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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s313515942 | p03836 | u611871297 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 279 | Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him. | sx, sy, tx, ty = map(int, input().split())
h = tx - sx
w = ty - sy
s = ''
s += 'U'
s += 'U' * (h - 1)
s += 'R' * w
s += 'D'
s += 'D' * (h - 1)
s += 'L' * w
s += 'L'
s += 'U' * (h + 1)
s += 'R' * (w + 1)
s += 'D'
s += 'R'
s += 'D' * (h + 1)
s += 'L' * (w + 1)
s += 'U'
print(s) | s079349316 | Accepted | 17 | 3,064 | 279 | sx, sy, tx, ty = map(int, input().split())
w = tx - sx
h = ty - sy
s = ''
s += 'U'
s += 'U' * (h - 1)
s += 'R' * w
s += 'D'
s += 'D' * (h - 1)
s += 'L' * w
s += 'L'
s += 'U' * (h + 1)
s += 'R' * (w + 1)
s += 'D'
s += 'R'
s += 'D' * (h + 1)
s += 'L' * (w + 1)
s += 'U'
print(s) |
s175614447 | p02408 | u641357568 | 1,000 | 131,072 | Wrong Answer | 20 | 5,608 | 305 | Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers). The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond. | import sys
cards_num = int(sys.stdin.readline())
cards =[ card[:-1] for card in sys.stdin.readlines() ]
print(cards_num)
print(cards)
all_cards = [ i + ' ' + str(j) for i in 'SHCD' for j in range(1,14) ]
missing_cards = [i for i in all_cards if not(i in cards) ]
for s in missing_cards:
print(s)
| s587747498 | Accepted | 30 | 5,604 | 274 | import sys
cards_num = int(sys.stdin.readline())
cards =[ card[:-1] for card in sys.stdin.readlines() ]
all_cards = [ i + ' ' + str(j) for i in 'SHCD' for j in range(1,14) ]
missing_cards = [i for i in all_cards if not(i in cards) ]
for s in missing_cards:
print(s)
|
s611236265 | p03737 | u478266845 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 123 | You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words. | s1,s2,s3 = [str(i) for i in input().split()]
s1.upper()
s2.upper()
s3.upper()
ans = s1[0] +s2[0] +s3[0]
print(str(ans)) | s501738683 | Accepted | 17 | 2,940 | 138 | s1,s2,s3 = [str(i) for i in input().split()]
S1 = s1.upper()
S2 = s2.upper()
S3 = s3.upper()
ans = S1[0] +S2[0] +S3[0]
print(str(ans)) |
s590458485 | p03455 | u546235346 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 114 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | l = input().split()
print(l)
a = int(l[0])
b = int(l[1])
if ((a*b)%2 == 0):
print ("Even")
else:
print ("Odd")
| s124065445 | Accepted | 17 | 2,940 | 115 | l = input().split()
# print(l)
a = int(l[0])
b = int(l[1])
if ((a*b)%2 == 0):
print ("Even")
else:
print ("Odd") |
s387538551 | p02613 | u938785734 | 2,000 | 1,048,576 | Wrong Answer | 158 | 16,240 | 344 | 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())
li=[]
for i in range(n):
s=input()
li.append(s)
ac=0
wa=0
tle=0
re=0
for i in li:
if i=="AC":
ac+=1
elif i =="TLE":
tle+=1
elif i=="wa":
wa+=1
else:
re+=1
print("AC × {}".format(ac))
print("WA × {}".format(wa))
print("TLE × {}".format(tle))
print("RE × {}".format(re)) | s424758726 | Accepted | 160 | 16,324 | 340 | n=int(input())
li=[]
for i in range(n):
s=input()
li.append(s)
ac=0
wa=0
tle=0
re=0
for i in li:
if i=="AC":
ac+=1
elif i =="TLE":
tle+=1
elif i=="WA":
wa+=1
else:
re+=1
print("AC x {}".format(ac))
print("WA x {}".format(wa))
print("TLE x {}".format(tle))
print("RE x {}".format(re)) |
s914343272 | p03377 | u570612423 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 195 | There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals. | # -*- coding: utf-8 -*-
A,B,X = map(int, input().split())
if X==A:
print("Yes")
elif A < X and X < (A+B):
print("Yes")
elif A < X and X == (A+B):
print("Yes")
else:
print("No") | s686856314 | Accepted | 17 | 3,060 | 195 | # -*- coding: utf-8 -*-
A,B,X = map(int, input().split())
if X==A:
print("YES")
elif A < X and X < (A+B):
print("YES")
elif A < X and X == (A+B):
print("YES")
else:
print("NO") |
s792511531 | p02233 | u007270338 | 1,000 | 131,072 | Wrong Answer | 20 | 5,600 | 236 | Write a program which prints $n$-th fibonacci number for a given integer $n$. The $n$-th fibonacci number is defined by the following recursive formula: \begin{equation*} fib(n)= \left \\{ \begin{array}{ll} 1 & (n = 0) \\\ 1 & (n = 1) \\\ fib(n - 1) + fib(n - 2) & \\\ \end{array} \right. \end{equation*} | #coding:utf-8
n = int(input())
F = [False for i in range(n+1)]
def fib(n):
if n == 1 or n == 2:
F[n] = 1
return 1
if F[n]:
return F[n]
F[n] = fib(n-2) + fib(n-1)
return F[n]
a = fib(n)
print(a)
| s860644117 | Accepted | 30 | 5,608 | 236 | #coding:utf-8
n = int(input())
F = [False for i in range(n+1)]
def fib(n):
if n == 0 or n == 1:
F[n] = 1
return 1
if F[n]:
return F[n]
F[n] = fib(n-2) + fib(n-1)
return F[n]
a = fib(n)
print(a)
|
s602915466 | p03067 | u085186789 | 2,000 | 1,048,576 | Wrong Answer | 27 | 9,080 | 105 | There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise. | A, B, C = map(int,input().split())
if C > min(A, B) and C < max(A, B):
print("YES")
else:
print("NO") | s431976017 | Accepted | 23 | 9,100 | 106 | A, B, C = map(int,input().split())
if C > min(A, B) and C < max(A, B):
print("Yes")
else:
print("No")
|
s668471251 | p03351 | u496687522 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 113 | Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate. | a, b, c, d = map(int, input().split())
print('Yes' if abs(a-c) <= d or abs(a-b) <= d and abs(a-c) <= d else 'No') | s051782070 | Accepted | 17 | 2,940 | 113 | a, b, c, d = map(int, input().split())
print('Yes' if abs(a-c) <= d or abs(a-b) <= d and abs(b-c) <= d else 'No') |
s242661380 | p03730 | u768896740 | 2,000 | 262,144 | Wrong Answer | 19 | 2,940 | 141 | We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`. | a, b, c = map(int, input().split())
for i in range(10000):
t = a * i
if t % b == c:
print('Yes')
exit()
print('No') | s005934310 | Accepted | 19 | 2,940 | 141 | a, b, c = map(int, input().split())
for i in range(10000):
t = a * i
if t % b == c:
print('YES')
exit()
print('NO') |
s715587746 | p03399 | u888933875 | 2,000 | 262,144 | Wrong Answer | 24 | 9,024 | 142 | You planned a trip using trains and buses. The train fare will be A yen (the currency of Japan) if you buy ordinary tickets along the way, and B yen if you buy an unlimited ticket. Similarly, the bus fare will be C yen if you buy ordinary tickets along the way, and D yen if you buy an unlimited ticket. Find the minimum total fare when the optimal choices are made for trains and buses. | l= sorted([int(input()) for _ in range(2)])
m=sorted([int(input()) for _ in range(2)])
print(l,m)
ans= min(x+y for x,y in zip(l,m))
print(ans) | s394903799 | Accepted | 29 | 9,024 | 149 | from itertools import product
l= [int(input()) for _ in range(2)]
m=[int(input()) for _ in range(2)]
ans= min(x+y for x,y in product(l,m))
print(ans) |
