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CodeNet4Repair

like 0

s839161517

p02264

u564464686

Wrong Answer

_n_ _q_ _name 1 time1_ _name 2 time2_ ... _name n timen_ In the first line the number of processes _n_ and the quantum _q_ are given separated by a single space. In the following _n_ lines, names and times for the _n_ processes are given. _name i_ and _time i_ are separated by a single space.

s=0 A=input().split() n=(int)(A[0]) t=(int)(A[1]) x=[0 for i in range(2*n)] for i in range(n): B=input().split() k=2*i x[k]=(B[0]) x[k+1]=(int)(B[1]) while len(x)>0: s=x[1]-t if s>0: sum=t x.append(x[0]) x.append(s) del x[0] del x[0] if s<=0: sum+=x[1] print( x[0],":",sum) del x[0] del x[0]

s926425072

Accepted

sum=0 s=0 A=input().split() n=(int)(A[0]) t=(int)(A[1]) x=[0 for i in range(2*n)] for i in range(n): B=input().split() k=2*i x[k]=(B[0]) x[k+1]=(int)(B[1]) while len(x)>0: s=x[1]-t if s>0: sum+=t x.append(x[0]) x.append(s) del x[0] del x[0] if s<=0: sum+=x[1] print( x[0],sum) del x[0] del x[0]

s545690776

p03129

u494151155

Wrong Answer

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

N,K = [int(_) for _ in input().split()] if N/2 >= K: print("YES") else: print("NO")

s855027244

Accepted

N,K = [int(_) for _ in input().split()] if N >= K*2 - 1: print("YES") else: print("NO")

s765367054

p02389

u623275452

Wrong Answer

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

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

s365816041

Accepted

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

s113999284

p03473

u672898046

Wrong Answer

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

i = int(input()) print(i+24)

s526824128

Accepted

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

s208610730

p03503

u282228874

Wrong Answer

Joisino is planning to open a shop in a shopping street. Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods. There are already N stores in the street, numbered 1 through N. You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2. Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}. Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.

from itertools import product n = int(input()) F = [list(map(int,input().split())) for i in range(n)] P = [list(map(int,input().split())) for i in range(n)] ans = [] for p in product([0,1],repeat = 10): if sum(p) == 0: continue res = 0 for i in range(n): cnt = 0 for j in range(10): if p[j] == 1: cnt += F[i][j] res += P[i][cnt] ans.append(res) print(max(ans))

s742959438

Accepted

from itertools import product n = int(input()) F = [list(map(int,input().split())) for i in range(n)] P = [list(map(int,input().split())) for i in range(n)] ans = [] for p in product([0,1],repeat = 10): if sum(p) == 0: continue res = 0 for i in range(n): cnt = 0 for j in range(10): if p[j]: cnt += F[i][j] res += P[i][cnt] ans.append(res) print(max(ans))

s644064577

p03473

u928784113

Wrong Answer

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

#!/usr/bin/env python # -*- coding utf-8 -*- x = int(input()) print("{}".format(x + 24))

s098666345

Accepted

# -*- coding: utf-8 -*- M = int(input()) print(48-M)

s347277026

p03447

u617659131

Wrong Answer

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

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

s180279205

Accepted

x = int(input()) a = int(input()) b = int(input()) x -= a print(x % b)

s487891121

p00042

u319725914

Wrong Answer

宝物がたくさん収蔵されている博物館に、泥棒が大きな風呂敷を一つだけ持って忍び込みました。盗み出したいものはたくさんありますが、風呂敷が耐えられる重さが限られており、これを超えると風呂敷が破れてしまいます。そこで泥棒は、用意した風呂敷を破らず且つ最も価値が高くなるようなお宝の組み合わせを考えなくてはなりません。 風呂敷が耐えられる重さ W、および博物館にある個々のお宝の価値と重さを読み込んで、重さの総和が W を超えない範囲で価値の総和が最大になるときの、お宝の価値総和と重さの総和を出力するプログラムを作成してください。ただし、価値の総和が最大になる組み合わせが複数あるときは、重さの総和が小さいものを出力することとします。

from sys import stdin case = 0 while(True): case += 1 W = int(stdin.readline()) if not W: break print("Case %d"%case) N = int(stdin.readline()) v = [0] * N w = [0] * N dp = [[0]*(W + 1) for _ in range(N + 1)] for k in range(N): v[k], w[k] = map(int, stdin.readline().split(",")) for i in range(N-1,-1,-1): for j in range(W+1): if j < w[i]: dp[i][j] = dp[i + 1][j] else: dp[i][j] = max(dp[i + 1][j], dp[i + 1][j - w[i]] + v[i]) print(dp[0][W]) for i in range(W+1): if dp[0][i] == dp[0][W]: print(i); break

s828116728

Accepted

from sys import stdin case = 0 while(True): case += 1 W = int(stdin.readline()) if not W: break print("Case %d:"%case) N = int(stdin.readline()) v = [0] * N w = [0] * N dp = [[0]*(W + 1) for _ in range(N + 1)] for k in range(N): v[k], w[k] = map(int, stdin.readline().split(",")) for i in range(N-1,-1,-1): for j in range(W+1): if j < w[i]: dp[i][j] = dp[i + 1][j] else: dp[i][j] = max(dp[i + 1][j], dp[i + 1][j - w[i]] + v[i]) print(dp[0][W]) for i in range(W+1): if dp[0][i] == dp[0][W]: print(i); break

s939723107

p04043

u422242631

Wrong Answer

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

A,B,C = list(input().split()) count = 0 if A == "7": count += 1 if B == "7": count += 1 if C == "7": count += 1 if count == 1: print("Yes") else: print("No")

s209764526

Accepted

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

s226755694

p03623

u023077142

Wrong Answer

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

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

s077556362

Accepted

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

s407942528

p02843

u963915126

Wrong Answer

AtCoder Mart sells 1000000 of each of the six items below: * Riceballs, priced at 100 yen (the currency of Japan) each * Sandwiches, priced at 101 yen each * Cookies, priced at 102 yen each * Cakes, priced at 103 yen each * Candies, priced at 104 yen each * Computers, priced at 105 yen each Takahashi wants to buy some of them that cost exactly X yen in total. Determine whether this is possible. (Ignore consumption tax.)

ni = lambda: int(input()) nm = lambda: map(int, input().split()) nl = lambda: list(map(int, input().split())) from math import ceil x=ni() p=x%100 if ceil(p/5) <= x//100: print('Yes') else: print('No')

s299353071

Accepted

ni = lambda: int(input()) nm = lambda: map(int, input().split()) nl = lambda: list(map(int, input().split())) from math import ceil x=ni() p=x%100 if ceil(p/5) <= x//100: print(1) else: print(0)

s867551671

p00049

u078042885

Wrong Answer

ある学級の生徒の出席番号と ABO 血液型を保存したデータを読み込んで、おのおのの血液型の人数を出力するプログラムを作成してください。なお、ABO 血液型には、A 型、B 型、AB 型、O 型の４種類の血液型があります。

a={'A':0,'B':0,'O':0,'AB':0} while 1: try: _,b=input().split(",") a[b]+=1 except: for i in a:print(a[i]) break

s984294896

Accepted

a={'A':0,'B':0,'O':0,'AB':0} while 1: try: _,b=input().split(",") a[b]+=1 except: for i in ['A','B','AB','O']:print(a[i]) break

s605148981

p03645

u662613022

Wrong Answer

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

N,M = map(int,input().split()) li = [] numb = [] ans = '' for i in range(M): li.append(tuple(map(int,input().split()))) if li[i][0] == 1: numb.append(i) for i in range(M): if li[i][0] in numb and li[i][1] == N: ans = 'POSSIBLE' break if ans == '': print('IMPOSSIBLE') else: print(ans)

s271211059

Accepted

N,M = map(int,input().split()) li = [] numb = [] for i in range(M): a,b = map(int,input().split()) if a == 1: li.append(b) elif b == N: numb.append(a) ans = set(li) & set(numb) if len(ans) >= 1: print('POSSIBLE') else: print('IMPOSSIBLE')

s305956571

p03795

u138486156

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*200//15)

s535306651

Accepted

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

s859935135

p02534

u667469290

Wrong Answer

You are given an integer K. Print the string obtained by repeating the string `ACL` K times and concatenating them. For example, if K = 3, print `ACLACLACL`.

