

def main(): class MyClass: def __init__(self): self.strings = "" def rpl(self, int1, int2, str1): self.strings = self.strings[:int1] + str1 + self.strings[int2+1:] def rvr(self, int1, int2): if int1 >= 1: revstr = self.strings[int2:int1-1:-1] else: revstr = self.strings[int2::-1] self.strings = self.strings[:int1] + revstr + self.strings[int2+1:] def pri(self, int1, int2): print(self.strings[int1:int2+1]) cls = MyClass() cls.strings = input() n = int(input()) for _ in range(n): li = input().split() if li[0] == "replace": cls.rpl(int(li[1]), int(li[2]), li[3]) elif li[0] == "reverse": cls.rvr(int(li[1]), int(li[2])) elif li[0] == "print": cls.pri(int(li[1]), int(li[2])) if __name__ == "__main__": main()

s089282074

p02401

u436634575

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: s = input() a, op, b = s.split() if op == '?': break print(eval(s))

s710572482

Accepted

while True: s = input() a, op, b = s.split() if op == '?': break print(int(eval(s)))

s674038420

p03155

u225388820

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()) print((n - int(input())) * (n - int(input())))

s876365269

Accepted

n = int(input()) print((n - int(input()) + 1) * (n - int(input()) + 1))

s783341500

p03555

u951601135

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.

list_1=list(input()) list_2=list(input()) print(list_1[0],list_2[2]) if(list_1[0]==list_2[2] and list_1[1]==list_2[1] and list_1[2]==list_2[0]): print('YES') else:print('NO')

s794922632

Accepted

print(["NO","YES"][input()==input()[::-1]])

s306508982

p03470

u658113376

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?

import sys def main(lines): print(len(set(lines[1:-1]))) if __name__ == '__main__': lines = [] for l in sys.stdin: lines.append(l.rstrip('\r\n')) main(lines)

s205149417

Accepted

import sys def main(lines): print(len(set(lines[1:]))) if __name__ == '__main__': lines = [] for l in sys.stdin: lines.append(l.rstrip('\r\n')) main(lines)

s302542042

p04044

u074161135

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.

info = input().split() lst = [] for i in range(int(info[0])): lst.append(str(input())) lst = sorted(lst) print(lst)

s653077647

Accepted

info = input().split() lst = [] for i in range(int(info[0])): lst.append(str(input())) lst = "".join(sorted(lst)) print((lst))

s371658630

p03593

u925364229

Wrong Answer

We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter. Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition: * Every row and column in A' can be read as a palindrome. Determine whether he can create a matrix satisfying the condition.

def fourcount(List): num = 0 tmp = 1 l = len(List) for i in range(1,l): if (List[i-1] == List[i]) and (tmp < 4): tmp += 1 else: if tmp == 4: num += 1 tmp = 1 return num def twocount(List): num = 0 tmp = 1 l = len(List) for i in range(1,l): if (List[i-1] == List[i]) and (tmp < 2): tmp += 1 else: if tmp == 2: num += 1 tmp = 1 return num def decision(H,W,num4,num2): if H == 1: if (W % 2==0) and (num2 == W/2): print("Yes") elif (W % 2 == 1) and (num2 == (W-1)/2): print("Yes") else: print("No") elif W == 1: if (H % 2==0) and (num2 == H/2): print("Yes") elif (H % 2 == 1) and (num2 == (H-1)/2): print("Yes") else: print("No") elif H > W: if (H % 2 == 0) and (H / 2 <= num4): same = H/2 decision(H,W-2,num4-same,num2) elif (H % 2 == 1) and ((H-1) / 2 <= num4 ) and (num2 >= 1): same = (H-1) /2 decision(H,W-2,num4-same,num2-1) else: print("No") elif W > H: if (W % 2 == 0) and (W / 2 <= num4): same = H/2 decision(H-2,W,num4-same,num2) elif (W % 2 == 1) and ((W-1) / 2 <= num4) and (num2 >= 1): same = (W-1) /2 decision(H-2,W,num4-same,num2-1) else: print("No") elif W == H: if (H % 2 == 0) and (H - 1) <= num4 : decision(H-2,W-2,num4-(H-1),num2) elif (H % 2 == 1) and (H - 2) <= num4 and (num2 >= 2): decision(H-2,W-2,num4-(H-2),num2-2) else : print("No") H,W = map(int,input().split(" ")) a = [] b = [] for i in range(H): a = list(input()) a = a[0:W] for string in a: b.append(string) b.sort() num4 = fourcount(b) num2 = twocount(b) - (num4*2) decision(H,W,num4,num2)