s110071700 | p01086 | u284260266 | 8,000 | 262,144 | Wrong Answer | 30 | 5,608 | 1,392 | A _Short Phrase_ (aka. Tanku) is a fixed verse, inspired by Japanese poetry Tanka and Haiku. It is a sequence of words, each consisting of lowercase letters 'a' to 'z', and must satisfy the following condition: > (The Condition for a Short Phrase) > The sequence of words can be divided into five sections such that the total > number of the letters in the word(s) of the first section is five, that of > the second is seven, and those of the rest are five, seven, and seven, > respectively. The following is an example of a Short Phrase. > > do the best and enjoy today at acm icpc In this example, the sequence of the nine words can be divided into five sections (1) "do" and "the", (2) "best" and "and", (3) "enjoy", (4) "today" and "at", and (5) "acm" and "icpc" such that they have 5, 7, 5, 7, and 7 letters in this order, respectively. This surely satisfies the condition of a Short Phrase. Now, _Short Phrase Parnassus_ published by your company has received a lot of contributions. By an unfortunate accident, however, some irrelevant texts seem to be added at beginnings and ends of contributed Short Phrases. Your mission is to write a program that finds the Short Phrase from a sequence of words that may have an irrelevant prefix and/or a suffix. | while True:
n = int(input())
if n == 0:
break
w = list()
for i in range(0, n):
w.append(input())
for j in range(1, n+1):
ck = 0
co = 0
p = 0
t = ""
for k in range(j-1,50):
#print(k)
if co == 0 or co == 2:
if ck < 5:
#print("<")
ck = ck + len(w[k+p])
elif ck == 5:
#print("==")
co = co + 1
ck = 0
p = p-1
else:
#print(">")
break
else:
if ck < 7:
#print("<7")
ck = ck+len(w[k+p])
#print(" " ,ck)
elif ck == 7:
#print("==7")
co = co+1
ck = 0
p = p-1
if co == 5:
t = "fin"
break
else:
#print(">7")
break
print(j)
if t == "fin":
break
#print("over")
| s732966864 | Accepted | 30 | 5,608 | 1,396 | while True:
n = int(input())
if n == 0:
break
w = list()
for i in range(0, n):
w.append(input())
for j in range(1, n+1):
ck = 0
co = 0
p = 0
t = ""
for k in range(j-1,50):
#print(k)
if co == 0 or co == 2:
if ck < 5:
#print("<")
ck = ck + len(w[k+p])
elif ck == 5:
#print("==")
co = co + 1
ck = 0
p = p-1
else:
#print(">")
break
else:
if ck < 7:
#print("<7")
ck = ck+len(w[k+p])
#print(" " ,ck)
elif ck == 7:
#print("==7")
co = co+1
ck = 0
p = p-1
if co == 5:
t = "fin"
break
else:
#print(">7")
break
if t == "fin":
print(j)
break
#print("over")
|
s696220175 | p03455 | u780675733 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 108 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | a, b = map(int, input().split())
if (a % 2 == 0) and (b % 2 == 0):
print("Even")
else:
print("Odd") | s896248752 | Accepted | 17 | 2,940 | 91 | a, b = map(int, input().split())
if a*b % 2 == 0:
print("Even")
else:
print("Odd") |
s655578451 | p02275 | u949517845 | 1,000 | 131,072 | Wrong Answer | 50 | 7,684 | 546 | Counting sort can be used for sorting elements in an array which each of the n input elements is an integer in the range 0 to k. The idea of counting sort is to determine, for each input element x, the number of elements less than x as C[x]. This information can be used to place element x directly into its position in the output array B. This scheme must be modified to handle the situation in which several elements have the same value. Please see the following pseudocode for the detail: Counting-Sort(A, B, k) 1 for i = 0 to k 2 do C[i] = 0 3 for j = 1 to length[A] 4 do C[A[j]] = C[A[j]]+1 5 /* C[i] now contains the number of elements equal to i */ 6 for i = 1 to k 7 do C[i] = C[i] + C[i-1] 8 /* C[i] now contains the number of elements less than or equal to i */ 9 for j = length[A] downto 1 10 do B[C[A[j]]] = A[j] 11 C[A[j]] = C[A[j]]-1 Write a program which sorts elements of given array ascending order based on the counting sort. | # -*- coding:utf-8 -*-
def countingSort(A, B, k, n):
C = [0] * (k+1)
for i in range(1, n + 1):
C[A[i]] += 1
for j in range(1, k + 1):
C[j] = C[j] + C[j - 1]
print(C)
for j in range(n, 0, -1):
B[C[A[j]]] = A[j]
C[A[j]] -= 1
if __name__ == "__main__":
n = int(input())
lst = [int(val) for val in input().split()]
lst.insert(0, 0)
result = [0] * (n + 1)
k = max(lst)
countingSort(lst, result, k, n)
result.pop(0)
print(" ".join([str(val) for val in result])) | s224292442 | Accepted | 1,620 | 242,152 | 465 | # -*- coding:utf-8 -*-
def countingSort(lst, n):
cnt_lst = [0] * (max(lst)+1)
for i in range(0, n):
cnt_lst[lst[i]] += 1
i = 0
result = []
for j, cnt in enumerate(cnt_lst):
for _ in range(0, cnt):
result.append(j)
return result
if __name__ == "__main__":
n = int(input())
lst = [int(val) for val in input().split()]
result = countingSort(lst, n)
print(" ".join([str(val) for val in result])) |
s485183102 | p02842 | u478417863 | 2,000 | 1,048,576 | Wrong Answer | 26 | 9,232 | 380 | Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact. | n=int(input())
a=n//108
B=n%108
if B<13:
b=B
elif 14<=B<24:
b=B-1
print(100*a+b)
elif 27<=B<38:
b=B-2
print(100*a+b)
elif 41<=B<50:
b=B-3
print(100*a+b)
elif 54<=B<63:
b=B-4
print(100*a+b)
elif 68<=B<75:
b=B-5
print(100*a+b)
elif 81<=B<88:
b=B-6
print(100*a+b)
elif 95<=B<100:
b=B-7
print(100*a+b)
else :
print(":(") | s905671305 | Accepted | 36 | 9,176 | 133 | N = int(input())
ans = -1
for i in range(1, N + 1):
if int(i * 1.08) == N:
ans = i
if ans != -1:
print(ans)
else:
print(":(")
|
s024856896 | p03455 | u806855121 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 137 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | a, b = map(str, input().split())
n = int(a+b)
for i in range(1, 318):
if i**2 == n:
print('Yes')
quit()
print('No')
| s838529767 | Accepted | 17 | 2,940 | 89 | a, b = map(int, input().split())
if a*b%2 == 0:
print('Even')
else:
print('Odd') |
s960738442 | p02866 | u587589241 | 2,000 | 1,048,576 | Wrong Answer | 85 | 24,248 | 257 | Given is an integer sequence D_1,...,D_N of N elements. Find the number, modulo 998244353, of trees with N vertices numbered 1 to N that satisfy the following condition: * For every integer i from 1 to N, the distance between Vertex 1 and Vertex i is D_i. | from collections import Counter
N = int(input())
D = list(map(int, input().split()))
mod = 998244353
c = Counter(D)
m = max(c.keys())
if D[0] == 0 and c[0] == 1:
ans = 1
for i in D[1:]:
ans *= c[i-1]
ans %= mod
else:
print(0)
| s309122428 | Accepted | 91 | 24,316 | 272 | from collections import Counter
N = int(input())
D = list(map(int, input().split()))
mod = 998244353
c = Counter(D)
m = max(c.keys())
if D[0] == 0 and c[0] == 1:
ans = 1
for i in D[1:]:
ans *= c[i-1]
ans %= mod
print(ans)
else:
print(0)
|
s787814002 | p03543 | u205700197 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 160 | We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**? | n=input()
count = 0
for i in range(0,2):
if n[i] == n[i+1]:
count = count + 1
print(count)
if count >= 2:
print("Yes")
else:print("No")
| s266538957 | Accepted | 17 | 2,940 | 149 | n=input()
count = 0
for i in range(0,2):
if n[i] == n[i+1] == n[i+2]:
count = count + 1
if count >= 1:
print("Yes")
else:print("No")
|
s193541832 | p02389 | u684306364 | 1,000 | 131,072 | Wrong Answer | 30 | 7,520 | 51 | Write a program which calculates the area and perimeter of a given rectangle. | ab = input().split()
