wrong_submission_id
stringlengths 10
10
| problem_id
stringlengths 6
6
| user_id
stringlengths 10
10
| time_limit
float64 1k
8k
| memory_limit
float64 131k
1.05M
| wrong_status
stringclasses 2
values | wrong_cpu_time
float64 10
40k
| wrong_memory
float64 2.94k
3.37M
| wrong_code_size
int64 1
15.5k
| problem_description
stringlengths 1
4.75k
| wrong_code
stringlengths 1
6.92k
| acc_submission_id
stringlengths 10
10
| acc_status
stringclasses 1
value | acc_cpu_time
float64 10
27.8k
| acc_memory
float64 2.94k
960k
| acc_code_size
int64 19
14.9k
| acc_code
stringlengths 19
14.9k
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s235582470 | p02928 | u363207161 | 2,000 | 1,048,576 | Wrong Answer | 477 | 3,188 | 365 | We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}. Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2. Find the inversion number of B, modulo 10^9 + 7. Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j. | n, k = map(int, input().split())
a = list(map(int, input().split()))
const = pow(10, 9) + 7
c = 0
for i in range(n):
for j in range(n - i - 1):
# print(i, 10 - j - 1)
if a[i] > a[n - j - 1]:
c += 1
print("c", c)
p = len(set(a))
q = p * (p - 1) / 2
print("q", q)
ans = 0 + c * k + q * k * (k - 1) / 2
print(int(ans) % const)
| s266957482 | Accepted | 1,294 | 3,700 | 421 | import copy
n, k = map(int, input().split())
a = list(map(int, input().split()))
const = 1000000007
c = 0
d = 0
for i in range(n):
for j in range(n - i - 1):
if a[i] > a[n - j - 1]:
c += 1
b = copy.deepcopy(a)
for i in range(n):
for j in range(n):
if b[i] > a[j]:
d += 1
# print("c", c)
# print("d", d)
ans = c * k + d * (k * (k - 1) // 2)
print(int(ans) % const)
|
s953911743 | p03795 | u871841829 | 2,000 | 262,144 | Wrong Answer | 25 | 9,080 | 38 | 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) | s755804786 | Accepted | 25 | 9,004 | 39 | N=int(input())
print(N*800 - N//15*200) |
s187027017 | p02647 | u692453235 | 2,000 | 1,048,576 | Wrong Answer | 2,206 | 51,488 | 435 | We have N bulbs arranged on a number line, numbered 1 to N from left to right. Bulb i is at coordinate i. Each bulb has a non-negative integer parameter called intensity. When there is a bulb of intensity d at coordinate x, the bulb illuminates the segment from coordinate x-d-0.5 to x+d+0.5. Initially, the intensity of Bulb i is A_i. We will now do the following operation K times in a row: * For each integer i between 1 and N (inclusive), let B_i be the number of bulbs illuminating coordinate i. Then, change the intensity of each bulb i to B_i. Find the intensity of each bulb after the K operations. | import numpy as np
N, K = map(int, input().split())
A = np.array(list(map(int, input().split())))
#A = np.array([0] * N)
#print(A)
cnt = 0
for _ in range(K):
B = np.array([0] * N)
stop = 0
for i in range(N):
p = A[i]
L = max(i-p, 0)
R = min(i+p, N-1)
for j in range(L, R+1):
B[j] += 1
if B[j] == N:
stop += 1
cnt += 1
A = B
#print(A)
if stop == N:
#print(cnt)
break
print(A) | s638259020 | Accepted | 913 | 123,384 | 397 | import numpy as np
N, K = map(int, input().split())
A = np.array(list(map(int, input().split())))
from numba import jit
@jit
def imos(x):
global N
B = np.zeros_like(x)
for i in range(N):
p = x[i]
L = max(i-p, 0)
R = min(i+p, N-1)
B[L] += 1
if R < N-1:
B[R+1] -= 1
B = np.cumsum(B)
return B
for i in range(K):
if i >= 50:
break
A = imos(A)
print(*A) |
s818990826 | p02645 | u920103253 | 2,000 | 1,048,576 | Wrong Answer | 21 | 9,076 | 23 | When you asked some guy in your class his name, he called himself S, where S is a string of length between 3 and 20 (inclusive) consisting of lowercase English letters. You have decided to choose some three consecutive characters from S and make it his nickname. Print a string that is a valid nickname for him. | s=input()
print(s[:4]) | s514330681 | Accepted | 22 | 8,944 | 23 | s=input()
print(s[:3]) |
s785889612 | p03485 | u093033848 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 118 | 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(round(((a + b) / 2) + 1)) | s639142321 | Accepted | 17 | 2,940 | 113 | a, b = map(int, input().split())
if (a + b) % 2 == 0:
print((a + b) // 2)
else:
print(((a + b) // 2 + 1)) |
s082309219 | p02255 | u231136358 | 1,000 | 131,072 | Wrong Answer | 20 | 7,344 | 287 | 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. | A = input().split()
N = A.pop()
# print(N)
# print(A)
# A = [5, 2, 4, 6, 1, 3]
print(' '.join(map(str, A)))
for i in range(1, len(A)):
key = A[i]
j = i - 1
while j >= 0 and A[j] > key:
A[j+1] = A[j]
j = j - 1
A[j+1] = key
print(' '.join(map(str, A))) | s227897178 | Accepted | 40 | 7,728 | 286 | N = int(input())
A = [int(s) for s in input().split()]
print(' '.join([str(item) for item in A]))
for i in range(1, N):
key = A[i]
j = i - 1
while j >= 0 and A[j] > key:
A[j+1] = A[j]
j = j - 1
A[j+1] = key
print(' '.join([str(item) for item in A])) |
s358271119 | p03583 | u801049006 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 225 | You are given an integer N. Find a triple of positive integers h, n and w such that 4/N = 1/h + 1/n + 1/w. If there are multiple solutions, any of them will be accepted. | N = int(input())
for h in range(1, 3501):
for n in range(1, 3501):
if 2 * h * n - N:
w = (h + n) // (2 * h * n - N)
if 0 < w <= 3500:
print(h, n, w)
exit()
| s779986166 | Accepted | 1,727 | 2,940 | 249 | N = int(input())
for h in range(1, 3501):
for n in range(h, 3501):
if 4 * h * n - N * n - N * h > 0 and N * h * n % (4 * h * n - N * n - N * h) == 0:
print(h, n, N * h * n // (4 * h * n - N * n - N * h))
exit()
|
s249359306 | p02393 | u613534067 | 1,000 | 131,072 | Wrong Answer | 20 | 5,576 | 52 | Write a program which reads three integers, and prints them in ascending order. | a = list(map(int, input().split()))
print(a.sort())
| s795276308 | Accepted | 20 | 5,584 | 74 | a = list(map(int, input().split()))
a.sort()
print(" ".join(map(str, a)))
|
s074782433 | p03351 | u021916304 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,064 | 317 | 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. | def ii():return int(input())
def iim():return map(int,input().split())
def iil():return list(map(int,input().split()))
def ism():return map(str,input().split())
def isl():return list(map(str,input().split()))
a,b,c,d = iim()
if abs(a-c) < d or (abs(a-b) < d and abs(b-c) < d):
print('Yes')
else:
print('No') | s646625426 | Accepted | 17 | 3,060 | 153 | def iim():return map(int,input().split())
a,b,c,d = iim()
if abs(a-c) <= d or (abs(a-b) <= d and abs(b-c) <= d):
print('Yes')
else:
print('No') |
s647735185 | p03251 | u242031676 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 147 | Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out. | a, x, y = [list(map(int, input().split()))for i in range(3)]
if max(a[2], sorted(x)[-1])<min(a[3],sorted(y)[0]): print("No war")
else: print("War") | s641459310 | Accepted | 18 | 2,940 | 147 | a, x, y = [list(map(int, input().split()))for i in range(3)]
if max(a[2], sorted(x)[-1])<min(a[3],sorted(y)[0]): print("No War")
else: print("War") |
s122621846 | p03637 | u427344224 | 2,000 | 262,144 | Wrong Answer | 57 | 14,252 | 398 | We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective. | N = int(input())
a_list = list(map(int, input().split()))
multi4 = len([a for a in a_list if a % 4 == 0])
odd_num = len([a for a in a_list if a % 2 != 0])
even_num = len(a_list) - odd_num
not4 = even_num - multi4
if odd_num <= multi4:
if not4 >= 1:
if not4 % 2 == 0:
print("Yes")
else:
print("No")
else:
print("Yes")
else:
print("No")
| s190027648 | Accepted | 57 | 15,020 | 394 | N = int(input())
a_list = list(map(int, input().split()))
multi4 = len([a for a in a_list if a % 4 == 0])
odd_num = len([a for a in a_list if a % 2 != 0])
even_num = len(a_list) - odd_num
not4 = even_num - multi4
if not4 >0 :
if odd_num <= multi4:
print("Yes")
else:
print("No")
else:
if odd_num <= multi4 + 1:
print("Yes")
else:
print("No")
|
s049285553 | p03369 | u843318346 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 35 | In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen. | s = input()
print(700+s.count('o')) | s781346810 | Accepted | 17 | 2,940 | 40 | s = input()
print(700+s.count('o')*100)
|
s969286476 | p03494 | u702786238 | 2,000 | 262,144 | Time Limit Exceeded | 2,108 | 9,328 | 99 | 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. | lis = list(map(int, input().split()))