# -*- coding: utf-8 -*- def solve(): return 'ALC'*int(input()) if __name__ == '__main__': print(solve())

s670311584

Accepted

# -*- coding: utf-8 -*- def solve(): return 'ACL'*int(input()) if __name__ == '__main__': print(solve())

s740624801

p02279

u637322311

Wrong Answer

A graph _G_ = ( _V_ , _E_ ) is a data structure where _V_ is a finite set of vertices and _E_ is a binary relation on _V_ represented by a set of edges. Fig. 1 illustrates an example of a graph (or graphs). **Fig. 1** A free tree is a connnected, acyclic, undirected graph. A rooted tree is a free tree in which one of the vertices is distinguished from the others. A vertex of a rooted tree is called "node." Your task is to write a program which reports the following information for each node _u_ of a given rooted tree _T_ : * node ID of _u_ * parent of _u_ * depth of _u_ * node type (root, internal node or leaf) * a list of chidlren of _u_ If the last edge on the path from the root _r_ of a tree _T_ to a node _x_ is ( _p_ , _x_ ), then _p_ is the **parent** of _x_ , and _x_ is a **child** of _p_. The root is the only node in _T_ with no parent. A node with no children is an **external node** or **leaf**. A nonleaf node is an **internal node** The number of children of a node _x_ in a rooted tree _T_ is called the **degree** of _x_. The length of the path from the root _r_ to a node _x_ is the **depth** of _x_ in _T_. Here, the given tree consists of _n_ nodes and evey node has a unique ID from 0 to _n_ -1. Fig. 2 shows an example of rooted trees where ID of each node is indicated by a number in a circle (node). The example corresponds to the first sample input. **Fig. 2**

from sys import stdin class Node(object): def __init__(self, parent=None, children=None, name=None, depth=None): self.parent = parent self.children = children self.name = name self.depth = depth def set_depth(nodes, u): p = nodes[u].parent d = nodes[p].depth if d != None: nodes[u].depth = d + 1 def read_rooted_tree(n, nodes): for _ in range(n): i = [int(i) for i in stdin.readline().strip().split()] c = i[2:len(i)] nodes[i[0]].children = c if c: nodes[i[0]].name = "internal node" else: nodes[i[0]].name = "leaf" for j in nodes[i[0]].children: nodes[j].parent = i[0] set_depth(nodes, i[0]) n = int(input()) nodes = [Node() for _ in range(n)] nodes[0].parent = -1 nodes[0].depth = 0 nodes[0].name = "root" read_rooted_tree(n, nodes) for i in range(n): print("node {0}: parent = {1}, depth = {2}, {3}, {4}".format( i, nodes[i].parent, nodes[i].depth, nodes[i].name, nodes[i].children))

s090739079

Accepted

from sys import stdin class Node(object): def __init__(self, parent=None, children=None, name=None): self.parent = parent self.children = children self.name = name def print_nodes(nodes, n): def get_depth(nodes, u): p = nodes[u].parent nonlocal d if p != None and p != -1: get_depth(nodes, p) d += 1 return d for i in range(n): d = 0 if nodes[i].parent == None: nodes[i].parent = -1 nodes[i].name = "root" print("node {0}: parent = {1}, depth = {2}, {3}, {4}".format( i, nodes[i].parent, get_depth(nodes, i), nodes[i].name, nodes[i].children)) def read_rooted_tree(nodes, n): for _ in range(n): i = [int(i) for i in stdin.readline().strip().split()] c = i[2:len(i)] nodes[i[0]].children = c if c: nodes[i[0]].name = "internal node" else: nodes[i[0]].name = "leaf" for j in nodes[i[0]].children: nodes[j].parent = i[0] n = int(input()) nodes = [Node() for _ in range(n)] read_rooted_tree(nodes, n) print_nodes(nodes, n)

s419106254

p03130

u519939795

Wrong Answer

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

a1,b1=map(int,input().split()) a2,b2=map(int,input().split()) a3,b3=map(int,input().split()) l=[] l.append(a1) l.append(b1) l.append(a2) l.append(b2) l.append(a3) l.append(b3) print(l) if l.count(1)==1 and l.count(4)==1 and l.count(2)==2 and l.count(3)==2: print('YES') else: print('NO')

s285324704

Accepted

li=[0]*4 for i in range(3): a,b=map(int,input().split()) li[a-1]+=1 li[b-1]+=1 if li.count(1)==2 and li.count(2)==2: print('YES') else: print('NO')

s743153657

p03155

u433195318

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?

3 2 3

s317787236

Accepted

import sys import math import fractions import bisect import queue import heapq from collections import deque sys.setrecursionlimit(4100000) MOD = int(1e9+7) PI = 3.14159265358979323846264338327950288 INF = 1e18 N = int(input()) H = int(input()) W = int(input()) print((N-H+1)*(N-W+1))

s373193677

p02844

u408375121

Wrong Answer

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

n = int(input()) s = input() ans = 0 for a in range(10): for b in range(10): for c in range(10): for i in range(n-2): if s[i] == str(a): for j in range(i+1, n-1): if s[j] == str(b): for k in range(j+1, n): if s[k] == str(c): ans += 1 print(ans)

s657570992

Accepted

n = int(input()) s = input() before = [[-1] * 10] for i in range(n): c = s[i] num = int(c) hoge = before[-1][::] hoge[num] = i before.append(hoge) ans = 0 for i in range(10): now = before[-1][i] if now == -1: continue for j in range(10): now2 = before[now][j] if now2 == -1: continue for k in range(10): now3 = before[now2][k] if now3 == -1: continue ans += 1 print(ans)

s127765739

p03163

u693173434

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) for i in range(N): w, v = map(int, input().split()) dp[w:] = np.maximum(dp[w:], dp[:-w] + v) print(dp) print(int(dp.max()))

s934932469

Accepted

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

s757384374

p02665

u801701525

Wrong Answer

Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.

import math N = int(input()) leaf = list(map(int,input().split())) end_flag = False count = sum(leaf) ans = 0 node = 1 for i in range(N+1): if end_flag: break ans += node count -= leaf[i] node = min(count, (node-leaf[i])*2) if node < 0: end_flag = True if end_flag: print('-1') else: print(1+sum(leaf)-leaf[N]+N*2)

s516962379

Accepted

import math N = int(input()) leaf = list(map(int,input().split())) end_flag = False count = sum(leaf) ans = 0 node = 1 for i in range(N+1): if end_flag: break ans += node count -= leaf[i] node = min(count, (node-leaf[i])*2) if node < 0: end_flag = True if end_flag: print('-1') else: print(ans)

s416512864

p02612

u962024525

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.

m=int(input()) print(m%1000)

s892842237

Accepted

m=int(input()) print((1000-m%1000)%1000)

s736442952

p02402

u921038488

Wrong Answer

Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.

def bubble_sort(l): N = len(l) isChange = True while isChange: isChange = False for i in range(N-1): if(l[i] > l[i+1]): l[i], l[i+1] = l[i+1], l[i] isChange = True return l N = int(input()) l = input().split() L = bubble_sort(l) sum = 0 for k in L: n = int(k) sum += n print("{} {} {}".format(L[0], L[-1], sum))

s348922162

Accepted

N = input() l = map(int, input().split()) l = list(l) print("{} {} {}".format(min(l), max(l), sum(l)))

s221472186

p03997

u545666277

Wrong Answer

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

# -*- coding: utf-8 -*- a = int(input()) b = int(input()) h = int(input()) print( ((a+b)*h)/2)

s050406628

Accepted

# -*- coding: utf-8 -*- a = int(input()) b = int(input()) h = int(input()) print( int(((a+b)*h)/2) )

s580637257

p03353

u644972721

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}.