print(int(ab[0]) * int(ab[1])) | s597013772 | Accepted | 30 | 7,480 | 96 | ab = input().split()
a = int(ab[0])
b = int(ab[1])
print("{0} {1}".format(a * b, a * 2 + b * 2)) |
s289021440 | p02415 | u075836834 | 1,000 | 131,072 | Wrong Answer | 20 | 7,328 | 112 | Write a program which converts uppercase/lowercase letters to lowercase/uppercase for a given string. | import sys
A=input().split()
print(A)
for i in A:
sys.stdout.write(i.swapcase())
sys.stdout.write(" ")
print() | s472899264 | Accepted | 30 | 7,428 | 118 | import sys
A=input().split()
for i in A:
sys.stdout.write(i.swapcase())
if i!=A[-1]:
sys.stdout.write(" ")
print() |
s739209255 | p03997 | u209619667 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 71 | 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())
c = int(input())
s = (a+b)/2
print(s) | s556569726 | Accepted | 17 | 2,940 | 84 | a = int(input())
b = int(input())
h = int(input())
s = (a+b)*h
s = s/2
print(int(s)) |
s048068179 | p04031 | u703214333 | 2,000 | 262,144 | Wrong Answer | 21 | 2,940 | 160 | Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective. | n=int(input())
a=list(map(int,input().split()))
ans=10**9
for i in a:
tmp=0
for j in range(n):
tmp+=(i-a[j])**2
ans=min(tmp,ans)
print(ans)
| s131304641 | Accepted | 26 | 3,060 | 173 | n=int(input())
a=list(map(int,input().split()))
ans=10**9
for i in range(-100,101):
tmp=0
for j in range(n):
tmp+=(i-a[j])**2
ans=min(tmp,ans)
print(ans) |
s314435554 | p03160 | u858033561 | 2,000 | 1,048,576 | Wrong Answer | 204 | 13,924 | 330 | There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N. | import sys
n = int(sys.stdin.readline().strip())
hs = list(map(int, sys.stdin.readline().strip().split(' ')))
print(n,hs)
dp = [0]*n
for i in range(1, n):
x = 10**10
if i-1 >= 0:
x = min(x, dp[i-1]+abs(hs[i]-hs[i-1]))
if i-2 >= 0:
x = min(x, dp[i-2]+abs(hs[i]-hs[i-2]))
dp[i] = x
print(dp[n-1]) | s133684043 | Accepted | 173 | 13,976 | 319 | import sys
n = int(sys.stdin.readline().strip())
hs = list(map(int, sys.stdin.readline().strip().split(' ')))
dp = [0]*n
for i in range(1, n):
x = 10**10
if i-1 >= 0:
x = min(x, dp[i-1]+abs(hs[i]-hs[i-1]))
if i-2 >= 0:
x = min(x, dp[i-2]+abs(hs[i]-hs[i-2]))
dp[i] = x
print(dp[n-1]) |
s160760490 | p03997 | u316603606 | 2,000 | 262,144 | Wrong Answer | 31 | 8,976 | 83 | 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 ())
S = ((a+b)*h)/2
print (S)
| s382787308 | Accepted | 25 | 9,048 | 84 | a = int (input ())
b = int (input ())
h = int (input ())
print ((a+b)*(round(h/2)))
|
s167945352 | p03407 | u811202694 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 108 | 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())
total = 2 * (a + b)
if total >= c:
print("YES")
else:
print("NO") | s764955364 | Accepted | 21 | 2,940 | 104 | a,b,c = map(int,input().split())
total = (a + b)
if total >= c:
print("Yes")
else:
print("No") |
s409719624 | p03251 | u307585631 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,064 | 327 | 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())
list_x = list(map(int,input().split(' ')))
list_y = list(map(int,input().split(' ')))
min_range = max(list_x)
max_range = min(list_y)
ans = 0
for i in range(X+1, Y):
if(i > min_range and i <= max_range):
ans = 1
if (ans == 1):
print("No war")
elif (ans == 0):
print("War")
| s085892465 | Accepted | 17 | 3,064 | 333 | N, M, X, Y = map(int, input().split())
list_x = list(map(int,input().split(' ')))
list_y = list(map(int,input().split(' ')))
min_range = max(list_x)
max_range = min(list_y)
ans = 0
for i in range(X+1, Y+1):
if(i > min_range and i <= max_range):
ans = 1
if (ans == 1):
print("No War")
elif (ans == 0):
print("War") |
s776882013 | p03024 | u366541443 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 129 | Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility. | import math
n = list(map(str, input().split()))
n=n[0]
yes=0
no=0
print(yes,no)
if(no>=8):
print("NO")
else:
print("YES") | s851775081 | Accepted | 18 | 3,060 | 170 | import math
n = list(map(str, input().split()))
n=n[0]
yes=0
no=0
for i in range(len(n)):
if(n[i]=='x'):
no+=1
if(no>=8):
print("NO")
else:
print("YES") |
s030482843 | p03563 | u317440328 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 44 | Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it. | R=int(input())
G=float(input())
print(2*G-R) | s904084452 | Accepted | 17 | 2,940 | 42 | R=int(input())
G=int(input())
print(2*G-R) |
s496508877 | p03379 | u026155812 | 2,000 | 262,144 | Wrong Answer | 311 | 25,220 | 156 | When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N. | N = int(input())
X = [int(i) for i in input().split()]
for i in range(N):
if i < N/2:
print(X[int(N/2)])
else:
print(X[int(N/2)-1]) | s260571720 | Accepted | 326 | 25,344 | 266 | N = int(input())
X = [int(i) for i in input().split()]
a = X[:]
a.sort()
l = a[N//2-1]
r = a[N//2]
if l == r:
for i in range(N):
print(l)
else:
for i in range(N):
if X[i] <= l:
print(r)
elif X[i] >= r:
print(l) |
s802619017 | p02846 | u968846084 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,192 | 673 | Takahashi and Aoki are training for long-distance races in an infinitely long straight course running from west to east. They start simultaneously at the same point and moves as follows **towards the east** : * Takahashi runs A_1 meters per minute for the first T_1 minutes, then runs at A_2 meters per minute for the subsequent T_2 minutes, and alternates between these two modes forever. * Aoki runs B_1 meters per minute for the first T_1 minutes, then runs at B_2 meters per minute for the subsequent T_2 minutes, and alternates between these two modes forever. How many times will Takahashi and Aoki meet each other, that is, come to the same point? We do not count the start of the run. If they meet infinitely many times, report that fact. | t=list(map(int,input().split()))
a=list(map(int,input().split()))
b=list(map(int,input().split()))
if t[0]*a[0]+t[1]*a[1]==t[0]*b[0]+t[1]*b[1]:
print("infinity")
else:
if t[0]*a[0]>t[0]*b[0]:
if t[0]*a[0]+t[1]*a[1]>t[0]*b[0]+t[1]*b[1]:
print(0)
else:
d=(t[0]*b[0]+t[1]*b[1])-(t[0]*a[0]+t[1]*a[1])
c=t[0]*a[0]-t[0]*b[0]
if c//d==0:
print(c//d*2)
else:
print(c//d*2+1)
else:
if t[0]*a[0]+t[1]*a[1]<t[0]*b[0]+t[1]*b[1]:
print(0)
else:
d=(t[0]*a[0]+t[1]*a[1])-(t[0]*b[0]+t[1]*b[1])
c=t[0]*b[0]-t[0]*a[0]
if c//d==0:
print(c//d*2)
else:
print(c//d*2+1)
| s171888966 | Accepted | 17 | 3,064 | 671 | t=list(map(int,input().split()))
a=list(map(int,input().split()))
b=list(map(int,input().split()))
if t[0]*a[0]+t[1]*a[1]==t[0]*b[0]+t[1]*b[1]:
print("infinity")
else:
if t[0]*a[0]>t[0]*b[0]:
if t[0]*a[0]+t[1]*a[1]>t[0]*b[0]+t[1]*b[1]:
print(0)
else:
d=(t[0]*b[0]+t[1]*b[1])-(t[0]*a[0]+t[1]*a[1])
c=t[0]*a[0]-t[0]*b[0]
if c%d==0:
print(c//d*2)
else:
print(c//d*2+1)
else:
if t[0]*a[0]+t[1]*a[1]<t[0]*b[0]+t[1]*b[1]:
print(0)
else:
d=(t[0]*a[0]+t[1]*a[1])-(t[0]*b[0]+t[1]*b[1])
c=t[0]*b[0]-t[0]*a[0]
if c%d==0:
print(c//d*2)
else:
print(c//d*2+1)
|