count = 0
while(1):
if sum(lis) % 2 == 1:
print(count) | s096519124 | Accepted | 150 | 12,468 | 212 | import numpy as np
N = int(input())
lis = np.array(list(map(int, input().split())))
count = 0
while(1):
if np.mod(lis, 2).sum() != 0:
print(count)
break
else:
lis = lis // 2
count = count + 1 |
s213940868 | p03992 | u766407523 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 38 | This contest is `CODE FESTIVAL`. However, Mr. Takahashi always writes it `CODEFESTIVAL`, omitting the single space between `CODE` and `FESTIVAL`. So he has decided to make a program that puts the single space he omitted. You are given a string s with 12 letters. Output the string putting a single space between the first 4 letters and last 8 letters in the string s. | S = input()
print(S[:5] + ' ' + S[5:]) | s937356107 | Accepted | 17 | 2,940 | 38 | S = input()
print(S[:4] + ' ' + S[4:]) |
s723039490 | p02821 | u311379832 | 2,000 | 1,048,576 | Wrong Answer | 1,186 | 18,460 | 691 | Takahashi has come to a party as a special guest. There are N ordinary guests at the party. The i-th ordinary guest has a _power_ of A_i. Takahashi has decided to perform M _handshakes_ to increase the _happiness_ of the party (let the current happiness be 0). A handshake will be performed as follows: * Takahashi chooses one (ordinary) guest x for his left hand and another guest y for his right hand (x and y can be the same). * Then, he shakes the left hand of Guest x and the right hand of Guest y simultaneously to increase the happiness by A_x+A_y. However, Takahashi should not perform the same handshake more than once. Formally, the following condition must hold: * Assume that, in the k-th handshake, Takahashi shakes the left hand of Guest x_k and the right hand of Guest y_k. Then, there is no pair p, q (1 \leq p < q \leq M) such that (x_p,y_p)=(x_q,y_q). What is the maximum possible happiness after M handshakes? | import bisect
from itertools import accumulate
def func(x):
cnt = 0
for i in a:
index = bisect.bisect_left(a, x - i)
cnt += N - index
if cnt >= M:
return True
else:
return False
N, M = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
ar = sorted(a, reverse=True)
b = [0] + list(accumulate(ar))
print(ar)
print(b)
MIN = 0
MAX = 2 * 10 ** 5 + 1
while MAX - MIN > 1:
MID = (MIN + MAX) // 2
if func(MID):
MIN = MID
else:
MAX = MID
ans = 0
cnt = 0
for i in ar:
index = bisect.bisect_left(a, MIN - i)
ans += i * (N - index) + b[N - index]
cnt += N - index
print(ans - (cnt - M) * MIN) | s574109062 | Accepted | 1,176 | 14,256 | 673 | import bisect
from itertools import accumulate
def func(x):
cnt = 0
for i in a:
index = bisect.bisect_left(a, x - i)
cnt += N - index
if cnt >= M:
return True
else:
return False
N, M = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
ar = sorted(a, reverse=True)
b = [0] + list(accumulate(ar))
MIN = 0
MAX = 2 * 10 ** 5 + 1
while MAX - MIN > 1:
MID = (MIN + MAX) // 2
if func(MID):
MIN = MID
else:
MAX = MID
ans = 0
cnt = 0
for i in ar:
index = bisect.bisect_left(a, MIN - i)
ans += i * (N - index) + b[N - index]
cnt += N - index
print(ans - (cnt - M) * MIN) |
s402773360 | p03471 | u974935538 | 2,000 | 262,144 | Wrong Answer | 2,104 | 3,064 | 360 | The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough. | N,Y = map(int,input().split())
a = 0
b = 0
c = 0
for i in range(N):
for j in range(N-i):
for k in range(N-i-j):
if 10000*i+5000*j+1000*k == Y:
a=i
b=j
c=k
break
if(a>0)|(b>0)|(c>0):
print(a,b,c)
elif (a==0)&(b ==0)&(c == 0):
print(-1,-1,-1) | s746929827 | Accepted | 754 | 3,064 | 321 | N,Y = map(int,input().split())
a = 0
b = 0
c = 0
for i in range(N+1):
for j in range(N-i+1):
if 10000*i+5000*j+1000*(N-i-j) == Y:
a=i
b=j
c=N-i-j
if(a>0)|(b>0)|(c>0):
print(a,b,c)
elif (a==0)&(b ==0)&(c == 0):
print(-1,-1,-1) |
s227882569 | p03089 | u631277801 | 2,000 | 1,048,576 | Wrong Answer | 21 | 3,316 | 1,164 | Snuke has an empty sequence a. He will perform N operations on this sequence. In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1). You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it. | import sys
stdin = sys.stdin
sys.setrecursionlimit(10 ** 7)
def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x) - 1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())
from collections import defaultdict
from operator import itemgetter
n = ni()
b = list(li())
lower = [0]*n
upper = [0]*n
filled = [False]*n
a = [0]*n
nums = defaultdict(int)
for i, bi in enumerate(b[::-1]):
lower[n-i-1] = bi + nums[bi] - 1
nums[bi] += 1
for i, bi in enumerate(b):
upper[i] = n - (i + 1 - bi) - 1
blu = [(bi, lowi, upi) for bi, lowi, upi in zip(b, lower, upper)]
blu.sort(key=itemgetter(2))
for bi, lowi, upi in blu:
for pos in range(lowi, upi+1):
if 0 > pos or pos >= n:
break
if not filled[pos]:
a[pos] = bi
filled[pos] = True
break
if not all(filled):
print("No")
else:
print("Yes")
print("\n".join(map(str, a)))
| s704914891 | Accepted | 18 | 3,064 | 975 | import sys
stdin = sys.stdin
sys.setrecursionlimit(10 ** 7)
def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x) - 1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())
n = ni()
b = list(li())
num_ofst = [[bi, (i+1) - bi] for i, bi in enumerate(b)]
ans = []
ok = True
while ok and num_ofst:
cur = -1
cur_idx = -1
for i in range(len(num_ofst)-1, -1, -1):
if num_ofst[i][1] == 0:
cur = num_ofst[i][0]
cur_idx = i
break
else:
num_ofst[i][1] -= 1
if cur != -1:
num_ofst = num_ofst[:i] + num_ofst[i+1:]
ans.append(cur)
else:
ok = False
if ok:
print("\n".join(map(str, ans[::-1])))
else:
print(-1) |
s455311523 | p04029 | u121732701 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 84 | There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total? | N = int(input())
count = 0
for i in range(N+1):
count += i
print(i)
| s763948227 | Accepted | 18 | 2,940 | 90 | N = int(input())
count = 0
for i in range(N+1):
count += i
print(count)
|
s776100725 | p03110 | u848654125 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 235 | Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total? | N = int(input())
gift = []
for i in range(N):
gift.append((input().split()))
sum = 0
for i in range(N):
if gift[i][1] == "JPN":
sum += float(gift[i][0])
else:
sum += float(gift[i][0]) * 380000
print(sum)
| s769325679 | Accepted | 17 | 2,940 | 235 | N = int(input())
gift = []
for i in range(N):
gift.append((input().split()))
sum = 0
for i in range(N):
if gift[i][1] == "JPY":
sum += float(gift[i][0])
else:
sum += float(gift[i][0]) * 380000
print(sum)
|
s763776843 | p03624 | u351914601 | 2,000 | 262,144 | Wrong Answer | 20 | 3,188 | 160 | You are given a string S consisting of lowercase English letters. Find the lexicographically (alphabetically) smallest lowercase English letter that does not occur in S. If every lowercase English letter occurs in S, print `None` instead. | s = sorted(set(input()))
print(*s)
ans = "None"
for i in range(len(s)):
if ord(s[i]) != ord("a")+i:
ans = chr(ord(s[i])-1)
break
print(ans)
| s358354850 | Accepted | 19 | 3,188 | 122 | letter = "abcdefghijklmnopqrstuvwxyz"
ans = sorted(set(letter) ^ set(input()))
print(ans[0] if len(ans) != 0 else "None")
|
s651951184 | p02257 | u500396695 | 1,000 | 131,072 | Wrong Answer | 20 | 7,540 | 339 | A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7. Write a program which reads a list of _N_ integers and prints the number of prime numbers in the list. | N = int(input())
A=[]
for i in range(N):
A.append(int(input()))
counter = 0
for i in range(0, N):
j = A[i] - 2
if j == 0 or j == 1:
counter += 1
else:
while j > 1:
if i % j == 0:
break
else:
j -= 1
else:
counter += 1
print(counter) | s670848956 | Accepted | 770 | 8,072 | 440 | def isprime(x):
import math
if x == 2:
return True
elif x < 2 or x % 2 == 0:
return False
else:
i = 3
while i <= math.sqrt(x):
if x % i == 0:
return False
i += 2
return True
N = int(input())
A=[]
for i in range(N):
A.append(int(input()))
counter = 0
for i in range(0, N):
if isprime(A[i]):
counter += 1
print(counter) |
s488377792 | p02392 | u298413451 | 1,000 | 131,072 | Wrong Answer | 20 | 7,612 | 163 | Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No". | # coding: utf-8
# Your code here!
s = input().rstrip().split(' ')
a = int(s[0])
b = int(s[1])
c = int(s[2])
if a < b < c:
print("yes")
else:
print("no") | s391990409 | Accepted | 20 | 7,692 | 163 | # coding: utf-8
# Your code here!
s = input().rstrip().split(' ')
a = int(s[0])
b = int(s[1])
c = int(s[2])
if a < b < c:
print("Yes")
else:
print("No") |
s300802255 | p04012 | u448659077 | 2,000 | 262,144 | Wrong Answer | 21 | 3,316 | 213 | Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful. | #ABC044
import collections
w = input()
a = list(collections.Counter(w).values())
b = [a[i] for i in range(len(a)) if int(a[i]) % 2 == 0]
print(a,b)
if len(a) == len(b):
print("Yes")
else:
print("No")
| s726814228 | Accepted | 21 | 3,316 | 203 | #ABC044
import collections
w = input()
a = list(collections.Counter(w).values())
b = [a[i] for i in range(len(a)) if int(a[i]) % 2 == 0]
if len(a) == len(b):
print("Yes")
else:
print("No")
|
s118142694 | p03546 | u506287026 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 9 | Joisino the magical girl has decided to turn every single digit that exists on this world into 1. Rewriting a digit i with j (0≤i,j≤9) costs c_{i,j} MP (Magic Points). She is now standing before a wall. The wall is divided into HW squares in H rows and W columns, and at least one square contains a digit between 0 and 9 (inclusive). You are given A_{i,j} that describes the square at the i-th row from the top and j-th column from the left, as follows: * If A_{i,j}≠-1, the square contains a digit A_{i,j}. * If A_{i,j}=-1, the square does not contain a digit. Find the minimum total amount of MP required to turn every digit on this wall into 1 in the end. | s079869073 | Accepted | 33 | 3,316 | 373 | h, _ = map(int, input().split())
d = [list(map(int, input().split())) for _ in range(10)]
A = [list(map(int, input().split())) for _ in range(h)]
N = 10
for k in range(N):
for i in range(N):
for j in range(N):
d[i][j] = min(d[i][j], d[i][k]+d[k][j])
ans = 0
for aa in A:
for a in aa:
if a != -1:
ans += d[a][1]
print(ans)
|
|
s469272006 | p02398 | u731896389 | 1,000 | 131,072 | Wrong Answer | 20 | 7,652 | 122 | Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b. | a,b,c = map(int,input().split())
count=0
for i in range(a,b+1):
if i%c==0:
count+=1
print(count)
| s776146102 | Accepted | 20 | 7,660 | 105 | a,b,c=map(int,input().split())
cnt=0
for i in range(a,b+1):
if c%i == 0:
cnt += 1
print(cnt) |
s483821783 | p03861 | u586639900 | 2,000 | 262,144 | Wrong Answer | 30 | 9,140 | 142 | You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x? | a, b, x = map(int, input().split())
p = a // x
q = b // x
res = p - q
print(res)
| s152683334 | Accepted | 29 | 9,160 | 153 | a, b, x = map(int, input().split())
p = a // x
r = a % x
q = b // x
s = b % x
res = q - p
if r == 0:
res += 1
print(res) |
s090870806 | p03657 | u982591663 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 133 | Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies. | A, B = map(int, input().split())
if A % 3 == 0 or B % 3 == 0 or (A+B) & 3 == 0:
print("Possible")
else:
print("Impossible")
| s582494500 | Accepted | 17 | 2,940 | 133 | A, B = map(int, input().split())
if A % 3 == 0 or B % 3 == 0 or (A+B) % 3 == 0:
print("Possible")
else:
print("Impossible")
|
s261207964 | p04012 | u656120612 | 2,000 | 262,144 | Wrong Answer | 28 | 9,000 | 189 | Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful. | flag=True
a=list(input())
a.sort
while len(a)>0 and flag:
n=a.count(a[0])
if n%2==0:
a=a[n:]
else:
flag=False
if flag:
print("Yes")
else:
print("No") | s606693631 | Accepted | 26 | 8,896 | 194 | flag=True
a=list(input())
a=sorted(a)
while len(a)>0 and flag:
n=a.count(a[0])
if n%2==0:
a=a[n:]
else:
flag=False
if flag:
print("Yes")
else:
print("No") |
s002391650 | p03730 | u002459665 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 360 | 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`. | def func():
pass
def main():
a, b, c = map(int, input().split())
l = []
i = 1
while True:
x = a * i
left = x % b
if left == c:
print("Yes")
return
if left in l:
print("No")
return
l.append(left)
i += 1
if __name__ == '__main__':
main()
| s641168955 | Accepted | 19 | 3,060 | 360 | def func():
pass
def main():
a, b, c = map(int, input().split())
l = []
i = 1
while True:
x = a * i
left = x % b
if left == c:
print("YES")
return
if left in l:
print("NO")
return
l.append(left)
i += 1
if __name__ == '__main__':
main()
|
s577560627 | p03471 | u136395536 | 2,000 | 262,144 | Wrong Answer | 1,864 | 3,064 | 287 | The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough. | N,Y = (int(i) for i in input().split())
x = -1
y = -1
z = -1
for i in range(N+1):
for j in range(N+1):
money = Y - 10000*i - 5000*j - 1000*(N-i-j)
if (N-i-j)==0 and money == 0:
x = i
y = j
z = N-i-j
break
print(x,y,z) | s679236224 | Accepted | 1,891 | 3,064 | 287 | N,Y = (int(i) for i in input().split())
x = -1
y = -1
z = -1
for i in range(N+1):
for j in range(N+1):
money = Y - 10000*i - 5000*j - 1000*(N-i-j)
if (N-i-j)>=0 and money == 0:
x = i
y = j
z = N-i-j
break
print(x,y,z) |
s271058092 | p02239 | u659034691 | 1,000 | 131,072 | Wrong Answer | 30 | 7,724 | 594 | Write a program which reads an directed graph $G = (V, E)$, and finds the shortest distance from vertex $1$ to each vertex (the number of edges in the shortest path). Vertices are identified by IDs $1, 2, ... n$. | # your code goes here
#Breadth First Search
n=int(input())
V=[]
a=[]
for i in range(n):
V.append([int(i) for i in input().split()])
a.append(0)
V=sorted(V,key=lambda x:x[0])
print('1 0')
d=1
c=0
w=[0]
while c<n and w!=[]:
i=0
y=[]
# print(c)
while c<n and i<len(w):
j=2
while j<2+V[w[i]][1] and c<n:
if a[V[w[i]][j]-1]==0:
# print(a)
a[V[w[i]][j]-1]=1
print(str(V[w[i]][j])+' '+str(d))
c+=1
y.append(V[w[i]][j]-1)
j+=1
i+=1
w=y
d+=1 | s009791814 | Accepted | 20 | 7,828 | 807 | #Breadth First Search
n=int(input())
V=[]
a=[1]
for i in range(n):
V.append([int(i) for i in input().split()])
a.append(0)
V=sorted(V,key=lambda x:x[0])
print('1 0')
d=1
c=0
w=[0]
p=[]
while c<n and w!=[]:
i=0
y=[]
# print(c)
while c<n and i<len(w):
j=2
while j<2+V[w[i]][1] and c<n:
if a[V[w[i]][j]-1]==0:
# print(a)
a[V[w[i]][j]-1]=1
p.append([V[w[i]][j],d])
c+=1
y.append(V[w[i]][j]-1)
j+=1
i+=1
w=y
d+=1
p=sorted(p, key=lambda x:x[0])
#print(p)
i=2
j=0
while i <=n and j<len(p):
while p[j][0]>i:
print(str(i) +" -1")
i+=1
print(str(p[j][0])+" "+str(p[j][1]))
i+=1
j+=1
while i<=n:
print(str(i)+" -1")
i+=1 |
s098597689 | p03759 | u636290142 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 94 | Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful. | a, b, c = map(int, input().split())
if a - b == c - b:
print('YES')
else:
print('NO')
| s788760198 | Accepted | 17 | 2,940 | 94 | a, b, c = map(int, input().split())
if b - a == c - b:
print('YES')
else:
print('NO')
|
s918653470 | p03407 | u894694822 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 79 | An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan. | a, b, c=map(int, input().split())
if a+b<=c:
print("Yes")
else:
print("No") | s603517285 | Accepted | 17 | 2,940 | 80 | a, b, c=map(int, input().split())
if a+b>=c:
print("Yes")
else:
print("No")
|
s551141753 | p02678 | u212328220 | 2,000 | 1,048,576 | Wrong Answer | 1,254 | 33,844 | 1,139 | There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists. | N, M = map(int,input().split())
graph = [[] for n in range(N+1)]
for _ in range(M):
a,b = map(int,input().split())
graph[a].append(b)
graph[b].append(a)
q = [1]
ans = ['No' for n in range(N+1)]
ans[1] = 1 #ans[1]
while q:
now = q.pop(0)
for edge in graph[now]:
if ans[edge] != 'No':
continue
ans[edge] = now
q.append(edge)
if not 'No' in ans:
print('Yes')
for n in range(2,N+1):
print(ans[n])
else:
print('No')
| s621137797 | Accepted | 1,452 | 33,836 | 1,385 | N, M = map(int, input().split())
graph = [[] for n in range(N + 1)]
for _ in range(M):
a, b = map(int, input().split())
graph[a].append(b)
graph[b].append(a)
q = [1]
ans = ['No' for n in range(N + 1)]
ans[0] = 1
ans[1] = 1
while q:
now = q.pop(0)
for edge in graph[now]:
if ans[edge] != 'No':
continue
ans[edge] = now
q.append(edge)
if not 'No' in ans:
print('Yes')
for n in range(2, N + 1):
print(ans[n])
else:
print('No')
|
s129354051 | p03228 | u750278664 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 383 | In the beginning, Takahashi has A cookies, and Aoki has B cookies. They will perform the following operation alternately, starting from Takahashi: * If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other person. Find the numbers of cookies Takahashi and Aoki respectively have after performing K operations in total. | # -*- coding: utf-8 -*-
A, B, K = map(int, input().split())
print(A, B, K)