s = list(input()) k = int(input()) t = sorted(s) u = [] v = [] for i in range(min(len(s), 5)): u.append(t[i]) for i in range(len(s)): if s[i] in u: if not [s[i]] in v: v.append([s[i]]) l = [s[i]] for j in range(1, 5): if i + j == len(s): break l = l + [s[i + j]] if not l in v: v.append(l) for i in range(len(v)): v[i] = "".join(v[i]) v.sort() print(v[k - 1]) print(v)

s686266044

Accepted

s = list(input()) k = int(input()) t = sorted(s) u = [] v = [] for i in range(min(len(s), 5)): u.append(t[i]) for i in range(len(s)): if s[i] in u: if not [s[i]] in v: v.append([s[i]]) l = [s[i]] for j in range(1, 5): if i + j == len(s): break l = l + [s[i + j]] if not l in v: v.append(l) for i in range(len(v)): v[i] = "".join(v[i]) v.sort() print(v[k - 1])

s949965535

p03049

u673338219

Wrong Answer

Snuke has N strings. The i-th string is s_i. Let us concatenate these strings into one string after arranging them in some order. Find the maximum possible number of occurrences of `AB` in the resulting string.

n = int(input()) c = 0 xa = 0 bx = 0 ba = 0 for _ in range(n): si = str(input()) l = len(si) s1 = si[0] s2 = si[-1] if s1=="B": if s2=="A": ba += 1 else: bx += 1 elif s2 == "A": xa += 1 for i in range(l-1): if si[i]=="A" and si[i+1]=="B": c += 1 print(c,ba,bx,xa) if ba == 0: print(c+min(bx,xa)) exit() if bx == 0: if xa == 0: print(ba-1+c) exit() else: print(ba+c) exit() if xa == 0: print(ba+c) exit() print(min(bx,xa)+ba+c)

s412883371

Accepted

n = int(input()) c = 0 xa = 0 bx = 0 ba = 0 for _ in range(n): si = str(input()) l = len(si) s1 = si[0] s2 = si[-1] if s1=="B": if s2=="A": ba += 1 else: bx += 1 elif s2 == "A": xa += 1 for i in range(l-1): if si[i]=="A" and si[i+1]=="B": c += 1 if ba == 0: print(c+min(bx,xa)) exit() if bx == 0: if xa == 0: print(ba-1+c) exit() else: print(ba+c) exit() if xa == 0: print(ba+c) exit() print(min(bx,xa)+ba+c)

s218529321

p03486

u134519179

Wrong Answer

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

s = input() t = input() s = sorted(s) t = sorted(t)[::-1] if s < t: print('YES') else: print('NO')

s195522725

Accepted

s = input() t = input() s = sorted(s) t = sorted(t)[::-1] if s < t: print('Yes') else: print('No')

s402129136

p03433

u789791593

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_yen = int(input()) A_1mai = int(input()) answer = 'YES' if N_yen % 500 <= A_1mai else 'NO' print(answer)

s806408161

Accepted

# coding: utf-8 N_yen = int(input()) A_1mai = int(input()) answer = 'Yes' if N_yen % 500 <= A_1mai else 'No' print(answer)

s375425310

p03149

u410118019

Wrong Answer

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

n = set(map(int,input().split())) print('YNeos'[n!={1,9,7,4}::2])

s693535243

Accepted

n = set(map(int,input().split())) print('YNEOS'[n!={1,9,7,4}::2])

s166382357

p03815

u870518235

Wrong Answer

Snuke has decided to play with a six-sided die. Each of its six sides shows an integer 1 through 6, and two numbers on opposite sides always add up to 7. Snuke will first put the die on the table with an arbitrary side facing upward, then repeatedly perform the following operation: * Operation: Rotate the die 90° toward one of the following directions: left, right, front (the die will come closer) and back (the die will go farther). Then, obtain y points where y is the number written in the side facing upward. For example, let us consider the situation where the side showing 1 faces upward, the near side shows 5 and the right side shows 4, as illustrated in the figure. If the die is rotated toward the right as shown in the figure, the side showing 3 will face upward. Besides, the side showing 4 will face upward if the die is rotated toward the left, the side showing 2 will face upward if the die is rotated toward the front, and the side showing 5 will face upward if the die is rotated toward the back. Find the minimum number of operation Snuke needs to perform in order to score at least x points in total.

x = int(input()) if x % 11 == 0: ans = (x // 11)*2 else: ans = (x // 11)*2 + 1 print(ans)

s106859526

Accepted

x = int(input()) X = x % 11 N = x // 11 ans = 0 if 1 <= X <= 6: ans = N*2 + 1 elif 7 <= X <= 10: ans = N*2 + 2 else: ans = N*2 print(ans)

s716757690

p03910

u879309973

Wrong Answer

The problem set at _CODE FESTIVAL 20XX Finals_ consists of N problems. The score allocated to the i-th (1≦i≦N) problem is i points. Takahashi, a contestant, is trying to score exactly N points. For that, he is deciding which problems to solve. As problems with higher scores are harder, he wants to minimize the highest score of a problem among the ones solved by him. Determine the set of problems that should be solved.

def solve(n): score = 0 res = [] for p in range(1, n+1): res.append(p) score += p if score >= n: break return "\n".join(map(str, res)) n = int(input()) print(solve(n))

s001581030

Accepted

def solve(n): score = 0 for p in range(1, n+1): score += p if score >= n: break x = score - n res = list(range(1,x)) + list(range(x+1,p+1)) return "\n".join(map(str, res)) n = int(input()) print(solve(n))

s524161621

p02578

u186397299

Wrong Answer

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

N = int(input('')) A = input('') A_N = A.split() num = 0 for i in range(1,N): if int(A_N[i]) >= int(A_N[i-1]): num += 0 elif (A_N[i]) < (A_N[i-1]): num += int(A_N[i-1]) - int(A_N[i]) print(num)

s611402676

Accepted

N = int(input('')) A = input('') a_n = A.split() A_N = list(map(int, a_n)) num = 0 for i in range(1,N): if A_N[i] >= A_N[i-1]: num += 0 elif A_N[i] < A_N[i-1]: num += A_N[i-1] - A_N[i] A_N[i] = A_N[i-1] print(num)

s611419958

p03434

u290187182

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.

if __name__ == '__main__': n = int(input()) k = [int(i) for i in input().split()] a = 0 b = 0 k.sort(reverse=True) print(k) flag = True for i in k: if flag == True: a+=i flag=False else: b+=i flag=True print(a-b)

s294369289

Accepted

if __name__ == '__main__': n = int(input()) k = [int(i) for i in input().split()] a = 0 b = 0 k.sort(reverse=True) flag = True for i in k: if flag == True: a+=i flag=False else: b+=i flag=True print(a-b)

s225823716

p03352

u460468647

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.

x = int(input()) lst = [i**2 for i in range(1,32)]+[8,27,125,216,343,512,1000]+[32,243]+[128]+[1001] lst.sort() print(lst) for i in range(len(lst)): if lst[i]<=x<lst[i+1]: print(lst[i])

s240307795

Accepted

x = int(input()) lst = [i**2 for i in range(1,32)]+[8,27,125,216,343,512,1000]+[32,243]+[128]+[1001] lst.sort() for i in range(len(lst)): if lst[i]<=x<lst[i+1]: print(lst[i])

s288906819

p03861

u893931781

Wrong Answer

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

li=list(map(int,input().split())) min=li[0]//li[2] max=li[1]//li[2] count=max-min if li[0]==0: count+=1 print(count)

s717466995

Accepted

li=list(map(int,input().split())) min=li[0]//li[2] max=li[1]//li[2] count=max-min if li[0]%li[2]==0: count+=1 print(count)

s227380923

p04029

u036363750

Wrong Answer

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

#!/usr/bin/env python3 def main(): a = int(input()) if a%2==0: print((a+1)*a/2) else: print((a+1)*a/2+(a+1)/2) if __name__=='__main__':main()

s245796174

Accepted

#!/usr/bin/env python3 def main(): a = int(input()) print(int(((a+1)*a)/2)) if __name__=='__main__':main()

s918277165

p03502

u136869985

Wrong Answer

An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number.