s416502772 | p02612 | u604055101 | 2,000 | 1,048,576 | Wrong Answer | 26 | 9,132 | 31 | We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required. | n = int(input())
print(n%1000)
| s034735478 | Accepted | 30 | 9,140 | 76 | n = int(input())
if n%1000 == 0:
print(0)
else:
print(1000-n%1000)
|
s484836017 | p02392 | u885889402 | 1,000 | 131,072 | Wrong Answer | 20 | 7,540 | 84 | Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No". | s=input()
a,b,c=map(int,s.split())
if(a<b<c):
print("yes")
else:
print("No") | s177839143 | Accepted | 20 | 7,680 | 90 | s=input()
a,b,c=map(int,s.split())
if(a<b and b<c):
print("Yes")
else:
print("No") |
s640815225 | p03777 | u234631479 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 103 | Two deer, AtCoDeer and TopCoDeer, are playing a game called _Honest or Dishonest_. In this game, an honest player always tells the truth, and an dishonest player always tell lies. You are given two characters a and b as the input. Each of them is either `H` or `D`, and carries the following information: If a=`H`, AtCoDeer is honest; if a=`D`, AtCoDeer is dishonest. If b=`H`, AtCoDeer is saying that TopCoDeer is honest; if b=`D`, AtCoDeer is saying that TopCoDeer is dishonest. Given this information, determine whether TopCoDeer is honest. | a,b = input().split()
print(a,b)
if a == "H":
print(b)
elif b == "H":
print("D")
else:
print("H") | s531674962 | Accepted | 17 | 2,940 | 92 | a,b = input().split()
if a == "H":
print(b)
elif b == "H":
print("D")
else:
print("H") |
s154960453 | p03779 | u870518235 | 2,000 | 262,144 | Wrong Answer | 34 | 9,140 | 117 | There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X. | # ABC056_C
X = int(input())
sum = 0
for i in range(1,X+1):
sum = sum + i
if sum > X:
break
print(i) | s039298929 | Accepted | 35 | 9,140 | 118 | # ABC056_C
X = int(input())
sum = 0
for i in range(1,X+1):
sum = sum + i
if sum >= X:
break
print(i) |
s234530260 | p03457 | u112523623 | 2,000 | 262,144 | Wrong Answer | 1,020 | 5,876 | 730 | AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan. | if __name__ == "__main__":
n = int(input())
cur_x = 0
cur_y = 0
cur_t = 0
can_travel = True
for _ in range(n):
next_t, next_x, next_y = list(map(int, input().split(" ")))
time_diff = abs(next_t - cur_t)
x_diff = abs(next_x - cur_x)
y_diff = abs(next_y - cur_y)
print("time_diff:{}, shortest_diff:{}".format(time_diff, (x_diff + y_diff)))
if time_diff < (x_diff + y_diff):
can_travel = False
elif time_diff > (x_diff + y_diff):
can_travel = ((time_diff - (x_diff + y_diff)) % 2 == 0)
cur_t, cur_x, cur_y = next_t, next_x, next_y
if not can_travel:
break
print("Yes" if can_travel else "No")
| s481123756 | Accepted | 415 | 3,064 | 601 | if __name__ == "__main__":
n = int(input())
cur_x = 0
cur_y = 0
cur_t = 0
can_travel = True
for _ in range(n):
next_t, next_x, next_y = list(map(int, input().split(" ")))
time_diff = abs(next_t - cur_t)
xy_diff = abs(next_x - cur_x) + abs(next_y - cur_y)
if time_diff < xy_diff:
can_travel = False
elif time_diff > xy_diff:
can_travel = ((time_diff - xy_diff) % 2 == 0)
if not can_travel:
break
cur_t, cur_x, cur_y = next_t, next_x, next_y
print("Yes" if can_travel else "No")
|
s969163404 | p00003 | u150984829 | 1,000 | 131,072 | Wrong Answer | 30 | 5,612 | 101 | 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. | for _ in[0]*int(input()):
a,b,c=map(int,sorted(input().split()))
print(['No','Yes'][a*a+b*b==c*c])
| s228967488 | Accepted | 30 | 5,640 | 119 | import sys
input()
for a,b,c in map(lambda x:sorted(map(int,x.split())),sys.stdin):
print(['NO','YES'][a*a+b*b==c*c])
|
s624466578 | p03860 | u867848444 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 25 | Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name. | print('A'+input()[8]+'c') | s929781007 | Accepted | 17 | 2,940 | 49 | a,b,c=input().split()
print('A',b[0],'C',sep='')
|
s349187951 | p03386 | u200239931 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 917 | Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers. | import math
import sys
def getinputdata():
array_result = []
data = input()
array_result.append(data.split(" "))
flg = True
try:
while flg:
data = input()
array_temp = []
if(data != ""):
array_result.append(data.split(" "))
flg = True
else:
flg = False
finally:
return array_result
arr_data = getinputdata()
a = int(arr_data[0][0])
b = int(arr_data[0][1])
k = int(arr_data[0][2])
#print(a,b,k)
s1=set()
cnt=0
for i in range(a,b+1):
# print(i)
if cnt<k:
s1.add(i)
cnt+=1
else:
break
cnt=0
for i in range(b,a-1,-1):
if cnt<k:
s1.add(i)
cnt+=1
else:
break
print("\n".join(list(map(str, list(s1)))))
| s613082086 | Accepted | 18 | 3,064 | 1,009 | import math
import sys
def getinputdata():
array_result = []
data = input()
array_result.append(data.split(" "))
flg = True
try:
while flg:
data = input()
array_temp = []
if(data != ""):
array_result.append(data.split(" "))
flg = True
else:
flg = False
finally:
return array_result
arr_data = getinputdata()
a = int(arr_data[0][0])
b = int(arr_data[0][1])
k = int(arr_data[0][2])
#print(a,b,k)
s1=set()
cnt=0
for i in range(a,b+1):
# print(i)
if cnt<k:
s1.add(i)
cnt+=1
else:
break
cnt=0
for i in range(b,a-1,-1):
if cnt<k:
s1.add(i)
cnt+=1
else:
break
#print(s1)
list01=sorted(list(s1),reverse=False)
#print(list01)
#print(list(map(str, list01)))
print("\n".join(list(map(str, list01))))
|
s504467394 | p03599 | u101490607 | 3,000 | 262,144 | Wrong Answer | 148 | 12,456 | 828 | Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the beaker. * Operation 3: Put C grams of sugar into the beaker. * Operation 4: Put D grams of sugar into the beaker. In our experimental environment, E grams of sugar can dissolve into 100 grams of water. Snuke will make sugar water with the highest possible density. The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker. Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it. If there is more than one candidate, any of them will be accepted. We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent. Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water. | a,b,c,d,e,f = map(int, input().split())
x = []
y = []
k=0
i = 0
j = 0
#print(a,b,c,d,e,f)
while( 100*a*i + 100*b*j < f ):
while( 100*a*i + 100*b*j < f):
x.append(100*a*i + 100*b*j)
j +=1
j = 0
i +=1
i = 0
j = 0
while( c*i + d*j < f ):
while( c*i + d*j < f):
y.append(c*i + d*j)
j +=1
j = 0
i +=1
x = list(set(x))
y = list(set(y))
x.remove(0)
y.remove(0)
#print( x )
noudo = 0
reg_i = 0
reg_j = 0
for i in range(len(x)):
if x[i] > f:
break
for j in range(len(y)):
if x[i]+y[j] > f:
break
tmp_noudo = 100 * y[j] / (x[i] + y[j])
if tmp_noudo <= 100 * e / (100 + e):
if noudo < tmp_noudo:
noudo = tmp_noudo
reg_i = i
reg_j = j
print( x[reg_i]+y[reg_j] , y[reg_j] )
#print( eki, satou)
| s164041129 | Accepted | 135 | 12,476 | 758 | a,b,c,d,e,f = map(int, input().split())
x = []
y = []
k=0
i = 0
j = 0
#print(a,b,c,d,e,f)
while( 100*a*i + 100*b*j <= f ):
while( 100*a*i + 100*b*j <= f):
x.append(100*a*i + 100*b*j)
j +=1
j = 0
i +=1
i = 0
j = 0
while( c*i + d*j <= f ):
while( c*i + d*j <= f):
y.append(c*i + d*j)
j +=1