for i in range(K):
if i % 2 == 0:
if A % 2 != 0:
A -= 1
A //= 2
B += A
else:
if B % 2 != 0:
B -= 1
B //= 2
A += B
print("{} {}".format(A, B))
| s445386264 | Accepted | 17 | 2,940 | 368 | # -*- coding: utf-8 -*-
A, B, K = map(int, input().split())
for i in range(K):
if i % 2 == 0:
if A % 2 != 0:
A -= 1
A //= 2
B += A
else:
if B % 2 != 0:
B -= 1
B //= 2
A += B
print("{} {}".format(A, B))
|
s592237267 | p02393 | u648117624 | 1,000 | 131,072 | Wrong Answer | 20 | 5,596 | 140 | Write a program which reads three integers, and prints them in ascending order. | #coding: UTF-8
#a,b,c=[int(x) for x in input().split()]
a,b,c=[int(x) for x in input().split()]
numlist = [a,b,c]
print(numlist.sort())
| s886249449 | Accepted | 20 | 5,576 | 86 |
print(*sorted(map(int, input().split())))
|
s933358991 | p02259 | u285980122 | 1,000 | 131,072 | Wrong Answer | 20 | 5,596 | 545 | Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode. | def bubbleSort(A, N):
print(A)
A = list(map(int, A.split()))
flag = 1
count = 0
while flag:
flag = 0
for j in range(N-1, 0, -1):
if A[j] < A[j-1]:
tmp = A[j]
A[j] = A[j-1]
A[j-1] = tmp
flag = 1
count += 1
print(count)
N = int(input())
A = input()
bubbleSort(A, N)
| s613040270 | Accepted | 20 | 5,600 | 564 | def bubbleSort(A, N):
A = list(map(int, A.split()))
flag = 1
count = 0
while flag:
flag = 0
for j in range(N-1, 0, -1):
if A[j] < A[j-1]:
tmp = A[j]
A[j] = A[j-1]
A[j-1] = tmp
flag = 1
count += 1
print(" ".join(map(str,A)))
print(count)
N = int(input())
A = input()
bubbleSort(A, N)
|
s215868105 | p03712 | u887207211 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 216 | 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)]
result = []
for a in A:
result.append('*' + a +'*')
waku = '*' * (W+2)
result.insert(0,waku)
result.insert(len(A)+1,waku)
for r in result:
print(r) | s192552617 | Accepted | 17 | 3,060 | 163 | H, W = map(int,input().split())
result = ["#"*(W+2)]
for _ in range(H):
c = input()
result.append("#"+c+"#")
result.append("#"*(W+2))
print("\n".join(result)) |
s613720979 | p02610 | u994935583 | 2,000 | 1,048,576 | Wrong Answer | 28 | 9,168 | 962 | We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row. The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise. Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel. Solve this problem for each of the T test cases given. |
from heapq import heappop, heappush
import sys
input = sys.stdin.readline
def resolve():
T = int(input())
for _ in range(T):
N = int(input())
ans = 0
camels0 = [[] for _ in range (n)]
camels1 = [[] for _ in range (n)]
for _ in range(N):
K,L,R = map(int, input().split())
if L>R:
camels0[K-1].append(L-R)
elif L<R and K<N:
camels1[N-K-1].append(R-L)
ans += min(L,R)
H = []
for i in range(N):
for j in camels0[i]:
heappush(H,j)
while len(H) > i+1:
heappop(H)
ans += sum(H)
H = []
for i in range(N):
for j in camels1[i]:
heappush(H,j)
while len(H) > i+1:
heappop(H)
ans += sum(H)
print(ans) | s244885776 | Accepted | 450 | 47,444 | 972 |
from heapq import heappop, heappush
import sys
input = sys.stdin.readline
def resolve():
T = int(input())
for _ in range(T):
N = int(input())
ans = 0
camels0 = [[] for _ in range (N)]
camels1 = [[] for _ in range (N)]
for _ in range(N):
K,L,R = map(int, input().split())
if L>R:
camels0[K-1].append(L-R)
elif L<R and K<N:
camels1[N-K-1].append(R-L)
ans += min(L,R)
H = []
for i in range(N):
for j in camels0[i]:
heappush(H,j)
while len(H) > i+1:
heappop(H)
ans += sum(H)
H = []
for i in range(N):
for j in camels1[i]:
heappush(H,j)
while len(H) > i+1:
heappop(H)
ans += sum(H)
print(ans)
resolve() |
s829035075 | p02613 | u734197299 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,128 | 173 | Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format. | N=input()
c0=N.count("AC")
c1=N.count("WA")
c2=N.count("TLE")
c3=N.count("RE")
print("AC x "+str(c0))
print("WA x "+str(c1))
print("TLE x "+str(c2))
print("RE x "+str(c3)) | s886788922 | Accepted | 150 | 16,288 | 239 | N=int(input())
inp=[]
for i in range(N):
inp.append((input()))
c0=inp.count("AC")
c1=inp.count("WA")
c2=inp.count("TLE")
c3=inp.count("RE")
print("AC x "+str(c0))
print("WA x "+str(c1))
print("TLE x "+str(c2))
print("RE x "+str(c3)) |
s115559096 | p03546 | u177411511 | 2,000 | 262,144 | Wrong Answer | 203 | 38,376 | 932 | Joisino the magical girl has decided to turn every single digit that exists on this world into 1. Rewriting a digit i with j (0≤i,j≤9) costs c_{i,j} MP (Magic Points). She is now standing before a wall. The wall is divided into HW squares in H rows and W columns, and at least one square contains a digit between 0 and 9 (inclusive). You are given A_{i,j} that describes the square at the i-th row from the top and j-th column from the left, as follows: * If A_{i,j}≠-1, the square contains a digit A_{i,j}. * If A_{i,j}=-1, the square does not contain a digit. Find the minimum total amount of MP required to turn every digit on this wall into 1 in the end. | import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time, copy,bisect
from operator import itemgetter
from heapq import heappush, heappop
import numpy as np
from scipy.sparse.csgraph import shortest_path, floyd_warshall, dijkstra, bellman_ford, johnson
from scipy.sparse import csr_matrix
sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7
stdin = sys.stdin
ni = lambda: int(ns())
nf = lambda: float(ns())
na = lambda: list(map(int, stdin.readline().split()))
nb = lambda: list(map(float, stdin.readline().split()))
ns = lambda: stdin.readline().rstrip() # ignore trailing spaces
H, W = na()
c = [na() for _ in range(10)]
A = [na() for _ in range(H)]
adj = [[] for _ in range(10)]
c = np.array(c)
d = shortest_path(c, indices=1)
#print(d)
ans = 0
for y in range(H):
for x in range(W):
if A[y][x] != -1:
#print(ans)
ans += d[A[y][x]]
print(int(ans))
| s071791266 | Accepted | 198 | 38,220 | 914 | import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time, copy,bisect
from operator import itemgetter
from heapq import heappush, heappop
import numpy as np
from scipy.sparse.csgraph import shortest_path, floyd_warshall, dijkstra, bellman_ford, johnson
from scipy.sparse import csr_matrix
sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7
stdin = sys.stdin
ni = lambda: int(ns())
nf = lambda: float(ns())
na = lambda: list(map(int, stdin.readline().split()))
nb = lambda: list(map(float, stdin.readline().split()))
ns = lambda: stdin.readline().rstrip() # ignore trailing spaces
H, W = na()
c = [na() for _ in range(10)]
A = [na() for _ in range(H)]
adj = [[] for _ in range(10)]
c = np.array(c)
d = shortest_path(c)
ans = 0
for y in range(H):
for x in range(W):
if A[y][x] != -1:
#print(ans)
ans += d[A[y][x]][1]
print(int(ans))
|
s943875938 | p04011 | u743354620 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 144 | There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee. | n = int(input())
k = int(input())
x = int(input())
y = int(input())
if n - k <0:
print(int(n * x))
else:
print(int((k * y) + (n - k) * y)) | s100784282 | Accepted | 17 | 3,060 | 144 | n = int(input())
k = int(input())
x = int(input())
y = int(input())
if n - k <0:
print(int(n * x))
else:
print(int((k * x) + (n - k) * y)) |
s726496350 | p03711 | u992759582 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 149 | Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group. | x,y = map(int,input().split())
if x ==2 and y == 2:
print('Yes')
elif x % 2 == y % 2 and x >=4 and y >= 4:
print('Yes')
else:
print('No') | s841225117 | Accepted | 17 | 3,060 | 274 | x,y = map(int,input().split())
a_group = [1,3,5,7,8,10,12]
b_group = [4,6,9,11]
c_group = [2]
if x in a_group and y in a_group:
print('Yes')
elif x in b_group and y in b_group:
print('Yes')
elif x in c_group and y in c_group:
print('Yes')
else:
print('No')
|
s870010030 | p02612 | u995062424 | 2,000 | 1,048,576 | Wrong Answer | 25 | 9,168 | 41 | 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-(N//1000)*1000) | s162828817 | Accepted | 28 | 9,156 | 138 | N = int(input())
if(N >= 1000):
if(N%1000 != 0):
print((N//1000+1)*1000-N)
else:
print(0)
else:
print(1000-N) |
s149205130 | p04043 | u588081069 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 150 | 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. | input_str = ''.join(list(map(str, input())))
if input_str == '755' or input_str == '575' or input_str == '557':
print('YES')
else:
print('NO') | s532953530 | Accepted | 17 | 2,940 | 372 | input_str = list(map(int, input().split()))
if input_str == [7, 5, 5] or input_str == [5, 7, 5] or input_str == [5, 5, 7]:
print('YES')
else:
print('NO')
|