s777378138

Accepted

def fourcount(List): num = 0 tmp = 1 l = len(List) for i in range(1,l): if (List[i-1] == List[i]) and (tmp < 4): tmp += 1 elif tmp == 4: num += 1 tmp = 1 else: tmp = 1 if tmp == 4: num += 1 return num def twocount(List): num = 0 tmp = 1 l = len(List) for i in range(1,l): if (List[i-1] == List[i]) and (tmp < 2): tmp += 1 elif tmp == 2: num += 1 tmp = 1 else: tmp = 1 if tmp == 2: num += 1 return num def decision(H,W,num4,num2): if H == 1: if (W % 2 == 0) and (num2 == W/2): print("Yes") elif (W % 2 == 1) and (num2 == (W-1)/2): print("Yes") else: print("No") elif W == 1: if (H % 2 == 0) and (num2 == H/2): print("Yes") elif (H % 2 == 1) and (num2 == (H-1)/2): print("Yes") else: print("No") elif H == 0 or W == 0: print("Yes") elif H > W: if (H % 2 == 0) and (H / 2 <= num4): same = H/2 decision(H,W-2,num4-same,num2-same*2) elif (H % 2 == 1) and ((H-1) / 2 <= num4 ) and (num2-(H-1) >= 1): same = (H-1) /2 decision(H,W-2,num4-same,num2-same*2-1) else: print("No") elif W > H: if (W % 2 == 0) and (W / 2 <= num4): same= W/2 decision(H-2,W,num4-same,num2-same*2) elif (W % 2 == 1) and ((W-1) / 2 <= num4) and (num2-(W-1) >= 1): same = (W-1) /2 decision(H-2,W,num4-same,num2-same*2-1) else: print("No") elif W == H: if (H % 2 == 0) and (H - 1) <= num4 : same = H-1 decision(H-2,W-2,num4-same,num2-same*2) elif (H % 2 == 1) and (H - 2) <= num4 and (num2-2*(H-2) >= 2): same = H-2 decision(H-2,W-2,num4-same,num2-same*2-2) else : print("No") H,W = map(int,input().split(" ")) a = [] b = [] for i in range(H): a = list(input()) a = a[0:W] for string in a: b.append(string) b.sort() num4 = fourcount(b) num2 = twocount(b) decision(H,W,num4,num2)

s663529475

p03448

u729217226

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

s701906852

Accepted

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

s828703862

p02608

u437990671

Wrong Answer

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

n = int(input()) dp = [0 for i in range(n+1)] maxn = 101 for x in range(1,maxn): for y in range(1,maxn): for z in range(1,maxn): p = (x+y+z)**2 p-=(x*y + y*z + z*x) if p<=n: dp[p]+=1 for i in dp: print(i)

s095440887

Accepted

n = int(input()) dp = [0 for i in range(n+1)] maxn = n**0.5 maxn = int(maxn)+1 for x in range(1,maxn): for y in range(1,maxn): for z in range(1,maxn): p = x*x + y*y + z*z + x*y + y*z +z*x if p<=n: dp[p]+=1 for i in dp[1::]: print(i)

s443631760

p03548

u495415554

Wrong Answer

We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person. At most how many people can sit on the seat?

a, b, c = map(int, input().split()) max = int((a + c) / (b + 2*c)) print(max)

s600718063

Accepted

import sys import math a, b, c = map(int, input().split()) if a < (b + 2*c): print("a >= b + 2*cとなるようにしてください") sys.exit() elif a == (b + 2*c): print("1") else: max = math.floor((a - c) / (b + c)) print(max)

s614152396

p03795

u319612498

Wrong Answer

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

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

s794665352

Accepted

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

s063050057

p02972

u748241164

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.