j = 0
i +=1
x = list(set(x))
y = list(set(y))
x.remove(0)
#y.remove(0)
#print( x )
noudo = 0
reg_i = 0
reg_j = 0
for i in range(len(x)):
for j in range(len(y)):
tmp_noudo = 100 * y[j] / (x[i] + y[j])
if tmp_noudo <= 100 * e / (100 + e) and (x[i] + y[j]) <= f:
if noudo < tmp_noudo:
noudo = tmp_noudo
reg_i = i
reg_j = j
print( x[reg_i]+y[reg_j] , y[reg_j] ) |
s739622853 | p02379 | u216425054 | 1,000 | 131,072 | Wrong Answer | 20 | 7,444 | 112 | Write a program which calculates the distance between two points P1(x1, y1) and P2(x2, y2). | import math
x1, x2, y1, y2 = map(float,input().split())
print("{0:8f}".format(math.sqrt((x1-x2)**2+(y1-y1)**2))) | s782047438 | Accepted | 20 | 7,524 | 110 | import math
x1,y1,x2,y2 = map(float,input().split())
print("{0:.8f}".format(math.sqrt((x1-x2)**2+(y1-y2)**2))) |
s283703959 | p03407 | u695429668 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 122 | 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. | # coding: utf-8
A, B, C = list(map(int, input().split()))
ans = A+B
if ans <= C:
print("Yes")
else:
print("No")
| s052495553 | Accepted | 17 | 2,940 | 122 | # coding: utf-8
A, B, C = list(map(int, input().split()))
ans = A+B
if ans >= C:
print("Yes")
else:
print("No")
|
s551192497 | p03605 | u077291787 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 153 | It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N? | # ABC073A - September 9
def main():
N = input().rstrip()
flg = "g" in N
print("Yes" if flg else "No")
if __name__ == "__main__":
main() | s327167228 | Accepted | 18 | 2,940 | 153 | # ABC073A - September 9
def main():
N = input().rstrip()
flg = "9" in N
print("Yes" if flg else "No")
if __name__ == "__main__":
main() |
s826748773 | p00067 | u093607836 | 1,000 | 131,072 | Wrong Answer | 40 | 6,760 | 965 | 地勢を示す縦 12, 横 12 のマスからなる平面図があります。おのおののマスは白か黒に塗られています。白は海を、黒は陸地を表します。二つの黒いマスが上下、あるいは左右に接しているとき、これらは地続きであるといいます。この平面図では、黒いマス一つのみ、あるいは地続きの黒いマスが作る領域を「島」といいます。例えば下図には、5 つの島があります。 ■■■■□□□□■■■■ ■■■□□□□□■■■■ ■■□□□□□□■■■■ ■□□□□□□□■■■■ □□□■□□□■□□□□ □□□□□□■■■□□□ □□□□□■■■■■□□ ■□□□■■■■■■■□ ■■□□□■■■■■□□ ■■■□□□■■■□□□ ■■■■□□□■□□□□ □□□□□□□□□□□□ マスのデータを読み込んで、島の数を出力するプログラムを作成してください。 | import itertools
def getPoint(m,x,y):
if not (0 <= x < 12) or not (0 <= y < 12):
return 0
return m[y][x]
def setPoint(m,x,y,v):
m[y][x] = v
while 1:
try:
m = [list(map(int,input().strip())) for _ in range(12)]
count = 0
empty = False
while not empty:
empty = True
for x,y in itertools.product(range(12),repeat=2):
if(getPoint(m,x,y)):
print(x,y)
count += 1
empty = False
q = []
q.append((x,y))
while q != []:
p = q.pop()
if(getPoint(m,p[0],p[1])):
setPoint(m,p[0],p[1],0)
for o in [(1,0),(-1,0),(0,1),(0,-1)]:
q.append((p[0] + o[0], p[1] + o[1]))
print(count)
input()
except:
exit() | s397588046 | Accepted | 40 | 6,756 | 934 | import itertools
def getPoint(m,x,y):
if not (0 <= x < 12) or not (0 <= y < 12):
return 0
return m[y][x]
def setPoint(m,x,y,v):
m[y][x] = v
while 1:
try:
m = [list(map(int,input().strip())) for _ in range(12)]
count = 0
empty = False
while not empty:
empty = True
for x,y in itertools.product(range(12),repeat=2):
if(getPoint(m,x,y)):
count += 1
empty = False
q = []
q.append((x,y))
while q != []:
p = q.pop()
if(getPoint(m,p[0],p[1])):
setPoint(m,p[0],p[1],0)
for o in [(1,0),(-1,0),(0,1),(0,-1)]:
q.append((p[0] + o[0], p[1] + o[1]))
print(count)
input()
except:
exit() |
s771613992 | p03476 | u785989355 | 2,000 | 262,144 | Wrong Answer | 1,029 | 8,704 | 524 | 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. | def sieve_of_eratosthenes(n):
pl = [i for i in range(2, n + 1)]
i=0
p = pl[0]
while p**2<=n:
pl = [pl[j] for j in range(len(pl)) if pl[j]%p!=0 or j<=i]
i+=1
p=pl[i]
return pl
pl=sieve_of_eratosthenes(100000)
p_list = [0 for i in range(100001)]
for p in pl:
p_list[p]+=1
ps_list = [0]
ps = 0
for i in range(1,100001):
ps+=p_list[i]
ps_list.append(ps)
Q = int(input())
for i in range(Q):
l,r = list(map(int,input().split()))
print(ps_list[r]-ps_list[l-1]) | s927994071 | Accepted | 1,038 | 8,320 | 591 | def sieve_of_eratosthenes(n):
pl = [i for i in range(2, n + 1)]
i=0
p = pl[0]
while p**2<=n:
pl = [pl[j] for j in range(len(pl)) if pl[j]%p!=0 or j<=i]
i+=1
p=pl[i]
return pl
pl=sieve_of_eratosthenes(100000)
p_list = [0 for i in range(100001)]
for p in pl:
if 2*p-1<=100000:
p_list[2*p-1]+=1
p_list[p]+=1
ps_list = [0]
ps = 0
for i in range(1,100001):
if p_list[i]>=2:
ps+=1
ps_list.append(ps)
Q = int(input())
for i in range(Q):
l,r = list(map(int,input().split()))
print(ps_list[r]-ps_list[l-1]) |
s552780697 | p02406 | u643542669 | 1,000 | 131,072 | Wrong Answer | 20 | 5,580 | 92 | In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; } | import sys
n = int(input())
for i in range(3, n + 1, 3):
sys.stdout.write("%d " % (i))
| s335551620 | Accepted | 20 | 5,640 | 127 | n = input()
out = ""
for i in range(1, int(n) + 1):
if i % 3 == 0 or "3" in str(i):
out += " " + str(i)
print(out)
|
s675340428 | p03476 | u993435350 | 2,000 | 262,144 | Wrong Answer | 2,104 | 4,636 | 491 | 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. | def isprime(n):
m = (n + 1) // 2
i = 2
if m in primedic and n in primedic:
return True
while i * i <= m:
if m % i == 0:
return False
i += 1
i -= 1
while i * i <= n:
if n % i == 0:
return False
i += 1
return True
primedic = {}
Q = int(input())
for i in range(Q):
l,r = map(int,input().split())
ans = 0
for i in range(l,r + 1):
if isprime(i):
primedic[i] = True
primedic[(i + 1) // 2] = True
ans += 1
print(ans) | s401195801 | Accepted | 533 | 5,432 | 432 | import sys
input = sys.stdin.readline
def isprime(x):
if x <= 1:
return False
for i in range(2, int(x**0.5) + 1):
if x % i == 0:
return False
return True
A = [0] * 10 ** 5
B = [0] * 10 ** 5
con = 0
for i in range(10 ** 5):
if isprime(i):
A[i] = 1
if A[i] and A[(i + 1) // 2]:
con += 1
B[i] = con
Q = int(input())
for _ in range(Q):
l, r = map(int, input().split())
print(B[r] - B[l - 1])
|
s221285737 | p02833 | u437860615 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 66 | For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N). | def func(n):
if n < 2:
return 1
return n * f(n-2)
func(1) | s458959888 | Accepted | 17 | 2,940 | 122 | n = int(input())
a = 0
i = 1
if n % 2 == 0:
while 2 * (5 ** i) <= n:
a += n // (2 * (5 ** i))
i += 1
print(a) |
s404226288 | p03556 | u840958781 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 37 | Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer. | n = int(input())
print(n ** 0.5 ** 2) | s108127660 | Accepted | 17 | 2,940 | 42 | n = int(input())
print(int(n ** 0.5) ** 2) |
s585889538 | p03477 | u693953100 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 126 | A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right. | a,b,c,d = map(int,input().split())
if a+b<c+d:
print('Left')
elif a+b==c+d:
print('Balanced')
else:
print('Right') | s200922843 | Accepted | 17 | 2,940 | 126 | a,b,c,d = map(int,input().split())
if a+b>c+d:
print('Left')
elif a+b==c+d:
print('Balanced')
else:
print('Right') |