s136365922 | p02612 | u689723321 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,000 | 88 | 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())
for i in range(0, 10):
if i*1000 <= m < (i+1)*1000:
print(m-i*1000) | s509148318 | Accepted | 31 | 9,008 | 99 | m=int(input())
for i in range(0, 11):
if i*1000 < m <= (i+1)*1000:
m=(i+1)*1000 - m
print(m)
|
s787641723 | p02612 | u481429249 | 2,000 | 1,048,576 | Wrong Answer | 26 | 9,152 | 72 | 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. | payment = int(input())
bucks = 1000
change = payment%bucks
print(change) | s871847570 | Accepted | 29 | 8,848 | 160 | payment = int(input())
bucks = 1000
if payment%bucks == 0:
change = 0
else:
receive = payment//bucks + 1
change = receive*bucks - payment
print(change) |
s725815365 | p03049 | u618107373 | 2,000 | 1,048,576 | Wrong Answer | 44 | 3,700 | 477 | 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())
l = []
cnt = 0
te = [0,0,0]
for i in range(n):
l.append(input())
for i in l:
if i[0] == "B" and i[-1] == "A":
te[0] += 1
elif i[0] == "B":
te[1] += 1
elif i[-1] == "A":
te[2] += 1
for j in range(len(i)-1):
if i[j] == "A" and i[j+1] == "B":
cnt += 1
if te[0] != 0:
te[0] -= 1
if te[1]>0:
te[0] += 1
te[1] -= 1
if te[2]>0:
te[0] += 1
te[2] -= 1
print(cnt+te[0]+min(te[1],te[2])) | s671722160 | Accepted | 43 | 3,700 | 501 | n = int(input())
l = []
cnt = 0
te = [0,0,0]
for i in range(n):
l.append(input())
for i in l:
if i[0] == "B" and i[-1] == "A":
te[0] += 1
elif i[0] == "B":
te[1] += 1
elif i[-1] == "A":
te[2] += 1
for j in range(len(i)-1):
if i[j] == "A" and i[j+1] == "B":
cnt += 1
if te[0] != 0:
te[0] -= 1
if te[1]>0:
te[0] += 1
te[1] -= 1
if te[2]>0:
te[0] += 1
te[2] -= 1
print(cnt+te[0]+min(te[1],te[2])) |
s809182040 | p03698 | u571999153 | 2,000 | 262,144 | Wrong Answer | 27 | 9,028 | 133 | You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different. | s = input()
letters = []
ans = 'no'
for i in s:
if i in letters:
ans = 'yes'
break
else:
letters.append(i)
print(ans) | s708751513 | Accepted | 29 | 8,848 | 134 | s = input()
letters = []
ans = 'yes'
for i in s:
if i in letters:
ans = 'no'
break
else:
letters.append(i)
print(ans)
|
s697725266 | p02612 | u465423770 | 2,000 | 1,048,576 | Wrong Answer | 30 | 9,104 | 44 | 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. | payment = int(input())
print(payment%1000)
| s458613580 | Accepted | 33 | 9,104 | 109 | payment = int(input())
num = int(payment//1000)
if payment%1000 != 0:
num += 1
print((num*1000)-payment)
|
s771679673 | p03371 | u425762225 | 2,000 | 262,144 | Wrong Answer | 2,104 | 2,940 | 143 | "Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas. | A, B, C, X, Y = map(int,input().split())
m = 10^9
for i in range(X):
for j in range(Y):
m = min(m,A*i+B*j+C*((X+Y)-(i+j)))
print(m) | s672032463 | Accepted | 675 | 49,716 | 200 | def solve(a,b,c,x,y):
l = []
for i in range(10**6):
s = a*max(x-i,0)+b*max(y-i,0)+c*i*2
l.append(s)
return min(l)
A,B,C,X,Y = map(int,input().split())
print(solve(A,B,C,X,Y))
|
s379075890 | p04011 | u278143034 | 2,000 | 262,144 | Wrong Answer | 26 | 9,024 | 272 | There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee. |
N = int(input())
K = int(input())
X = int(input())
Y = int(input())
account = 0
cnt = 1
for cnt in range(1,K + 1,1):
account = account + X
for cnt in range(K + 1,N,1):
account = account + Y
print(account) | s158797486 | Accepted | 27 | 9,176 | 220 |
N = int(input())
K = int(input())
X = int(input())
Y = int(input())
if N <= K:
account = N * X
else:
account = (K * X) + (max(N - K,0) * Y)
print(account) |
s665258791 | p03962 | u857673087 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 37 | 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. | x = list(input())
print(len(set(x))) | s157278783 | Accepted | 17 | 3,064 | 50 | abc = list(input().split())
print(len(set(abc)))
|
s858352911 | p03997 | u385244248 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 56 | You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid. | a,b,h = map(int,open(0).read().split())
print((a+b)+h/2) | s520731405 | Accepted | 17 | 2,940 | 61 | a,b,h = map(int,open(0).read().split())
print(int((a+b)*h/2)) |
s527785520 | p03545 | u695079172 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 464 | 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. | plus = lambda a,b:a+b
minus = lambda a,b:a-b
bit = 0
nums = [int(c) for c in input()]
for bit in range(8):
op_lst = [(plus if bit >> i & 1 else minus) for i in [2,1,0]]
temp = nums[0]
for i in range(len(op_lst)):
temp = op_lst[i](temp,nums[i+1])
if temp == 7:
break
op_strings = {plus:"+",minus:"-"}
s = str(nums[0]) + op_strings[op_lst[0]] + str(nums[1]) + op_strings[op_lst[1]] + str(nums[2]) + op_strings[op_lst[2]] + str(nums[3])
print(s) | s245567680 | Accepted | 17 | 3,064 | 469 | plus = lambda a,b:a+b
minus = lambda a,b:a-b
bit = 0
nums = [int(c) for c in input()]
for bit in range(8):
op_lst = [(plus if bit >> i & 1 else minus) for i in [2,1,0]]
temp = nums[0]
for i in range(len(op_lst)):
temp = op_lst[i](temp,nums[i+1])
if temp == 7:
break
op_strings = {plus:"+",minus:"-"}
s = str(nums[0]) + op_strings[op_lst[0]] + str(nums[1]) + op_strings[op_lst[1]] + str(nums[2]) + op_strings[op_lst[2]] + str(nums[3])
print(s+"=7") |
s493050488 | p02255 | u742013327 | 1,000 | 131,072 | Wrong Answer | 20 | 7,644 | 879 | 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. | #http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_1_A&lang=jp
def insertion_sort(target_list, n_list):
for focus_index in range(1, n_list):
print(*target_list)
target = target_list[focus_index]
if target < target_list[focus_index - 1]:
compare_index = focus_index
while compare_index > 0 and target_list[compare_index - 1] > target:
target_list[compare_index] = target_list[compare_index - 1]
compare_index -= 1;
target_list[compare_index] = target
return target_list
def main():
n_list = int(input())
target_list = [int(a) for a in input().split()]
insertion_sort(target_list, n_list)
if __name__ == "__main__":
main() | s759976826 | Accepted | 20 | 8,044 | 883 | #http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_1_A&lang=jp
def insertion_sort(target_list, n_list):
for focus_index in range(1, n_list):
print(*target_list)
target = target_list[focus_index]
if target < target_list[focus_index - 1]:
compare_index = focus_index
while compare_index > 0 and target_list[compare_index - 1] > target:
target_list[compare_index] = target_list[compare_index - 1]
compare_index -= 1;
target_list[compare_index] = target
return target_list
def main():
n_list = int(input())
target_list = [int(a) for a in input().split()]
print(*insertion_sort(target_list, n_list))
if __name__ == "__main__":
main() |
s718199737 | p03962 | u052066895 | 2,000 | 262,144 | Wrong Answer | 20 | 3,188 | 183 | AtCoDeer the deer recently bought three paint cans. The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c. Here, the color of each paint can is represented by an integer between 1 and 100, inclusive. Since he is forgetful, he might have bought more than one paint can in the same color. Count the number of different kinds of colors of these paint cans and tell him. | A, B, C = map(int, input().split())
if A==B and B==C :
print(1)
elif A==B and B!=C :
print(2)
elif B==C and B!=A :
print(2)
elif A==C and A!=B :
print(2)
else :
print(1) | s461639993 | Accepted | 17 | 2,940 | 176 | A = input().split(" ")
a = A[0]
b = A[1]
c = A[2]
if a == b and b == c: print(1)
elif a == b or a == c and b != c : print(2)
elif b == c and a != b : print(2)
else : print(3) |
s163967833 | p03854 | u920299620 | 2,000 | 262,144 | Wrong Answer | 18 | 3,316 | 937 | 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()
ind =0
len_s=len(s)
s = s+ "zzzzzzzzzz"
A = "dreamer"
B= "dream"
C= "erase"
D="eraser"
flag =True
while(1):
if(s[ind] =="d"):
if(s[ind + 7] !="a"):
if(s[ind:ind+7]!=A):
flag = False
break
else:
ind+=7
else:
if(s[ind:ind+5] !=B):
flag = False
break
else:
ind+=5
elif(s[ind]=="e"):
if(s[ind+5]=="r"):
if(s[ind:ind+6]!=D):
flag = False
break
else:
ind+=6
else:
if(s[ind:ind +5] !=C):
flag =False
break
else:
ind+=5
else:
flag=False
break
if(ind == len_s):
break
if(ind > len(s)):
flag=False
break
if(flag):
print("YES")
else:
print("NO")
| s964610938 | Accepted | 31 | 3,316 | 1,268 | s=input()
ind =0
len_s=len(s)
s = s+ "AAAAAAAAA"
A = "dreamer"
B= "dream"
C= "erase"
D="eraser"
flag =True
while(1):
if(s[ind] =="d"):
if(s[ind + 5] =="d"):
if(s[ind:ind+5]!=B):
flag = False
break
else:
ind+=5
else:
if(s[ind+5]!='e' and s[ind+5]!='A'):
flag=False
break
if(s[ind+7] == "a" or ind + 7 >len_s):
if(s[ind:ind+5]!=B):