N=input() t=sum(map(int, list(N))) print("Yes" if int(N)==t else "No")

s440117808

Accepted

X = input() fx = 0 for x in X: fx += int(x) if int(X) % fx == 0: print("Yes") else: print("No")

s771034287

p03576

u155687575

Wrong Answer

We have N points in a two-dimensional plane. The coordinates of the i-th point (1 \leq i \leq N) are (x_i,y_i). Let us consider a rectangle whose sides are parallel to the coordinate axes that contains K or more of the N points in its interior. Here, points on the sides of the rectangle are considered to be in the interior. Find the minimum possible area of such a rectangle.

n, k = map(int,input().split()) xs = [] ys = [] points = [] for _ in range(n): x, y = map(int, input().split()) xs.append(x) ys.append(y) points.append((x, y)) for leftx in xs: for rightx in xs: if leftx >= rightx: continue for downy in ys: for upy in ys: if downy >= upy: continue min_ = float('inf') for tx, ty in points: if (leftx <= tx <= rightx) and (downy <= ty <= upy): yoko = rightx - leftx tate = upy - downy men = yoko*tate if men < min_: min_ = men print(min_)

s834248909

Accepted

n, k = map(int,input().split()) ps = [tuple(map(int,input().split())) for i in range(n)] sx = sorted(ps,key = lambda x: x[0]) nx = list(enumerate(sx)) ans = 5e18 for f, (x1,y1) in nx[:n-k+1]: for e, (x2,y2) in nx[f+k-1:]: dx = x2 - x1 sy = sorted(y for x,y in sx[f:e+1]) for y3,y4 in zip(sy,sy[k-1:]): if y3 <= y1 and y4 >= y2: ans = min(ans, dx * (y4 - y3)) print(ans)

s527099700

p03494

u408375121

Wrong Answer

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

N = int(input()) A = list(map(int, input().split())) B = [] for a in A: i = 0 while True: if a % 2 ** i == 0: i += 1 else: B.append(i) break B.sort() print(B[0])

s922104332

Accepted

N = int(input()) A = list(map(int, input().split())) B = [] for a in A: i = 0 while a % 2 ** i == 0: i += 1 B.append(i - 1) B.sort() print(B[0])

s382148318

p03545

u240793404

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.

n = list(input()) for i in range(8): ans = n[0] ans += '+-'[i%2] + n[1] ans += '+-'[i//4] + n[2] ans += '+-'[(i%4)//2] + n[3] if eval(ans) == 7: print(ans) break

s851468430

Accepted

n = list(input()) for i in range(8): ans = n[0] ans += '+-'[i%2] + n[1] ans += '+-'[i//4] + n[2] ans += '+-'[(i%4)//2] + n[3] if eval(ans) == 7: print(ans+"=7") break

s172240036

p03139

u738037409

Wrong Answer

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

N, A, B = map(int, input().split()) print(min(A, B), max(0, N - (A + B)))

s024264408

Accepted

N, A, B = map(int, input().split()) print(min(A, B), max(0, (A + B) - N))

s341488543

p03049

u305422591

Wrong Answer

Snuke has N strings. The i-th string is s_i. Let us concatenate these strings into one string after arranging them in some order. Find the maximum possible number of occurrences of `AB` in the resulting string.

N = int(input()) res = 0 both = 0 head = 0 last = 0 for i in range(N): inp = input() if 'AB' in inp: res += 1 if 'B'==inp[0] and 'A'==inp[-1]: both+=1 elif 'B'==inp[0]: head+=1 elif 'A'==inp[-1]: last += 1 res += max(0,both -1) if head > 0 and last >0: res += 2 + min(head-1,last-1) elif head>0: res += 1 + min(head-1,last) elif last>0: res += 1 + min(head,last-1) print(res)

s374227876

Accepted

n=int(input()) s=[0]*n ct=0 a=0 b=0 c=0 for i in range(n): s[i]=input() for i in range(n): for j in range(len(s[i])): t=s[i] ts=list(t) ts.append(0) if ts[j]=='A' and ts[j+1]=='B': ct+=1 if ts[0]=='B' and ts[len(s[i])-1]!='A': a+=1 if ts[len(s[i])-1]=='A' and ts[0]!='B': b+=1 if ts[0]=='B' and ts[len(s[i])-1]=='A': c+=1 ct += max(0,c -1) if c !=0: if a > 0 and b >0: ct += 2 + min(a-1,b-1) elif a>0: ct += 1 + min(a-1,b) elif b>0: ct += 1 + min(a,b-1) else: ct += min(a,b) print(ct)

s008426649

p03760

u836737505

Time Limit Exceeded

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

o = list(input()) e = list(input()) c = "" for i in range(len(o)): c += o[i] while e: c += e[i] print(c)

s210674291

Accepted

o = list(input()) e = list(input()) c = "" for i in range(len(o)): c += o[i] if i < len(e): c += e[i] print(c)

s205416010

p03759

u962718741

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.

#!/usr/bin/env python3 def main(): a, b, c = map(int, input().split()) print("YES" if b - a == c - a else "NO") if __name__ == "__main__": main()

s772607588

Accepted

#!/usr/bin/env python3 def main(): a, b, c = map(int, input().split()) print("YES" if b - a == c - b else "NO") if __name__ == "__main__": main()

s556354746

p03971

u857673087

Wrong Answer

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

N,A,B = map(int,input().split()) S =input() for i in range(N): if S[i]=='a' and i < A+B: print('Yes') elif S[i]=='b' and i < B: print('Yes') else: print('No')

s327524429

Accepted

N,A,B = map(int,input().split()) S = input() passed = 0 fo = 1 for i in range(N): if(S[i] == 'a'): if(passed < A+B): print('Yes') passed += 1 else: print('No') elif(S[i] == 'b'): if((passed < A+B)and(fo <= B)): print('Yes') fo += 1 passed += 1 else: print('No') elif(S[i] == 'c'): print('No')

s103034875

p02264

u153665391

Wrong Answer

_n_ _q_ _name 1 time1_ _name 2 time2_ ... _name n timen_ In the first line the number of processes _n_ and the quantum _q_ are given separated by a single space. In the following _n_ lines, names and times for the _n_ processes are given. _name i_ and _time i_ are separated by a single space.