s214476912 | p03050 | u522726434 | 2,000 | 1,048,576 | Wrong Answer | 188 | 3,060 | 167 | Snuke received a positive integer N from Takahashi. A positive integer m is called a _favorite number_ when the following condition is satisfied: * The quotient and remainder of N divided by m are equal, that is, \lfloor \frac{N}{m} \rfloor = N \bmod m holds. Find all favorite numbers and print the sum of those. | import math
N = int(input())
max_m = math.floor(math.sqrt(N) - 1)
count = 0
for i in range(1, max_m+1):
if (N - i) % i == 0:
count += (N - i) // i
print(count) | s429604881 | Accepted | 525 | 3,064 | 182 | import math
N = int(input())
count = 0
m = 1
while m**2 < N:
if (N - m) % m == 0:
fav_num = (N - m) // m
if fav_num > m:
count += fav_num
m += 1
print(count) |
s098567241 | p03612 | u530844517 | 2,000 | 262,144 | Wrong Answer | 124 | 13,880 | 305 | You are given a permutation p_1,p_2,...,p_N consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero): Operation: Swap two **adjacent** elements in the permutation. You want to have p_i ≠ i for all 1≤i≤N. Find the minimum required number of operations to achieve this. | n = int(input())
p = list(map(int, input().split()))
swap = 0
flag = 0
while True:
flag = 0
for i in range(n-1):
if p[i] == i+1:
p[i], p[i+1] = p[i+1], p[i]
flag = 1
swap += 1
print(swap)
if flag == 0:
print(swap)
break | s331747545 | Accepted | 88 | 13,880 | 388 | n = int(input())
p = list(map(int, input().split()))
swap = 0
flag = 0
while True:
flag = 0
for i in range(n-1):
if p[i] == i+1:
p[i], p[i+1] = p[i+1], p[i]
flag = 1
swap += 1
if p[n-1] == n:
p[n-1], p[n-2] = p[n-2], p[n-1]
flag = 1
swap += 1
if flag == 0:
print(swap)
break |
s506176031 | p03434 | u652150585 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 119 | 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. | num=int(input())
n=list(map(int,input().split()))
n.sort(reverse=True)
print(n)
a=sum(n[::2])
b=sum(n[1::2])
print(a-b) | s604773880 | Accepted | 17 | 2,940 | 120 | num=int(input())
n=list(map(int,input().split()))
n.sort(reverse=True)
#print(n)
a=sum(n[::2])
b=sum(n[1::2])
print(a-b) |
s358518791 | p03079 | u927764913 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 66 | You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C. | a,b,c = map(int,input().split())
if a == b == c:
print("yes") | s756226492 | Accepted | 17 | 2,940 | 85 | a,b,c = map(int,input().split())
if a == b == c:
print("Yes")
else :
print("No") |
s037421120 | p03469 | u158225062 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 75 | On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it. | # coding: utf-8
N = input()
N_list = list(N)
N_list[3] = 8
print(N_list) | s758511323 | Accepted | 681 | 15,416 | 124 | # coding: utf-8
import numpy
N = input()
N_list = list(N)
map(str,N_list)
N_list[3] = "8"
ans = ''.join(N_list)
print(ans) |
s435431731 | p03386 | u550014122 | 2,000 | 262,144 | Wrong Answer | 29 | 9,116 | 154 | Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers. | a,b,k = map(int, input().split())
s = set()
for i in range(a,min(a+k,b+1)):
s.add(i)
for i in range(max(a,b-k+1),b+1):
s.add(i)
for i in s:
print(i) | s920701262 | Accepted | 25 | 9,168 | 119 | a,b,k=map(int,input().split())
b+=1
for i in range(a,min(a+k,b)):
print(i)
for i in range(max(b-k,a+k),b):
print(i) |
s502455768 | p02927 | u629540524 | 2,000 | 1,048,576 | Wrong Answer | 30 | 9,056 | 104 | Today is August 24, one of the five Product Days in a year. A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day): * d_1 \geq 2 * d_{10} \geq 2 * d_1 \times d_{10} = m Takahashi wants more Product Days, and he made a new calendar called Takahashi Calendar where a year consists of M month from Month 1 to Month M, and each month consists of D days from Day 1 to Day D. In Takahashi Calendar, how many Product Days does a year have? | m,d = map(int,input().split())
c = 0
for i in range(22,d+1):
if i//10*i%10<=m:
c+=1
print(c) | s936067555 | Accepted | 26 | 9,096 | 118 | m,d = map(int,input().split())
c = 0
for i in range(22,d+1):
if i//10*(i%10)<=m and i%10>=2:
c+=1
print(c) |
s478157315 | p03478 | u849622363 | 2,000 | 262,144 | Wrong Answer | 27 | 3,064 | 324 | Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive). | n,a,b = map(int, input().split())
counter = 0
for i in range(1,n+1):
thou = i // 1000
hund = (i - thou * 1000) // 100
ten = (i - thou * 1000 - hund * 100) // 10
one = i - thou * 1000 - hund * 100 - ten * 10
sum_all = thou + hund + ten + one
if sum_all >= a and sum_all <= b:
counter += 1
print(counter) | s695052194 | Accepted | 39 | 3,060 | 413 | n,a,b = map(int, input().split())
counter = 0
for i in range(1,n+1):
tent = i //10000
thou = (i - tent * 10000) // 1000
hund = (i - thou * 1000 - tent * 10000) // 100
ten = (i - thou * 1000 - hund * 100 - tent * 10000) // 10
one = i - thou * 1000 - hund * 100 - ten * 10 - tent * 10000
sum_all = tent + thou + hund + ten + one
if sum_all >= a and sum_all <= b:
counter += i
print(counter)
|
s299353971 | p03730 | u488884575 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 154 | We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`. | a,b,c = map(int, input().split())
ans = 0
for i in range(1, 200):
if i*a%b == c:
ans = 1
if ans == 1:
print('Yes')
else:
print('No') | s363023875 | Accepted | 17 | 3,060 | 154 | a,b,c = map(int, input().split())
ans = 0
for i in range(1, 200):
if i*a%b == c:
ans = 1
if ans == 1:
print('YES')
else:
print('NO') |
s640298132 | p03388 | u383196771 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 424 | 10^{10^{10}} participants, including Takahashi, competed in two programming contests. In each contest, all participants had distinct ranks from first through 10^{10^{10}}-th. The _score_ of a participant is the product of his/her ranks in the two contests. Process the following Q queries: * In the i-th query, you are given two positive integers A_i and B_i. Assuming that Takahashi was ranked A_i-th in the first contest and B_i-th in the second contest, find the maximum possible number of participants whose scores are smaller than Takahashi's. | import math
def worst_case(A, B):
c = min(A, B)
d = max(A, B)
if c == d or c + 1 == d:
return 2 * (c - 1)
root = int(math.sqrt(A * B))
return 2 * (root - 1) + 1 if root * (root + 1) < A * B else 2 * (root - 1)
Q = int(input())
A = []
B = []
for i in range(Q):
line = input().split()
A.append(int(line[0]))
B.append(int(line[1]))
for i in range(Q):
print(worst_case(A[i], B[i])) | s955271113 | Accepted | 18 | 3,064 | 469 | import math
def worst_case(A, B):
c = min(A, B)
d = max(A, B)
if c == d or c + 1 == d:
return 2 * (c - 1)
root = int(math.sqrt(A * B))
if root ** 2 == A * B:
root -= 1
return 2 * (root - 1) + 1 if root * (root + 1) < A * B else 2 * (root - 1)
Q = int(input())
A = []
B = []
for i in range(Q):
line = input().split()
A.append(int(line[0]))
B.append(int(line[1]))
for i in range(Q):
print(worst_case(A[i], B[i])) |
s593158690 | p02406 | u908651435 | 1,000 | 131,072 | Wrong Answer | 20 | 5,588 | 135 | In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; } | n=int(input())