flag = False
break
else:
ind+=5
else:
if(s[ind:ind+7] !=A):
flag = False
break
else:
ind+=7
elif(s[ind]=="e"):
if(s[ind+5]=="r"):
if(s[ind:ind+6]!=D):
flag = False
break
else:
ind+=6
else:
if(s[ind:ind +5] !=C):
flag =False
break
else:
ind+=5
else:
flag=False
break
if(ind == len_s):
break
if(ind > len_s):
flag=False
break
if(flag):
print("YES")
else:
print("NO")
|
s138250984 | p03759 | u434311880 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 88 | Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful. | a, b, c =map(int,input().split())
if b-a == c-b:
print("Yes")
else:
print("No")
| s223286462 | Accepted | 18 | 2,940 | 88 | a, b, c =map(int,input().split())
if b-a == c-b:
print("YES")
else:
print("NO")
|
s104266928 | p02260 | u022407960 | 1,000 | 131,072 | Wrong Answer | 30 | 7,632 | 724 | Write a program of the Selection Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: SelectionSort(A) 1 for i = 0 to A.length-1 2 mini = i 3 for j = i to A.length-1 4 if A[j] < A[mini] 5 mini = j 6 swap A[i] and A[mini] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 6 of the pseudocode in the case where i ≠ mini. | #!/usr/bin/env python
# encoding: utf-8
class Solution:
@staticmethod
def selection_sort():
# write your code here
array_length = int(input())
unsorted_array = [int(x) for x in input().split()]
count = 0
for i in range(array_length):
min_j = i
for j in range(i, array_length):
if unsorted_array[j] < unsorted_array[min_j]:
min_j = j
unsorted_array[i], unsorted_array[min_j] = unsorted_array[min_j], unsorted_array[i]
count += 1
print(" ".join(map(str, unsorted_array)))
print(str(count))
if __name__ == '__main__':
solution = Solution()
solution.selection_sort() | s250335006 | Accepted | 40 | 7,704 | 755 | #!/usr/bin/env python
# encoding: utf-8
class Solution:
@staticmethod
def selection_sort():
# write your code here
array_length = int(input())
unsorted_array = [int(x) for x in input().split()]
count = 0
for i in range(array_length):
min_j = i
for j in range(i, array_length):
if unsorted_array[j] < unsorted_array[min_j]:
min_j = j
unsorted_array[i], unsorted_array[min_j] = unsorted_array[min_j], unsorted_array[i]
if i != min_j:
count += 1
print(" ".join(map(str, unsorted_array)))
print(str(count))
if __name__ == '__main__':
solution = Solution()
solution.selection_sort() |
s100471820 | p03971 | u534308356 | 2,000 | 262,144 | Wrong Answer | 111 | 4,712 | 346 | 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())
data = list(input())
for i in range(len(data)):
if data[i] == "a":
if i <= a + b:
print("Yes")
else:
print("No")
elif data[i] == "b":
if i <= a + b and i <= b:
print("Yes")
else:
print("No")
else:
print("No")
| s469966248 | Accepted | 104 | 4,016 | 344 |
n, a, b=map(int,input().split())
for s in input():
if s=='a':
if a+b>0:
print('Yes')
a-=1
else:
print('No')
elif s=='b'and a+b>0 and b>0:
print('Yes')
b-=1
else:
print('No')
|
s281648586 | p03997 | u175217658 | 2,000 | 262,144 | Wrong Answer | 26 | 9,120 | 70 | 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) | s785960825 | Accepted | 27 | 9,088 | 76 | a = int(input())
b = int(input())
h = int(input())
print(int(((a+b)*h)/2))
|
s654029187 | p03337 | u374146618 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 66 | You are given two integers A and B. Find the largest value among A+B, A-B and A \times B. | a,b = [int(x) for x in input().split()]
ans = max([a+b, a-b, a*b]) | s270178864 | Accepted | 18 | 2,940 | 77 | a,b = [int(x) for x in input().split()]
ans = max([a+b, a-b, a*b])
print(ans) |
s870259414 | p02619 | u516554284 | 2,000 | 1,048,576 | Wrong Answer | 36 | 9,500 | 313 | Let's first write a program to calculate the score from a pair of input and output. You can know the total score by submitting your solution, or an official program to calculate a score is often provided for local evaluation as in this contest. Nevertheless, writing a score calculator by yourself is still useful to check your understanding of the problem specification. Moreover, the source code of the score calculator can often be reused for solving the problem or debugging your solution. So it is worthwhile to write a score calculator unless it is very complicated. | d=int(input())
a=[]
for x in range(d+1):
b=list(map(int,input().split()))
a.append(b)
print(a[1][1])
sum1=sum(a[0][1:d])
ans=[0]*d
for n in range(d):
ans[n]=(a[n+1][0])
ans[n]=int(int(ans[n])-(sum1))
try:
ans[n]=ans[n]+ans[n-1]
except:
True
for y in range(d):
print(ans[y])
| s981817429 | Accepted | 39 | 9,668 | 592 | d=int(input())
a=[]
for x in range(d+1):
b=list(map(int,input().split()))
a.append(b)
nittei=[]
for x in range(d):
nittei.append(int(input()))
sum1=sum(a[0][0:26])
ans=[0]*d
humann=a[0]*d
humann=humann+[0]*26
for n in range(d):
date=nittei[n]
ans[n]=(a[n+1][date-1])
try:
for z in range(26):
humann[n*26+z]=int(humann[(n-1)*26+z])+int(humann[n*26+z])
except:
True
humann[n*26+date-1]=0
sum2=sum(humann[n*26+0:n*26+26])
ans[n]=int(ans[n]) - int(sum2)
try:
ans[n]=ans[n]+ans[n-1]
except:
True
for y in range(d):
print(ans[y])
|
s892580108 | p02612 | u317485668 | 2,000 | 1,048,576 | Wrong Answer | 27 | 9,136 | 111 | 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. | #!/usr/bin/python
# -*- Coding: utf-8 -*-
n = int(input())
if n > 1000:
print(n%1000)
else:
print(n)
| s682710599 | Accepted | 32 | 9,192 | 176 | #!/usr/bin/python
# -*- Coding: utf-8 -*-
n = int(input())
if n % 1000 == 0:
print(0)
elif n > 1000:
a = (n//1000 + 1)*1000
print(a - n)
else:
print(1000 - n) |
s602483564 | p03476 | u786020649 | 2,000 | 262,144 | Wrong Answer | 275 | 24,904 | 466 | We say that a odd number N is _similar to 2017_ when both N and (N+1)/2 are prime. You are given Q queries. In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i. | q=int(input())
lr=[tuple(map(int,input().split())) for _ in range(q)]
def main():
n=10**5
ansl=[]
sieve=[1-i%2 for i in range(n+1)]
sieve[0]=0
for i in range(3,n,2):
if sieve[i]:
sieve[2*i::i]=[0 for _ in range(2*i,n+1,i)]
ans=[0]*(n+1)
for i in range(3,n):
if sieve[i]==1 and sieve[(i+1)//2]==1:
ans[i]=ans[i-1]+1
else:
ans[i]=ans[i-1]
for e in lr:
print(ans[e[1]]-ans[e[0]-1])
if __name__=='__main__':
main()
| s701456183 | Accepted | 133 | 32,504 | 462 | import sys
read=sys.stdin.read
def main():
q,*lr=map(int,read().split())
n=10**5
ansl=[]
sieve=[1]*(n+1)
sieve[0],sieve[1]=0,0
for i in range(2,n):
if sieve[i]:
sieve[2*i::i]=[0 for _ in range(2*i,n+1,i)]
ans=[0]*(n+1)
for i in range(3,n):
if sieve[i]==1 and sieve[(i+1)//2]==1:
ans[i]=ans[i-1]+1
else:
ans[i]=ans[i-1]
for l,r in zip(*[iter(lr)]*2):
print(ans[r]-ans[l-1])
if __name__=='__main__':
main()
|
s718123662 | p03997 | u276144769 | 2,000 | 262,144 | Wrong Answer | 22 | 3,064 | 61 | 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) | s983347293 | Accepted | 22 | 3,064 | 62 | a=int(input())
b=int(input())
h=int(input())
print((a+b)*h//2) |
s022848331 | p03556 | u382431597 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 170 | Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer. | import math
n = int(input())
print(math.floor(math.sqrt(n)))
| s783206303 | Accepted | 17 | 2,940 | 63 | import math
print(pow(math.floor(math.sqrt(int(input()))), 2))
|
s672401279 | p03130 | u562767072 | 2,000 | 1,048,576 | Wrong Answer | 30 | 3,572 | 532 | 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. | L = [set(map(int,input().split())) for i in range(3)]
L += [set([1,0]),set([2,0]),set([3,0]),set([4,0])]
import copy
def dfs(left_towns, current_town):
if len(left_towns) == 0:
return 1
ret = 0
for t in left_towns:
if set([t,current_town]) in L:
new_left_towns = copy.deepcopy(left_towns)
new_left_towns.remove(t)
ret += dfs(new_left_towns, t)
if ret>0:
return 1
else:
return 0
ans = 'Yes' if dfs([1,2,3,4],0)>0 else 'No'
print(ans)
| s979354745 | Accepted | 31 | 3,700 | 532 | L = [set(map(int,input().split())) for i in range(3)]
L += [set([1,0]),set([2,0]),set([3,0]),set([4,0])]
import copy
def dfs(left_towns, current_town):
if len(left_towns) == 0:
return 1
ret = 0
for t in left_towns:
if set([t,current_town]) in L:
new_left_towns = copy.deepcopy(left_towns)
new_left_towns.remove(t)
ret += dfs(new_left_towns, t)
if ret>0:
return 1
else:
return 0
ans = 'YES' if dfs([1,2,3,4],0)>0 else 'NO'
print(ans)
|
s425861116 | p03471 | u595064211 | 2,000 | 262,144 | Wrong Answer | 1,412 | 3,064 | 321 | The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough. | N , Y = map(int, input().split())
ans = []
for i in range(N):
for j in range(N):
k = N - i - j
if k >= 0:
num = i*10000 + j*5000 + (N-i-j)*1000
if num == Y:
ans.append([i, j, (N-i-j)])
if len(ans) == 0:
ans = [-1, -1, -1]
print(" ".join(map(str,ans))) | s992480785 | Accepted | 1,267 | 3,060 | 307 | N , Y = map(int, input().split())
ans = []
for i in range(N+1):
for j in range(N+1):
k = N - i - j
if k >= 0:
num = i*10000 + j*5000 + k*1000
if num == Y:
ans = [i, j, k]
if len(ans) == 0:
ans = [-1, -1, -1]
print(" ".join(map(str,ans))) |
s815210105 | p03474 | u732870425 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 222 | The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom. | a, b = map(int, input().split())
s = input()
if s[a+1] == "-":
s2 = s[:a+1] + s[a+2:]
for i in range(a+b):
if s[i] not in range(10):
print("No")
break
else:
print("Yes") | s748830498 | Accepted | 17 | 2,940 | 144 | a, b = map(int, input().split())
s = input()
if s[:a].isdecimal() and s[a]=="-" and s[a+1:].isdecimal():
print("Yes")
else:
print("No") |
s483170601 | p03479 | u329865314 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 95 | 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. | from math import log
n = list(map(int,input().split()))
x, y = n[0],n[1]
print(int(log(y/x,2))) | s942119975 | Accepted | 17 | 3,064 | 140 | from math import log
n = list(map(int,input().split()))
x, y = n[0],n[1]
for i in range(0,100):
if x * 2 ** i > y:
print(i)
quit() |
s568816705 | p02613 | u551425474 | 2,000 | 1,048,576 | Wrong Answer | 165 | 9,108 | 203 | 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())
word_dict = {}
for _ in range(n):
word = str(input())
if word in word_dict:
word_dict[word] += 1
else:
word_dict[word] = 1
for k, v in word_dict.items():
print(k, 'x', v) | s180486543 | Accepted | 164 | 9,100 | 238 | n = int(input())
word_dict = {'AC': 0, 'WA': 0, 'TLE': 0, 'RE': 0}
for _ in range(n):
word = str(input())
if word in word_dict:
word_dict[word] += 1
else:
word_dict[word] = 1
for k, v in word_dict.items():
print(k, 'x', v) |
s698646018 | p02694 | u394352233 | 2,000 | 1,048,576 | Wrong Answer | 22 | 9,164 | 123 | 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? | import math
X = int(input())
count = 0
res = 100
while res <= X:
count +=1
res = math.floor(1.01*res)
print(count) | s727200000 | Accepted | 23 | 9,164 | 122 | import math
X = int(input())
count = 0
res = 100
while res < X:
count +=1
res = math.floor(1.01*res)
print(count) |
s581804257 | p04012 | u496966444 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 120 | Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful. | s = input()
for i in s:
count = s.count(i)
if count % 2 != 0:
print('NO')
exit()
print('YES')
| s051281140 | Accepted | 17 | 3,064 | 120 | s = input()
for i in s:
count = s.count(i)
if count % 2 != 0:
print('No')
exit()
print('Yes')
|
s545477031 | p00470 | u150984829 | 1,000 | 131,072 | Wrong Answer | 90 | 7,872 | 277 | JOIさんが住むカナダのある都市は，南北方向にまっすぐに伸びる w 本の道路と，東西方向にまっすぐに伸びる h 本の道路により，碁盤の目の形に区分けされている． 南北方向の w 本の道路には，西から順に 1, 2, ... , w の番号が付けられている．また，東西方向の h 本の道路には，南から順に 1, 2, ... , h の番号が付けられている．西から i 番目の南北方向の道路と，南から j 番目の東西方向の道路が交わる交差点を (i, j) で表す． JOIさんは，交差点 (1, 1) の近くに住んでおり，交差点 (w, h) の近くの会社に車で通っている．車は道路に沿ってのみ移動することができる．JOIさんは，通勤時間を短くするため，東または北にのみ向かって移動して通勤している．また，この都市では，交通事故を減らすために，次のような交通規則が設けられている： * 交差点を曲がった車は，その直後の交差点で曲がることは出来ない． すなわち，交差点で曲がったあとに1ブロックだけ進んで再び曲がることは許されない．このとき，JOIさんの通勤経路は何通り考えられるだろうか． w と h が与えられたとき，JOIさんの通勤経路の個数を 100000 で割った余りを出力するプログラムを作成せよ． | for e in iter(input,'0 0'):
w,h=map(int,e.split())
M=[[[1,0]*2 for _ in[0]*h]for _ in[0]*w]
for i in range(1,w):
for j in range(1,h):
a,b=M[i-1][j][:2]
c,d=M[i][j-1][2:]
M[i][j]=[d,a+b,b,c+d]
for x in M:print(x)
print(sum(M[w-2][h-1][:2])+sum(M[w-1][h-2][2:]))
| s318561917 | Accepted | 50 | 7,836 | 286 | def s():
for e in iter(input,'0 0'):
w,h=map(int,e.split())
M=[[[1,0]*2 for _ in[0]*h]for _ in[0]*w]
for i in range(1,w):
for j in range(1,h):
a,b=M[i-1][j][:2]
c,d=M[i][j-1][2:]
M[i][j]=[d,a+b,b,c+d]
print((sum(M[w-2][h-1][:2])+sum(M[w-1][h-2][2:]))%10**5)
s()
|
s722601189 | p02924 | u094932051 | 2,000 | 1,048,576 | Time Limit Exceeded | 2,125 | 347,684 | 237 | For an integer N, we will choose a permutation \\{P_1, P_2, ..., P_N\\} of \\{1, 2, ..., N\\}. Then, for each i=1,2,...,N, let M_i be the remainder when i is divided by P_i. Find the maximum possible value of M_1 + M_2 + \cdots + M_N. | F = [0, 0, 1]
while True:
try:
N = int(input())
if N > len(F) - 1:
for i in range(len(F), N+1):
F.append(F[i-1] + i - 1)
print(F[N])
except:
break | s660513487 | Accepted | 17 | 2,940 | 193 | R = [0, 0, 1]
while True:
try:
N = int(input())
if N <= 2:
print(R[N])
else:
print((N-1)*(N-2)//2 + N - 1)
except:
break |
s066224244 | p03129 | u075155299 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 83 | Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1. | n,k=map(int,input().split())
if int((n+1)/2)>k:
print("YES")
else:
print("NO") | s572178111 | Accepted | 17 | 2,940 | 85 | n,k=map(int,input().split())
if int((n+1)/2)>=k:
print("YES")
else:
print("NO")
|
s742429884 | p02401 | u956645355 | 1,000 | 131,072 | Wrong Answer | 40 | 7,432 | 120 | 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:
a = input()
try:
i = a.index('?')
except:
print(eval(a))
else:
break | s313689540 | Accepted | 50 | 7,416 | 125 | while True:
a = input()
try:
i = a.index('?')
except:
print(int(eval(a)))
else:
break |
s585440083 | p03435 | u686036872 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 219 | We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left. According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j. Determine if he is correct. | a, b, c = map(int, input().split())
d, e, f = map(int, input().split())
g, h, i = map(int, input().split())
if a-d==b-e==c-f and a-g==b-h==c-i and a-b--d-e==g-h and a-c==d-f==g-i:
print("Yes")
else:
print("No") | s895442442 | Accepted | 17 | 3,060 | 219 | a, b, c = map(int, input().split())
d, e, f = map(int, input().split())
g, h, i = map(int, input().split())
if a-d==b-e==c-f and a-g==b-h==c-i and a-b==d-e==g-h and a-c==d-f==g-i:
print("Yes")
else:
print("No") |
s871928343 | p03251 | u974935538 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 210 | Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out. | N,M,X,Y = map(int,input().split())
li_x = map(int,input().split())
li_y = map(int,input().split())
x1 = set(li_x)
y1 = set(li_y)
if max(x1)>=min(y1):
print("War")
elif max(x1)<min(y1):
print("No war") | s929936971 | Accepted | 17 | 3,060 | 198 | N,M,X,Y = map(int,input().split())
lx = list(map(int,input().split()))
lx.append(X)
ly = list(map(int,input().split()))
ly.append(Y)
if max(lx) < min(ly):
print('No War')
else:
print('War')
|
s654405999 | p03380 | u941438707 | 2,000 | 262,144 | Wrong Answer | 109 | 14,028 | 145 | 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. | n,*a=map(int,open(0).read().split())
a.sort()
b=a[-1]
c=a[0]
d=b/2-c
for i in a:
if d>abs(b/2-i):
c=i
d=abs(b/2-i)
print(c,b) | s440162667 | Accepted | 68 | 14,028 | 87 | n,*a=map(int,open(0).read().split())
n=max(a)
print(n,min((abs(n/2-i),i)for i in a)[1]) |
s252131987 | p03712 | u166306121 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 180 | 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 = [""] * H
for i in range(H):
a[i] = input()
print("#" * (W * 2))
for i in range(H):
print('#', a[i], "#", sep="")
print("#" * (W * 2))
| s192172513 | Accepted | 18 | 3,060 | 180 | H, W = map(int, input().split())
a = [""] * H
for i in range(H):
a[i] = input()
print("#" * (W + 2))
for i in range(H):
print('#', a[i], "#", sep="")
print("#" * (W + 2))
|
s115293967 | p00003 | u623894175 | 1,000 | 131,072 | Wrong Answer | 40 | 7,520 | 189 | Write a program which judges wheather given length of three side form a right triangle. Print "YES" if the given sides (integers) form a right triangle, "NO" if not so. | deta_set_count = int(input())
for _ in range(deta_set_count):
k = list(map(int, input().split()))
if k[0]**2 + k[1]**2 == k[2]**2:
print(True)
else:
print(False) | s149823105 | Accepted | 40 | 7,540 | 206 | deta_set_count = int(input())
for _ in range(deta_set_count):
k = list(map(int, input().split()))
k= sorted(k)
if k[0]**2 + k[1]**2 == k[2]**2:
print('YES')
else:
print('NO') |
s674531507 | p03407 | u811528179 | 2,000 | 262,144 | Wrong Answer | 21 | 3,316 | 59 | An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan. | a,b,c=map(int,input().split())
print(['No','Yes'][a+b==c])
| s046326978 | Accepted | 17 | 2,940 | 59 | a,b,c=map(int,input().split())
print(['No','Yes'][a+b>=c])
|
s422964478 | p03129 | u036744414 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 128 | Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1. | # coding:utf-8
import math
N, K = map(int, input().split())