N, Q = map(int, input().strip().split()) A = [] D = {} for l in range(N): n, t = input().strip().split() D[n] = int(t) A.append(n) print(A) T = 0 d = 0 while True: n = A.pop(0) if D[n] > Q: D[n] -= Q T += Q A.append(n) else: T += D[n] print('{0} {1}'.format(n, T)) d += 1 if d == N: break

s681534583

Accepted

N, qms = map(int, input().split()) raw_proc = [input() for i in range(N)] queue = [{} for i in range(N+1)] limit = N+1 for i in range(N): name, time = raw_proc[i].split() time = int(time) queue[i] = {"name": name, "time": time} def enque(tail, p): queue[tail] = p tail = (tail+1) % limit return tail def deque(head): proc = queue[head] head = (head+1) % limit return proc, head if __name__ == '__main__': head = 0 tail = N total_time = 0 while head != tail: p, head = deque(head) t = min(qms, p["time"]) p["time"] = p["time"] - t total_time += t if p["time"] > 0: tail = enque(tail, p) else: print("{} {}".format(p["name"], total_time))

s182872341

p03494

u069129582

Time Limit Exceeded

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

n=int(input()) a=list(map(int,input().split())) ans=0 while all(j%2==0 for j in a): for k in range(len(a)): a[k]==a[k]//2 ans+=1 print(ans)

s473686405

Accepted

n=int(input()) A=list(map(int,input().split())) ans=0 while all(a%2==0 for a in A): A=[a/2 for a in A] ans+=1 print(ans)

s903510583

p03455

u883792993

Wrong Answer

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

a,b = map(int, input().split()) print(a) print(b) if a*b/2 == 1: print("Odd") else: print("Even")

s950282297

Accepted

a,b = map(int, input().split()) if a*b%2 == 1: print("Odd") else: print("Even")

s789601449

p03455

u350166773

Wrong Answer

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

a, b = map(int, input().split()) if a%2 == 0 and b%2 == 0: print('Even') else: print('Odd')

s405185905

Accepted

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

s419882634

p03385

u596536048

Wrong Answer

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

print('Yes' if 'abc' in input() else 'No')

s908504857

Accepted

string = input() if string[0] != string[1] and string[1] != string[2] and string[2] != string[0]: print("Yes") else: print("No")

s420584786

p02844

u102275718

Wrong Answer

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

import sys input = sys.stdin.readline import numpy as np from math import factorial import itertools def permutations_count(n, r): return factorial(n) // factorial(n - r) def combinations_count(n, r): return factorial(n) // (factorial(n - r) * factorial(r)) n = int(input()) *s, = map(int, list(input().split()[0])) ans = 0 for i in range(1000): num_str = str(i) if i < 100: num_str = '0'+num_str if i<10: num_str = '0'+num_str idx = 0 for j in s: # print(j, idx) if j == int(s[idx]): idx += 1 if idx==3: break if idx == 3: ans += 1 print(ans)

s884435200

Accepted

import sys input = sys.stdin.readline n = int(input()) s = input().split()[0] ans = 0 for i in range(1000): num_str = str(i).zfill(3) a = s.find(num_str[0]) if a==-1: continue b = s[a+1:].find(num_str[1]) if b==-1: continue b += a+1 c = s[b+1:].find(num_str[2]) if c!=-1: ans += 1 print(ans)

s456823729

p03378

u569776981

Wrong Answer

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

N, M, X = map(int,input().split()) A = list(map(int,input().split())) ans = [] z = 0 y = 0 for i in range(X + 1, N + 1): if i in A: z += 1 print(z) ans.append(z) for i in range(1,X): if i in A: y += 1 print(y) ans.append(y) ans = sorted(ans) print(ans) print(ans[0])

s955543475

Accepted

N, M, X = map(int,input().split()) A = list(map(int,input().split())) ans = [] z = 0 y = 0 for i in range(X + 1, N + 1): if i in A: z += 1 ans.append(z) for i in range(1,X): if i in A: y += 1 ans.append(y) ans = sorted(ans) print(ans[0])

s753131466

p04035

u608088992

Wrong Answer

We have N pieces of ropes, numbered 1 through N. The length of piece i is a_i. At first, for each i (1≤i≤N-1), piece i and piece i+1 are tied at the ends, forming one long rope with N-1 knots. Snuke will try to untie all of the knots by performing the following operation repeatedly: * Choose a (connected) rope with a total length of at least L, then untie one of its knots. Is it possible to untie all of the N-1 knots by properly applying this operation? If the answer is positive, find one possible order to untie the knots.

N, L = map(int, input().split()) A = [int(a) for a in input().split()] CanRemain = [] for i in range(N-1): if A[i] + A[i+1] >= L: CanRemain.append(i+1) if not CanRemain: print("impossible") else: print("possible") for i in range(1, CanRemain[-1]): if i not in CanRemain: print(i) for i in reversed(range(CanRemain[-1]+1, N)): print(i) for num in CanRemain: print(num)

s821157096

Accepted

import sys def solve(): input = sys.stdin.readline N, L = map(int, input().split()) A = [int(a) for a in input().split()] for i in range(N - 1): if A[i] + A[i+1] >= L: print("Possible") for k in range(i): print(k + 1) for k in reversed(range(i + 1, N - 1)): print(k + 1) print(i + 1) break else: print("Impossible") return 0 if __name__ == "__main__": solve()

s020461434

p03447

u959759457

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 i in range(3)] s=X-(((X-A)//B)*B+A) s

s430024806

Accepted

X,A,B=[int(input()) for i in range(3)] s=X-(((X-A)//B)*B+A) print(s)

s405112409

p03862

u714888375

Wrong Answer

There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies. Snuke can perform the following operation any number of times: * Choose a box containing at least one candy, and eat one of the candies in the chosen box. His objective is as follows: * Any two neighboring boxes contain at most x candies in total. Find the minimum number of operations required to achieve the objective.

N, x = list(map(int,input().split())) a = list(map(int,input().split())) result = 0 for i in range(N-1): n = a[i] + a[i+1] - x if n > 0: result += n a[i+1] -= n if a[i+1] < 0: a[i+1] = 0 print(result, a)

s930722673

Accepted

N, x = list(map(int,input().split())) a = list(map(int,input().split())) result = 0 for i in range(N-1): n = a[i] + a[i+1] - x if n > 0: result += n a[i+1] -= n if a[i+1] < 0: a[i+1] = 0 print(result)

s805346340

p03998

u018984506

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 sys import exit s = [input() for i in range(3)] turn = 0 card = s[0][0] sa,sb,sc = 0,0,0 while 1: if card == "a": if sa == len(s[0]): print("A") exit() card = s[0][sa] sa += 1 print(sa,sb,sc) if card == "b": if sb == len(s[1]): print("B") exit() card = s[1][sb] sb += 1 print(sa,sb,sc) if card == "c": if sc == len(s[2]): print("C") exit() card = s[2][sc] sc += 1 print(sa,sb,sc)

s336954114

Accepted

from sys import exit s = [input() for i in range(3)] turn = 0 card = s[0][0] sa,sb,sc = 0,0,0 while 1: if card == "a": if sa == len(s[0]): print("A") exit() card = s[0][sa] sa += 1 if card == "b": if sb == len(s[1]): print("B") exit() card = s[1][sb] sb += 1 if card == "c": if sc == len(s[2]): print("C") exit() card = s[2][sc] sc += 1

s965752644

p03679

u952130512

Wrong Answer

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

x,a,b=input().split() if int(b)-int(a)>=0: print("delicious") elif int(b)-int(a)<=int(x): print("safe") else: print("dangerous")

s641404167

Accepted

x,a,b=input().split() if int(b)-int(a)<=0: print("delicious") elif int(b)-int(a)<=int(x): print("safe") else: print("dangerous")

s576398456

p04044

u036190609

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 =list(map(int, input().split())) array = [] word = "" for i in range(N): s = input() array.append(s) print(array) new_array = sorted(array) print(new_array) for j in range(N): word += new_array[j] print(word)

s710351481

Accepted

N, L =list(map(int, input().split())) array = [] word = "" for i in range(N): s = input() array.append(s) new_array = sorted(array) for j in range(N): word += new_array[j] print(word)

s533977135

p02388

u067975558

Wrong Answer

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

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

s606184503

Accepted

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

s497886287

p03605

u518455500

Wrong Answer

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

N = input() if '9' in N: print("YES") else: print("NO")

s861727352

Accepted

N = input() if '9' in N: print("Yes") else: print("No")

s885045039

p03697

u540631540

Wrong Answer

You are given two integers A and B as the input. Output the value of A + B. However, if A + B is 10 or greater, output `error` instead.

a, b = map(int, input().split()) if a + b > 10: print(a + b) else: print("error")

s854950861

Accepted

a, b = map(int, input().split()) if a + b < 10: print(a + b) else: print("error")

s652250665

p02276

u742013327

Wrong Answer

Quick sort is based on the Divide-and-conquer approach. In QuickSort(A, p, r), first, a procedure Partition(A, p, r) divides an array A[p..r] into two subarrays A[p..q-1] and A[q+1..r] such that each element of A[p..q-1] is less than or equal to A[q], which is, inturn, less than or equal to each element of A[q+1..r]. It also computes the index q. In the conquer processes, the two subarrays A[p..q-1] and A[q+1..r] are sorted by recursive calls of QuickSort(A, p, q-1) and QuickSort(A, q+1, r). Your task is to read a sequence A and perform the Partition 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 Note that, in this algorithm, Partition always selects an element A[r] as a pivot element around which to partition the array A[p..r].