n = int(input()) a = list(map(int, input().split())) a = [0] + a c = [0] * (n+1) for i in range(n, 0, -1): c[i] = a[i] tmp = i * 2 while tmp <= n: c[i] -= c[tmp] tmp += i c[i] = c[i] % 2 for i in range(1, n+1): if c[i] == 1: print(i)

s225577870

Accepted

n = int(input()) a = list(map(int, input().split())) a = [0] + a c = [0] * (n+1) for i in range(n, 0, -1): c[i] = a[i] tmp = i * 2 while tmp <= n: c[i] -= c[tmp] tmp += i c[i] = c[i] % 2 print(sum(c)) for i in range(1, n+1): if c[i] == 1: print(i)

s462418882

p00007

u617990214

Wrong Answer

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

a=int(input()) for i in range(5): a*=1.05 if a%1000==0: pass else: a=(a+1000)-a%1000 print(int(a))

s859674330

Accepted

a=100000 j=int(input()) for i in range(j): a*=1.05 if a%1000==0: pass else: a=(a+1000)-a%1000 print(int(a))

s027730990

p02748

u052821962

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.

na,nb,M=map(int, input().split()) A = list(map(int, input().split())) B = list(map(int, input().split())) C = [[0 for i in range(3)] for j in range(M)] #C = [[0]*3]*M s=min(A)+min(B) t=0 #print(A) for i in range(M): C[i][0],C[i][1],C[i][2] = map(int, input().split()) #print(C) for j in range(M): print(j) t=A[C[j][0]-1]+B[C[j][1]-1]-C[j][2] if t<s: s=t print(s)

s186775341

Accepted

na,nb,M=map(int, input().split()) A = list(map(int, input().split())) B = list(map(int, input().split())) C = [[0 for i in range(3)] for j in range(M)] #C = [[0]*3]*M s=min(A)+min(B) t=0 #print(A) for i in range(M): C[i][0],C[i][1],C[i][2] = map(int, input().split()) #print(C) for j in range(M): #print(j) t=A[C[j][0]-1]+B[C[j][1]-1]-C[j][2] if t<s: s=t print(s)

s706130577

p02612

u695079172

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(): n = int(input()) print(n % 1000) if __name__ == '__main__': main()

s907696849

Accepted

def main(): n = int(input()) n = n % 1000 if n == 0: print(0) else: print(1000 - n) if __name__ == '__main__': main()

s390215517

p03353

u059210959

Wrong Answer

You are given a string s. Among the **different** substrings of s, print the K-th lexicographically smallest one. A substring of s is a string obtained by taking out a non-empty contiguous part in s. For example, if s = `ababc`, `a`, `bab` and `ababc` are substrings of s, while `ac`, `z` and an empty string are not. Also, we say that substrings are different when they are different as strings. Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.

#encoding utf-8 import copy import numpy as np import random s = list(str(input())) k = int(input()) substrings = [] for length in range(1,len(s)+1): for i in range(len(s)-length+1): word = "".join(s[i:i+length]) print(word) if word in substrings: pass else: substrings.append(str(word)) substrings.sort() # print(substrings) print(substrings[k-1])

s393202068

Accepted

#encoding utf-8 import copy import numpy as np import random s = list(str(input())) k = int(input()) substrings = [] for length in range(1,len(s)+1): if length > 5: break for i in range(len(s)-length+1): word = "".join(s[i:i+length]) # print(word) if word in substrings: pass else: substrings.append(str(word)) substrings.sort() # print(substrings) print(substrings[k-1])

s011587580

p03545

u174181999

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.

x = str(input()) l = [x[0], x[1], x[2], x[3]] m = [int(s) for s in l] N = 2 ** 3 for i in range(N): n = [m[0], 0, 0, 0] o = [] for j in range(3): if (i >> j) & 1: n[-j-1] = -m[-j-1] o.append('-') else: n[-j-1] = m[-j-1] o.append('+') if sum(n) == 7: print(str(m[0]) + o[0] + str(m[1]) + o[1] + str(m[2]) + o[2] + str(m[3])) exit()

s804561969

Accepted

x = str(input()) l = [x[0], x[1], x[2], x[3]] m = [int(s) for s in l] N = 2 ** 3 for i in range(N): n = [m[0], 0, 0, 0] o = [] for j in range(3): if (i >> j) & 1: n[-j-1] = -m[-j-1] o.append('-') else: n[-j-1] = m[-j-1] o.append('+') if sum(n) == 7: print(str(m[0]) + o[2] + str(m[1]) + o[1] + str(m[2]) + o[0] + str(m[3]) + '=7') exit()