if math.ceil(N / 2) >= K:
print("Yes")
else:
print("No")
| s134869562 | Accepted | 17 | 2,940 | 109 | # coding:utf-8
N, K = map(int, input().split())
if -(-N // 2) >= K:
print("YES")
else:
print("NO")
|
s608402191 | p03470 | u371467115 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 72 | 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())
d=[int(input()) for i in range(n)]
s=set(d)
print(len(d)) | s485878529 | Accepted | 18 | 2,940 | 72 | n=int(input())
d=[int(input()) for i in range(n)]
s=set(d)
print(len(s)) |
s224317219 | p02255 | u150711673 | 1,000 | 131,072 | Wrong Answer | 20 | 5,596 | 267 | 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_str = input()
A = []
for s in A_str.split():
A.append(int(s))
for i in range(1, len(A)):
key = A[i]
j = i - 1
while j >= 0 and A[j] > key:
A[j + 1] = A[j]
j -= 1
A[j + 1] = key
print(" ".join(map(str, A)))
| s935874957 | Accepted | 20 | 5,596 | 264 | N = int(input())
A = list(map(int, input().split()))
print(" ".join(map(str, A)))
for i in range(1, len(A)):
key = A[i]
j = i - 1
while j >= 0 and A[j] > key:
A[j + 1] = A[j]
j -= 1
A[j + 1] = key
print(" ".join(map(str, A)))
|
s490997856 | p04044 | u556589653 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 144 | 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 = []
ans = ""
for i in range(N):
S.append(input())
sorted(S)
for i in range(len(S)):
ans += S[i]
print(ans) | s163210481 | Accepted | 26 | 9,164 | 109 | N,L = map(int,input().split())
ls = []
for i in range(N):
ls.append(input())
ls.sort()
print("".join(ls)) |
s290034317 | p03471 | u128661070 | 2,000 | 262,144 | Wrong Answer | 792 | 3,316 | 266 | The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough. | n,y = map(int, input().strip().split())
a_x, a_y, a_z = -1, -1, -1
for xi in range(n + 1):
for yi in range(n - xi + 1):
zi = n - xi - yi
if 1000*xi + 5000*yi + 10000*zi == y:
a_x = xi
a_y = yi
a_z = zi
break
print(a_x, a_y, a_z)
| s382114648 | Accepted | 833 | 3,316 | 264 | n,y = map(int, input().strip().split())
a_x, a_y, a_z = -1, -1, -1
for xi in range(n + 1):
for yi in range(n - xi + 1):
zi = n - xi - yi
if 10000*xi + 5000*yi + 1000*zi == y:
a_x = xi
a_y = yi
a_z = zi
break
print(a_x, a_y, a_z) |
s994152877 | p03408 | u668726177 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 197 | Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards. | d = {}
N = int(input())
for _ in range(N):
c = input()
d[c] = d.get(c, 0)+1
M = int(input())
for _ in range(M):
c = input()
d[c] = d.get(c, 0)-1
print(max(d.items(), key=lambda x: x[1])[0]) | s363653543 | Accepted | 17 | 3,060 | 205 | d = {}
N = int(input())
for _ in range(N):
c = input()
d[c] = d.get(c, 0)+1
M = int(input())
for _ in range(M):
c = input()
d[c] = d.get(c, 0)-1
print(max(max(d.items(), key=lambda x: x[1])[1], 0)) |
s221351751 | p03576 | u761320129 | 2,000 | 262,144 | Wrong Answer | 1,408 | 3,188 | 828 | 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 = []
for i in range(N):
x,y = map(int,input().split())
xs.append((x,y,i))
ys.append((y,x,i))
xs.sort()
ys.sort()
x_order = {}
for xo,p in enumerate(xs):
x,y,i = p
x_order[i] = xo
cums = [[0 for j in range(N+1)] for i in range(N+1)]
for i in range(N):
y,x,pi = ys[i]
xo = x_order[pi]
for j in range(N+1):
cums[i+1][j] = cums[i][j] + (1 if j > xo else 0)
print(cums)
ans = float('inf')
for x1 in range(N-1):
for x2 in range(x1+1,N):
for y1 in range(N-1):
for y2 in range(y1+1,N):
if cums[y2+1][x2+1] - cums[y2+1][x1] - cums[y1][x2+1] + cums[y1][x1] < K:
continue
area = (xs[x2][0] - xs[x1][0]) * (ys[y2][0] - ys[y1][0])
ans = min(ans, area)
print(ans)
| s654765751 | Accepted | 1,421 | 3,188 | 815 | N,K = map(int,input().split())
xs = []
ys = []
for i in range(N):
x,y = map(int,input().split())
xs.append((x,y,i))
ys.append((y,x,i))
xs.sort()
ys.sort()
x_order = {}
for xo,p in enumerate(xs):
x,y,i = p
x_order[i] = xo
cums = [[0 for j in range(N+1)] for i in range(N+1)]
for i in range(N):
y,x,pi = ys[i]
xo = x_order[pi]
for j in range(N+1):
cums[i+1][j] = cums[i][j] + (1 if j > xo else 0)
ans = float('inf')
for x1 in range(N-1):
for x2 in range(x1+1,N):
for y1 in range(N-1):
for y2 in range(y1+1,N):
if cums[y2+1][x2+1] - cums[y2+1][x1] - cums[y1][x2+1] + cums[y1][x1] < K:
continue
area = (xs[x2][0] - xs[x1][0]) * (ys[y2][0] - ys[y1][0])
ans = min(ans, area)
print(ans) |
s551253293 | p03644 | u969211566 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 163 | Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times. | n = int(input())
cmax = 0
imax = 0
for i in range(1,n+1):
cnt = 0
ig = i
while i % 2 == 0:
i /= 2
cnt += 1
if cnt > cmax:
imax = ig
print(imax) | s948860478 | Accepted | 17 | 3,060 | 179 | n = int(input())
cmax = 0
imax = 1
for i in range(1,n+1):
cnt = 0
ig = i
while i % 2 == 0:
i /= 2
cnt += 1
if cnt > cmax:
cmax = cnt
imax = ig
print(imax) |
s288777311 | p02396 | u208157605 | 1,000 | 131,072 | Wrong Answer | 80 | 7,884 | 153 | 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. | list = []
while True:
n = int(input())
if n == 0:
break
list.append(n)
for i,n in enumerate(list):
print('case %i: %i' %(i+1, n)) | s088632366 | Accepted | 80 | 7,864 | 153 | list = []
while True:
n = int(input())
if n == 0:
break
list.append(n)
for i,n in enumerate(list):
print('Case %i: %i' %(i+1, n)) |
s733071250 | p02406 | u610816226 | 1,000 | 131,072 | Wrong Answer | 30 | 5,596 | 254 | 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; } | x = int(input())
k = []
for i in range(1, x + 1):
y = i
if i % 3 == 0:
k.append(i)
else:
while y >= 10:
y %= 10
if y == 3:
k.append(i)
for i in range(len(k)):
print(' '+ str(k[i]))
| s313260759 | Accepted | 30 | 6,116 | 373 | x = int(input())
k = []
for i in range(1, x + 1):
y = i
if i % 3 == 0:
k.append(i)
else:
while True:
if y % 10 == 3:
k.append(i)
break
else:
if y <= 10:
break
else:
y //= 10
print(' '+ ' '.join([str(x) for x in k]))
|
s463818478 | p03434 | u626337957 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 97 | 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. | N = int(input())
A = list(map(int, input().split()))
A.sort(key=lambda x: -x)
print(sum(A[0::2])) | s662804678 | Accepted | 17 | 2,940 | 111 | N = int(input())
A = list(map(int, input().split()))
A.sort(key=lambda x: -x)
print(sum(A[0::2])-sum(A[1::2]))
|
s885203501 | p03485 | u485566817 | 2,000 | 262,144 | Wrong Answer | 20 | 2,940 | 49 | 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())
x = round(a*b/2)
x | s558281701 | Accepted | 17 | 2,940 | 76 | import math
a,b = map(int,input().split())
x = (a + b)/2
print(math.ceil(x)) |
s685770793 | p03455 | u476155259 | 2,000 | 262,144 | Wrong Answer | 19 | 3,316 | 190 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | # coding: utf-8
# Your code here!
input_num = input().split()
num_int = [int(i) for i in input_num]
seki = num_int[0] + num_int[1]
if seki % 2 == 0:
print("Even")
else:
print("Odd") | s311028516 | Accepted | 17 | 2,940 | 190 | # coding: utf-8
# Your code here!
input_num = input().split()
num_int = [int(i) for i in input_num]
seki = num_int[0] * num_int[1]
if seki % 2 == 0:
print("Even")
else:
print("Odd") |
s429620348 | p03854 | u196697332 | 2,000 | 262,144 | Wrong Answer | 19 | 4,796 | 448 | 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`. | import sys
S = str(input())
while S != '':
print(S)
if (S[:5] == 'dream' and S[5:11] == 'eraser') or (S[:5] == 'dream' and S[5: 10] == 'erase') or (S[:5] == 'dream' and S[5:] == ''):
S = S[5:]
elif S[:5] == 'erase' and S[5] != 'r':
S = S[5:]
elif S[:6] == 'eraser':
S = S[6:]
elif S[:7] == 'dreamer':
S = S[7:]
else:
print('NO')
sys.exit()
break
print('YES')
| s746649108 | Accepted | 126 | 3,188 | 536 | import sys
S = str(input())
while S != '':
if (S[:5] == 'dream' and S[5:11] == 'eraser') or (S[:5] == 'dream' and S[5: 10] == 'erase') or (S[:5] == 'dream' and S[5:] == '') or (S[:5] == 'dream' and S[5:10] == 'dream'):
S = S[5:]
elif (S[:5] == 'erase' and S[5:] != '' and S[5] != 'r') or (S[:5] == 'erase' and S[5:] == ''):
S = S[5:]
elif S[:6] == 'eraser':
S = S[6:]
elif S[:7] == 'dreamer':
S = S[7:]
else:
print('NO')
sys.exit()
break
print('YES')
|
s876831835 | p02416 | u292012552 | 1,000 | 131,072 | Wrong Answer | 20 | 7,308 | 162 | Write a program which reads an integer and prints sum of its digits. | n = input()
while n != "0":
ans = 0
for i in range(len(n)):
print(ord(n[i:i+1]))
ans += ord(n[i:i+1]) - 48
print(ans)
n = input() | s940080985 | Accepted | 30 | 7,420 | 133 | n = input()
while n != "0":
ans = 0
for i in range(len(n)):
ans += ord(n[i:i+1]) - 48
print(ans)
n = input() |
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