#http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_6_B&lang=jp #partition def partition(target_list): x = target_list[-1] i = -1 for j in range(len(target_list))[:-1]: if target_list[j] <= x: i = i+1 tmp = target_list[i] target_list[i] = target_list[j] target_list[j] = tmp tmp = target_list[-1] target_list[-1] = target_list[i + 1] target_list[i + 1] = tmp return i + 1 def convert(l): return " ".join([str(n) for n in l]) if __name__ == "__main__": n_list = int(input()) target_list = [int(n) for n in input().split()] center_index = partition(target_list) print(convert(target_list[:center_index]) + " ["+ str(target_list[center_index]) + "] " + convert(target_list[center_index:]))

s732194928

Accepted

#http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_6_B&lang=jp #partition def partition(target_list): x = target_list[-1] i = -1 for j in range(len(target_list))[:-1]: if target_list[j] <= x: i = i+1 tmp = target_list[i] target_list[i] = target_list[j] target_list[j] = tmp tmp = target_list[-1] target_list[-1] = target_list[i + 1] target_list[i + 1] = tmp return i + 1 def convert(l): return " ".join([str(n) for n in l]) if __name__ == "__main__": n_list = int(input()) target_list = [int(n) for n in input().split()] center_index = partition(target_list) print(convert(target_list[:center_index]) + " ["+ str(target_list[center_index]) + "] " + convert(target_list[center_index + 1:]))

s201108161

p03548

u503228842

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?

x,y,z =map(int,input().split()) print((x-2*z)//(y+z))

s196137114

Accepted

x,y,z =map(int,input().split()) print((x-z)//(y+z))

s955230340

p03228

u085186789

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 A % 2 == 1: A -= 1 if B % 2 == 1: B -= 1 m = A / 2 n = B / 2 A += n B += m print(A, B)

s017045378

Accepted

A, B, K = map(int,input().split()) for i in range(K): if i % 2 == 0: A = A // 2 B += A else: B = B // 2 A += B print(int(A), int(B))

s956122625

p02255

u647766105

Wrong Answer

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

def insert(array, index): value = array[index] for i in range(index-1, -1, -1): if array[i] < value: array[i] = value break array[i + 1] = array[i] else: array[0] = value def insertion_sort(array, log=False): for i in range(len(array)): insert(array, i) if log: print(*array) if __name__ == "__main__": N = int(input()) A = list(map(int, input().split())) insertion_sort(A, True)

s672162213

Accepted

def insert(array, index): value = array[index] for i in range(index - 1, -1, -1): if array[i] < value: array[i+1] = value break array[i + 1] = array[i] else: array[0] = value def insertion_sort(array, log=False): for i in range(len(array)): insert(array, i) if log: print(*array) if __name__ == "__main__": N = int(input()) A = list(map(int, input().split())) insertion_sort(A, True)

s844459176

p03814

u459697504

Wrong Answer

Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.

#! /usr/bin/python3 # A to Z String s = input() list_s = list(s) test = True def length_AZ(l): len_l = len(l) index_A = [] index_Z = [] for i in range(len_l): if l[i] == 'A': index_A.append(i) elif l[i] == 'Z': index_Z.append(i) if test: print('index_A =', index_A) print('index_Z =', index_Z) return max(index_Z) - min(index_A) + 1 def main(): print(length_AZ(list_s)) if __name__ == '__main__': main()

s786947603

Accepted

#! /usr/bin/python3 # A to Z String s = input() list_s = list(s) test = False def length_AZ(l): len_l = len(l) index_A = [] index_Z = [] for i in range(len_l): if l[i] == 'A': index_A.append(i) elif l[i] == 'Z': index_Z.append(i) if test: print('index_A =', index_A) print('index_Z =', index_Z) return max(index_Z) - min(index_A) + 1 def main(): print(length_AZ(list_s)) if __name__ == '__main__': main()

s875393969

p03854

u383551754

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

s362276180

Accepted

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

s621036917

p03485

u692632484

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()) if (a+b)%2==0 : print((a+b)/2) else: print((a+b)/2+1)

s414844222

Accepted

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

s682576715

p02273

u196653484

Wrong Answer

Write a program which reads an integer _n_ and draws a Koch curve based on recursive calles of depth _n_. The Koch curve is well known as a kind of You should start (0, 0), (100, 0) as the first segment.

class Dot: x=0.0 y=0.0 def __init__(self,x,y): self.x=x self.y=y def __lt__(self,other): return self.x < other.x def print(self): print("{} {}".format(self.x,self.y)) def divide_line_n(a,b,dots,n): dx=abs(a.x-b.x) dy=abs(a.y-b.y) for i in range(1,n): dots.append(Dot((dx*i)/n,(dy*i)/n)) if __name__ == "__main__": dots=[] dots.append(Dot(0,0)) dots.append(Dot(100,0)) n=int(input()) for i in range(n): divide_line_n(dots[i],dots[i+1],dots,3) divide_line_n(dots[-1],dots[-2],dots,2) dots.sort() for i in dots: i.print()

s077169253

Accepted

import math class Dot: x=0.0 y=0.0 def __init__(self,x,y): self.x=x self.y=y def __lt__(self,other): return self.x < other.x def print(self): x=self.x y=self.y print("{:.8f} {:.8f}".format(x,y)) def kock(n,a,b,dots): if n == 0: return s=Dot((2*a.x+b.x)/3,(2*a.y+b.y)/3) t=Dot((a.x+2*b.x)/3,(a.y+2*b.y)/3) rad=math.radians(60) x=(t.x-s.x)*math.cos(rad)-(t.y-s.y)*math.sin(rad)+s.x y=(t.x-s.x)*math.sin(rad)+(t.y-s.y)*math.cos(rad)+s.y u=Dot(x,y) kock(n-1,a,s,dots) s.print() dots.append(s) kock(n-1,s,u,dots) u.print() dots.append(u) kock(n-1,u,t,dots) t.print() dots.append(t) kock(n-1,t,b,dots) if __name__ == "__main__": n=int(input()) dots=[] dot1=Dot(0,0) dot2=Dot(100,0) dots.append(dot1) dots.append(dot2) dot1.print() kock(n,dot1,dot2,dots) dot2.print()

s124760540

p02274

u148628801

Wrong Answer

For a given sequence $A = \\{a_0, a_1, ... a_{n-1}\\}$, the number of pairs $(i, j)$ where $a_i > a_j$ and $i < j$, is called the number of inversions. The number of inversions is equal to the number of swaps of Bubble Sort defined in the following program: bubbleSort(A) cnt = 0 // the number of inversions for i = 0 to A.length-1 for j = A.length-1 downto i+1 if A[j] < A[j-1] swap(A[j], A[j-1]) cnt++ return cnt For the given sequence $A$, print the number of inversions of $A$. Note that you should not use the above program, which brings Time Limit Exceeded.