for i in range(n+1):
if i%3==0:
print(' '+str(i),end='')
elif i%10==3:
print(' '+str(i),end='')
| s264079542 | Accepted | 20 | 5,624 | 244 | n = int(input())
ans = ""
for i in range(1, n + 1):
if(i % 3 == 0):
ans += " " + str(i)
continue
x = i
while(x):
if(x % 10 == 3):
ans += " " + str(i)
break
x //= 10
print(ans)
|
s385470811 | p03407 | u514118270 | 2,000 | 262,144 | Wrong Answer | 39 | 9,976 | 534 | 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. | import sys
import math
import itertools
import bisect
from copy import copy
from collections import deque,Counter
from decimal import Decimal
def s(): return input()
def i(): return int(input())
def S(): return input().split()
def I(): return map(int,input().split())
def L(): return list(input().split())
def l(): return list(map(int,input().split()))
def lcm(a,b): return a*b//math.gcd(a,b)
sys.setrecursionlimit(10 ** 9)
INF = 10**9
mod = 10**9+7
A,B,C = I()
if C == A+B or C == A or C == B:
print('Yes')
else:
print('No') | s083772421 | Accepted | 34 | 9,952 | 514 | import sys
import math
import itertools
import bisect
from copy import copy
from collections import deque,Counter
from decimal import Decimal
def s(): return input()
def i(): return int(input())
def S(): return input().split()
def I(): return map(int,input().split())
def L(): return list(input().split())
def l(): return list(map(int,input().split()))
def lcm(a,b): return a*b//math.gcd(a,b)
sys.setrecursionlimit(10 ** 9)
INF = 10**9
mod = 10**9+7
A,B,C = I()
if C <= A+B:
print('Yes')
else:
print('No') |
s001820328 | p02255 | u630566146 | 1,000 | 131,072 | Wrong Answer | 20 | 5,604 | 359 | Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step. | def insertion_sort(l):
for i in range(1, len(l)):
tmp = l[i]
j = i - 1
while j >= 0 and l[j] > tmp:
l[j+1] = l[j]
j -= 1
l[j+1] = tmp
print(' '.join(map(str, l)))
return l
if __name__ == '__main__':
N = int(input())
l = list(map(int, input().split()))
insertion_sort(l)
| s747664855 | Accepted | 20 | 5,604 | 392 | def insertion_sort(l):
for i in range(1, len(l)):
tmp = l[i]
j = i - 1
while j >= 0 and l[j] > tmp:
l[j+1] = l[j]
j -= 1
l[j+1] = tmp
print(' '.join(map(str, l)))
return l
if __name__ == '__main__':
N = int(input())
l = list(map(int, input().split()))
print(' '.join(map(str, l)))
insertion_sort(l)
|
s688641368 | p03493 | u105210954 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 67 | Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble. | s = input()
n= 0
for c in s:
if s == '1':
c += 1
print(c) | s153547741 | Accepted | 17 | 2,940 | 67 | s = input()
n= 0
for c in s:
if c == '1':
n += 1
print(n) |
s147990820 | p03023 | u735975757 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 97 | Given an integer N not less than 3, find the sum of the interior angles of a regular polygon with N sides. Print the answer in degrees, but do not print units. | S = input()
num = S.count("○")
print(num)
if num >= 8 :
print("YES")
else :
print("NO") | s484151791 | Accepted | 18 | 2,940 | 39 | N = int(input())
r = (N-2)*180
print(r) |
s278141187 | p03448 | u528124741 | 2,000 | 262,144 | Wrong Answer | 50 | 3,060 | 210 | You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different. | a = int(input())
b = int(input())
c = int(input())
x = int(input())
ret = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if (i*500 + i*100 + k*50) == x:
ret += 1
print(ret) | s499655117 | Accepted | 47 | 3,060 | 210 | a = int(input())
b = int(input())
c = int(input())
x = int(input())
ret = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if (i*500 + j*100 + k*50) == x:
ret += 1
print(ret) |
s355570440 | p03305 | u281610856 | 2,000 | 1,048,576 | Wrong Answer | 1,163 | 58,912 | 1,347 | Kenkoooo is planning a trip in Republic of Snuke. In this country, there are n cities and m trains running. The cities are numbered 1 through n, and the i-th train connects City u_i and v_i bidirectionally. Any city can be reached from any city by changing trains. Two currencies are used in the country: yen and snuuk. Any train fare can be paid by both yen and snuuk. The fare of the i-th train is a_i yen if paid in yen, and b_i snuuk if paid in snuuk. In a city with a money exchange office, you can change 1 yen into 1 snuuk. However, when you do a money exchange, you have to change all your yen into snuuk. That is, if Kenkoooo does a money exchange when he has X yen, he will then have X snuuk. Currently, there is a money exchange office in every city, but the office in City i will shut down in i years and can never be used in and after that year. Kenkoooo is planning to depart City s with 10^{15} yen in his pocket and head for City t, and change his yen into snuuk in some city while traveling. It is acceptable to do the exchange in City s or City t. Kenkoooo would like to have as much snuuk as possible when he reaches City t by making the optimal choices for the route to travel and the city to do the exchange. For each i=0,...,n-1, find the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i years. You can assume that the trip finishes within the year. | from heapq import heappush, heappop
INF = float("inf")
def Dijkstra(n, s, g, mode):
dist = [INF] * n
dist[s] = 0
hq = [(0, s)]
while hq:
d, v = heappop(hq)
if dist[v] < d:
continue
for w, *a in g[v]:
nd = d + a[mode]
if dist[w] > nd:
dist[w] = nd
heappush(hq, (nd, w))
return dist
def main():
n, m, s, t = map(int, input().split())
s -= 1
t -= 1
g = [[] for _ in range(n)]
for _ in range(m):
u, v, a, b = map(int, input().split())
u -= 1
v -= 1
g[u].append((v, a, b))
g[v].append((u, a, b))
d_yen = Dijkstra(n, s, g, 0)
d_sunnk = Dijkstra(n, t, g, 1)
ans = [0] * n
cost = float("inf")
for i in range(n-1, -1, -1):
cost = min(cost, d_yen[i] + d_sunnk[i])
ans[i-1] = 10 ** 15 - cost
print(*ans, sep='\n')
if __name__ == "__main__":
main()
| s868674758 | Accepted | 1,152 | 58,936 | 1,345 | from heapq import heappush, heappop
INF = float("inf")
def Dijkstra(n, s, g, mode):
dist = [INF] * n
dist[s] = 0
hq = [(0, s)]
while hq:
d, v = heappop(hq)
if dist[v] < d:
continue
for w, *a in g[v]:
nd = d + a[mode]
if dist[w] > nd:
dist[w] = nd
heappush(hq, (nd, w))
return dist
def main():
n, m, s, t = map(int, input().split())
s -= 1
t -= 1
g = [[] for _ in range(n)]
for _ in range(m):
u, v, a, b = map(int, input().split())
u -= 1
v -= 1
g[u].append((v, a, b))
g[v].append((u, a, b))
d_yen = Dijkstra(n, s, g, 0)
d_sunnk = Dijkstra(n, t, g, 1)
ans = [0] * n
cost = float("inf")
for i in range(n-1, -1, -1):
cost = min(cost, d_yen[i] + d_sunnk[i])
ans[i] = 10 ** 15 - cost
print(*ans, sep='\n')
if __name__ == "__main__":
main()
|
s729597653 | p03494 | u256106029 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 265 | There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform. |
n = int(input())
a = list(map(int, input().strip().split()))
count = 0
for i in range(n):
if i + 1 != n:
if a[i] % 2 == 0:
continue
else:
break
else:
if a[i] % 2 == 0:
count += 1
print(count)
| s533514900 | Accepted | 19 | 3,064 | 512 |
n = int(input())
a = list(map(int, input().strip().split()))