import sys a_max = 10000000000 def merge_sort(A): n = len(A) num_swap = 0 if n > 1: m = n // 2 num_swap0, A0 = merge_sort(A[0:m]) num_swap1, A1 = merge_sort(A[m:]) num_swap += num_swap0 + num_swap1 n1 = len(A0) A0.append(a_max) A1.append(a_max) i = 0 j = 0 for k in range(n): if A0[i] < A1[j]: A[k] = A0[i] i += 1 else: A[k] = A1[j] num_swap += n1 - i j += 1 return num_swap, A def bubble_sort(A): n = len(A) num_swap = 0 for i in range(n): for j in range(n - 1, i, -1): if A[j] < A[j - 1]: A[j], A[j - 1] = A[j - 1], A[j] num_swap += 1 return num_swap, A #fin = open("test.txt", "r") fin = sys.stdin fin.readline() A = list(map(int, fin.readline().split())) num_swap, B = merge_sort(A) print(num_swap) print(B)

s928562454

Accepted

import sys a_max = 10000000000 def merge_sort(A): n = len(A) num_swap = 0 if n > 1: m = n // 2 num_swap0, A0 = merge_sort(A[0:m]) num_swap1, A1 = merge_sort(A[m:]) num_swap += num_swap0 + num_swap1 n1 = len(A0) A0.append(a_max) A1.append(a_max) i = 0 j = 0 for k in range(n): if A0[i] < A1[j]: A[k] = A0[i] i += 1 else: A[k] = A1[j] num_swap += n1 - i j += 1 return num_swap, A def bubble_sort(A): n = len(A) num_swap = 0 for i in range(n): for j in range(n - 1, i, -1): if A[j] < A[j - 1]: A[j], A[j - 1] = A[j - 1], A[j] num_swap += 1 return num_swap, A #fin = open("test.txt", "r") fin = sys.stdin fin.readline() A = list(map(int, fin.readline().split())) num_swap, B = merge_sort(A) print(num_swap)

s439228915

p00014

u354053070

Wrong Answer

Write a program which computes the area of a shape represented by the following three lines: $y = x^2$ $y = 0$ $x = 600$ It is clear that the area is $72000000$, if you use an integral you learn in high school. On the other hand, we can obtain an approximative area of the shape by adding up areas of many rectangles in the shape as shown in the following figure: $f(x) = x^2$ The approximative area $s$ where the width of the rectangles is $d$ is: area of rectangle where its width is $d$ and height is $f(d)$ $+$ area of rectangle where its width is $d$ and height is $f(2d)$ $+$ area of rectangle where its width is $d$ and height is $f(3d)$ $+$ ... area of rectangle where its width is $d$ and height is $f(600 - d)$ The more we decrease $d$, the higer-precision value which is close to $72000000$ we could obtain. Your program should read the integer $d$ which is a divisor of $600$, and print the area $s$.

import sys for line in sys.stdin: d = int(line) S = 0 for x in range(1, 600, d): S += (x ** 2) * d print(S)

s733353726

Accepted

import sys for line in sys.stdin: d = int(line) S = 0 for x in range(0, 600, d): S += (x ** 2) * d print(S)

s459640225

p03997

u951750034

Wrong Answer

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

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

s354853276

Accepted

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

s461623341

p03860

u557523358

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.

print(''.join(c for c in input() if 'A' <= c <= 'Z'))

s905812000

Accepted

print('A{}C'.format(input()[8]))

s069616537

p03469

u617225232

Wrong Answer

On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.

a = input() print('2018'+a[4:6])

s917020901

Accepted

a = input() print('2018'+a[4:])

s974071764

p03050

u500297289

Wrong Answer

Snuke received a positive integer N from Takahashi. A positive integer m is called a _favorite number_ when the following condition is satisfied: * The quotient and remainder of N divided by m are equal, that is, \lfloor \frac{N}{m} \rfloor = N \bmod m holds. Find all favorite numbers and print the sum of those.

N = int(input()) import math ans = 0 for k in range(1, int(math.sqrt(N) + 1)): if N % k == 0 and k ** 2 != N: ans += N // k - 1 print(ans - 1)

s871970598

Accepted

N = int(input()) import math ans = 0 for k in range(1, int(math.sqrt(N) + 1)): m = N // k - 1 if m != 0: if N % k == 0 and N // m == N % m: ans += m print(ans)

s728873555

p04045

u614984376

Wrong Answer

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

n,k=map(int,input().split()) d=list(map(int,input().split())) print(d) for i in range(n,10001): for x in d: if str(x) in str(i): break else: print(i) break

s212798172

Accepted

n,k=map(int,input().split()) d=list(map(int,input().split())) for i in range(n,100001): for x in d: if str(x) in str(i): break else: print(i) break

s801508744

p03997

u456416458

Wrong Answer

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

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

s618222969

Accepted

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

s012757036

p03997

u416758623

Wrong Answer

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

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

s580163129

Accepted

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

s495703995

p03962

u288948615

Wrong Answer

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

nums = list(input().split()) print(set(nums))

s136777614

Accepted

nums = list(input().split()) print(len(set(nums)))

s974907804

p03399

u345136423

Wrong Answer

You planned a trip using trains and buses. The train fare will be A yen (the currency of Japan) if you buy ordinary tickets along the way, and B yen if you buy an unlimited ticket. Similarly, the bus fare will be C yen if you buy ordinary tickets along the way, and D yen if you buy an unlimited ticket. Find the minimum total fare when the optimal choices are made for trains and buses.

a,b,c,d = [int(input()) for _ in range(4)] print(max(a,b) + max(c,d))

s969588492

Accepted

a,b,c,d = [int(input()) for _ in range(4)] print(min(a,b) + min(c,d))

s679178733

p03485

u519721530

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.

# ABC082_A a, b = map(int, input().split()) print((a+b)/2 if (a+b)%2==0 else (a+b+1)/2)

s588367573

Accepted

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

s719676858

p03712

u642528832

Wrong Answer

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

H,W = map(int,(input().split())) a = [input() for _ in range(H)] print('#'*(W+1)) for i in a: print('#'+str(i)+"#") print('#'*(W+1))

s828780408

Accepted

H,W = map(int,(input().split())) a = [input() for _ in range(H)] print('#'*(W+2)) for i in a: print('#'+str(i)+"#") print('#'*(W+2))

s193658415

p03720

u835482198

Wrong Answer

There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?

n, m = map(int, input().split()) g = [[]] * n for _ in range(m): a, b = map(int, input().split()) a -= 1 b -= 1 g[a] += [b] g[b] += [a] for i in range(n): print(len(set(g[i])))

s761983335

Accepted

n, m = map(int, input().split()) ret = [[] for i in range(n + 1)] for i in range(m): a, b = map(int, input().split()) ret[a] += [b] ret[b] += [a] for i in range(1, n + 1): print(len(ret[i]))

s856429256

p02326

u837811962

Wrong Answer

Given a matrix (H × W) which contains only 1 and 0, find the area of the largest square matrix which only contains 0s.

H,W = map(int,input().split()) tile = [[0 for i in range(W)] for i in range(H)] buf = [[1 for i in range(W)] for i in range(H)] maxsquare = 0 for i in range(H): tile[i] = list(map(int,input().split())) for i in range(1,H): for j in range(1,W): if tile[i-1][j]==1 or tile[i][j-1]==1 or tile[i-1][j-1]==1: buf[i][j] = 1 elif tile[i-1][j]==tile[i][j-1]: buf[i][j] = buf[i-1][j-1]+1 else: buf[i][j] = min(buf[i-1][j] , buf[i][j-1])+1 if buf[i][j] > maxsquare: maxsquare = buf[i][j] for i in buf: print(i) print(maxsquare**2)

s049669724

Accepted

H,W = map(int,input().split()) tile = [[0 for i in range(W)] for i in range(H)] buf = [[0 for i in range(W+1)] for i in range(H+1)] maxsquare = 0 for i in range(H): tile[i] = list(map(int,input().split())) for i in range(1,H+1): for j in range(1,W+1): if tile[i-1][j-1]!=1: buf[i][j] = min(buf[i-1][j],buf[i][j-1],buf[i-1][j-1]) + 1 if buf[i][j] > maxsquare: maxsquare = buf[i][j] print(maxsquare**2)

s147442981

p03352

u342051078

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 sys x = int(input()) mx = 0 for i in range(2,33): if i*i <= x: mx = max(mx,i*i) else : sys.exit(0) print(mx)

s291230544

Accepted

x = int(input()) ans = 1 for i in range(2,41): for j in range(2,11): if x < i**j: break ans = max(ans,i**j) print(ans)

s676906098

p03971

u197300773

Wrong Answer

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

n,a,b=map(int,input().split()) ca,cb=0,0 s=input() for i in range(n): if s[i]=="a": if ca<a+b: print("Yes") ca+=1 elif s[i]=="b": if ca<a+b and cb<b: print("Yes") ca+=1;cb+=1 else: print("No") else: print("No")

s509991100

Accepted

n,a,b=map(int,input().split()) ca,cb=0,0 s=input() for i in range(n): if s[i]=="a": if ca<a+b: print("Yes") ca+=1 else: print("No") elif s[i]=="b": if ca<a+b and cb<b: print("Yes") ca+=1;cb+=1 else: print("No") else: print("No")

s500099403

p02612

u962050670

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)

s974135804

Accepted

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

s630113738

p03608

u029169777

Wrong Answer

There are N towns in the State of Atcoder, connected by M bidirectional roads. The i-th road connects Town A_i and B_i and has a length of C_i. Joisino is visiting R towns in the state, r_1,r_2,..,r_R (not necessarily in this order). She will fly to the first town she visits, and fly back from the last town she visits, but for the rest of the trip she will have to travel by road. If she visits the towns in the order that minimizes the distance traveled by road, what will that distance be?

from itertools import permutations from scipy.sparse.csgraph import dijkstra as di N,M,R=map(int,input().split()) r=list(map(int,input().split())) dp=[[0 for _ in range(N)] for _ in range(N)] for _ in range(M): A,B,C=map(int,input().split()) if dp[A-1][B-1]==0 or dp[A-1][B-1]>C: dp[A-1][B-1]=C dp[B-1][A-1]=C print(dp) distance=di(dp) ans=10**11 for tmp in permutations(r): tmpans=0 for i in range(len(tmp)-1): tmpans+=distance[tmp[i]-1][tmp[i+1]-1] if ans>tmpans: ans=tmpans print(int(ans))

s574354431

Accepted

from itertools import permutations from scipy.sparse.csgraph import dijkstra as di N,M,R=map(int,input().split()) r=list(map(int,input().split())) dp=[[0 for _ in range(N)] for _ in range(N)] for _ in range(M): A,B,C=map(int,input().split()) if dp[A-1][B-1]==0 or dp[A-1][B-1]>C: dp[A-1][B-1]=C dp[B-1][A-1]=C distance=di(dp) ans=10**11 for tmp in permutations(r): tmpans=0 for i in range(len(tmp)-1): tmpans+=distance[tmp[i]-1][tmp[i+1]-1] if ans>tmpans: ans=tmpans print(int(ans))

s922288931

p03007

u159228113

Wrong Answer

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

n = int(input()) x = list(map(int,input().split())) #nmax = max(x) #ans = 2*nmax-sum(x) lst = [] while len(x) > 2: mi = min(x) ma = max(x) lst.append(str(mi)+" "+str(ma)) del x[x.index(min(x))] del x[x.index(max(x))] x.append(mi - ma) print(max(x) - min(x)) for i in range(len(lst)): print(lst[i]) print(max(x),min(x))

s624934844

Accepted

n = int(input()) x = list(map(int,input().split())) #nmax = max(x) #ans = 2*nmax-sum(x) lst = [] nmi = min(x) nma = max(x) del x[x.index(nmi)] del x[x.index(nma)] mainus = [i for i in x if i < 0] plus = [i for i in x if i >= 0] for i in mainus: lst.append(str(nma)+" "+str(i)) nma -= i for i in plus: lst.append(str(nmi)+" "+str(i)) nmi -= i lst.append(str(nma)+" "+str(nmi)) print(nma-nmi) for i in lst: print(i)

s717562808

p02397

u382316013

Wrong Answer

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

while True: x, y = [int(i) for i in input().split()] if x == y == 0: break if x > y: x, y = y, x print(x, y)

s872511289

Accepted

while True: x, y = [int(i) for i in input().split()] if x == y == 0: break if x > y: x, y = y, x print(x, y)

s129188521

p03861

u314089899

Wrong Answer

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

#48b import math a,b,x = map(int, input().split()) first = math.ceil(a/x) end = b//x

s463404539

Accepted

#48b a,b,x = map(int, input().split()) down = a//x if down*x == a: down -= 1 up = b//x #print(up,down) print(up - down)

s005307444

p03547

u449555432

Wrong Answer

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

a=list(input());print('>' if max(a)==a[0] else '<' if max(a)==a[1] else '=')

s437357706

Accepted

a,b=input().split();print('>' if a>b else '<' if a<b else '=')

s529592237

p03816

u759412327

Wrong Answer

Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written. He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept. Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.

N = int(input()) A = sorted(list(map(int,input().split()))) print(len(set(A)))

s202113689

Accepted

N = int(input()) K = len(set(input().split())) print(K+K%2-1)

s672757674

p02613

u410943190

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.

li = [0] * 4 N = int(input()) for i in range(N): result = input() if result == "AC": li[0] += 1 if result == "WA": li[1] += 1 if result == "TLE": li[2] += 1 if result == "RE": li[3] += 1 line = "AC × {0}\nWA × {1}\nTLE × {2}\nRE × {3}".format(li[0], li[1], li[2], li[3]) print(line)

s842865281

Accepted

li = [0] * 4 N = int(input()) for i in range(N): result = input() if result == "AC": li[0] += 1 if result == "WA": li[1] += 1 if result == "TLE": li[2] += 1 if result == "RE": li[3] += 1 line = "AC x {0}\nWA x {1}\nTLE x {2}\nRE x {3}".format(li[0], li[1], li[2], li[3]) print(line)

s360860520

p03479

u023229441

Wrong Answer

As a token of his gratitude, Takahashi has decided to give his mother an integer sequence. The sequence A needs to satisfy the conditions below: * A consists of integers between X and Y (inclusive). * For each 1\leq i \leq |A|-1, A_{i+1} is a multiple of A_i and strictly greater than A_i. Find the maximum possible length of the sequence.

x,y=map(int,input().split()) import bisect as bi ans=1 for i in range(80): if x*2<=y: ans+=1 print(ans)

s802325633

Accepted

x,y=map(int,input().split()) import bisect as bi ans=1 for i in range(80): if x*2<=y: ans+=1 x*=2 print(ans)

s273273129

p04043

u766407523

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.

ABC = list(map(int, input().split())) if ABC in [5, 5, 7]: print('YES') else: print('NO')

s249447901

Accepted

a = list(map(int, input().split())) if sum(a) == 17 and 5 in a and 7 in a: print('YES') else: print('NO')

s198549105

p03861

u252964975

Wrong Answer

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

import math a,b,x=map(int, input().split()) A=math.ceil(a/x) B=math.floor(b/x) print(A-B+1)

s813182614

Accepted

a,b,x=map(int, input().split()) print(b//x-(a-1)//x)

s533540294

p03407

u995914787

Wrong Answer

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

S = str(input()) X = S.split() A = X[0] B = X[1] C = X[2] if (A + B) < C: print("Yes") else: print("No")

s641453460

Accepted

S = str(input()) X = S.split() A = int(X[0]) B = int(X[1]) C = int(X[2]) if C <= (A + B): print("Yes") else: print("No")

s147598151

p04044

u300778480

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.

try: N, L = map(int, input().split()) l = [] for i in range(N): l.append(input()) l_ = sorted(l) print(l_) except EOFError: pass

s666156279

Accepted

try: N, L = map(int, input().split()) l = [] for i in range(N): l.append(input()) l_ = sorted(l) print(''.join(l_)) except EOFError: pass

s965529458

p02612

u382169090

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()) if n%1000==0: print(0) else: print(n%1000)

s228840913

Accepted

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