c = 0
def func(num: int, array: [int], count) -> int:
for i in range(num):
if i + 1 != num:
if array[i] % 2 == 0:
array[i] = array[i] / 2
continue
else:
break
else:
if array[i] % 2 == 0:
count += 1
array[i] = array[i] / 2
count = func(num, array, count)
return count
print(func(n, a, c))
|
s547214093 | p03963 | u901598613 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 51 | There are N balls placed in a row. AtCoDeer the deer is painting each of these in one of the K colors of his paint cans. For aesthetic reasons, any two adjacent balls must be painted in different colors. Find the number of the possible ways to paint the balls. | n,m=1,10
for i in range(n-1):
m*=(m-1)
print(m) | s690187385 | Accepted | 18 | 2,940 | 56 | n,m=map(int,input().split())
k=m*((m-1)**(n-1))
print(k) |
s426179236 | p03089 | u952467214 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 231 | 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. | n = int(input())
b = list( map(int, input().split()))
ans = []
while b:
for i in reversed(range(len(b))):
if b[i] == i+1:
ans.append(b.pop(i))
break
else:
print(-1)
exit()
| s945905663 | Accepted | 19 | 3,060 | 333 | import sys
sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline
n = int(input())
b = list( map(int, input().split()))
ans = []
while b:
for i in reversed(range(len(b))):
if b[i] == i+1:
ans.append(b.pop(i))
break
else:
print(-1)
exit()
print(*reversed(ans),sep='\n')
|
s871185829 | p03448 | u096983897 | 2,000 | 262,144 | Wrong Answer | 53 | 3,064 | 220 | You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different. | a = int(input())
b = int(input())
c = int(input())
x = int(input())
cnt = 0
for i in range(a):
for j in range(b):
for k in range(c):
sum = i*500 + j*100 + k*50
if sum == x:
cnt += 1
print(cnt) | s207516444 | Accepted | 55 | 3,064 | 226 | a = int(input())
b = int(input())
c = int(input())
x = int(input())
cnt = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
sum = i*500 + j*100 + k*50
if sum == x:
cnt += 1
print(cnt) |
s167724875 | p02284 | u569960318 | 2,000 | 131,072 | Wrong Answer | 30 | 7,740 | 1,740 | Write a program which performs the following operations to a binary search tree $T$ by adding the find operation to A: Binary Search Tree I. * insert $k$: Insert a node containing $k$ as key into $T$. * find $k$: Report whether $T$ has a node containing $k$. * print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively. | class BinarySearchTree:
def __init__(self):
self.root = None
def insert(self, z):
p = None
x = self.root
while x != None:
p = x
if z.key < x.key : x = x.left
else : x = x.right
z.p = p
if p == None : self.root = z
elif z.key < p.key: p.left = z
else : p.right = z
def get_inorder_list(self):
def _get_inorder_list(root):
l = root.left
r = root.right
if l: yield from _get_inorder_list(l)
yield root.key
if r: yield from _get_inorder_list(r)
yield from _get_inorder_list(self.root)
def get_preorder_list(self):
def _get_preorder_list(root):
l = root.left
r = root.right
yield root.key
if l: yield from _get_preorder_list(l)
if r: yield from _get_preorder_list(r)
yield from _get_preorder_list(self.root)
def find(self, k):
def _find(k, x):
if x == None : return False
if x.key == k: return True
return _find(k, x.left) or _find(k, x.right)
return _find(k, self.root)
class Node:
def __init__(self, k):
self.key = k
self.p = None
self.left = None
self.right = None
if __name__=='__main__':
m = int(input())
T = BinarySearchTree()
for _ in range(m):
op = input().split()
if op[0] == 'insert': T.insert( Node(int(op[1])) )
elif op[0] == 'find':
print("Yes" if T.find(int(op[1])) else "No")
else:
print("",*T.get_inorder_list())
print("",*T.get_preorder_list()) | s938981115 | Accepted | 8,780 | 130,768 | 1,729 | class BinarySearchTree:
def __init__(self):
self.root = None
def insert(self, z):
p = None
x = self.root
while x != None:
p = x
if z.key < x.key : x = x.left
else : x = x.right
z.p = p
if p == None : self.root = z
elif z.key < p.key: p.left = z
else : p.right = z
def get_inorder_list(self):
def _get_inorder_list(root):
l = root.left
r = root.right
if l: yield from _get_inorder_list(l)
yield root.key
if r: yield from _get_inorder_list(r)
yield from _get_inorder_list(self.root)
def get_preorder_list(self):
def _get_preorder_list(root):
l = root.left
r = root.right
yield root.key
if l: yield from _get_preorder_list(l)
if r: yield from _get_preorder_list(r)
yield from _get_preorder_list(self.root)
def find(self, k):
x = self.root
while x:
if k == x.key: return True
if k < x.key: x = x.left
else : x = x.right
return False
class Node:
def __init__(self, k):
self.key = k
self.p = None
self.left = None
self.right = None
if __name__=='__main__':
m = int(input())
T = BinarySearchTree()
for _ in range(m):
op = input().split()
if op[0] == 'insert': T.insert( Node(int(op[1])) )
elif op[0] == 'find':
print("yes" if T.find(int(op[1])) else "no")
else:
print("",*T.get_inorder_list())
print("",*T.get_preorder_list()) |
s211970337 | p02833 | u798768533 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 8 | For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N). | print(0) | s074790374 | Accepted | 17 | 2,940 | 140 | N = int(input())
if N % 2 == 1:
print(0)
exit(0)
count = 0
base = 10
while base <= N:
count += N // base
base *= 5
print(count) |
s535119678 | p03644 | u600215903 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 136 | 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())
result = 1
for i in range(n):
if result <= 2:
result = result * 2
result = result / 2
print(int(result)) | s838071958 | Accepted | 17 | 2,940 | 83 | n = int(input())
result = 1
while result <= n:
result *= 2
print(result // 2) |
s197329945 | p02663 | u859059120 | 2,000 | 1,048,576 | Wrong Answer | 22 | 9,156 | 65 | In this problem, we use the 24-hour clock. Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.) He has decided to study for K consecutive minutes while he is up. What is the length of the period in which he can start studying? | h1,m1,h2,m2,k=map(int,input().split())
print(h1*60+m1-k-m2-60*h2) | s047911432 | Accepted | 24 | 9,148 | 66 | h1,m1,h2,m2,k=map(int,input().split())
print(h2*60+m2-k-m1-60*h1)
|
s041494663 | p00001 | u756242733 | 1,000 | 131,072 | Wrong Answer | 20 | 5,596 | 122 | There is a data which provides heights (in meter) of mountains. The data is only for ten mountains. Write a program which prints heights of the top three mountains in descending order. | hills = [input() for _ in range(10)]
hills = list(map(int, hills))
hills.sort
for i in range(3):
print(hills.pop())
| s081257433 | Accepted | 20 | 5,604 | 124 | hills = [input() for _ in range(10)]
hills = list(map(int, hills))
hills.sort()
for i in range(3):
print(hills.pop())
|
s835949358 | p00045 | u964040941 | 1,000 | 131,072 | Wrong Answer | 20 | 7,608 | 237 | 販売単価と販売数量を読み込んで、販売金額の総合計と販売数量の平均を出力するプログラムを作成してください。 | A = []
B = []
while True:
try:
x,y = map(int,input().split(','))
A.append(x)
B.append(y)
except EOFError:
break
N = len(A)
print(sum(A [i] * B [i] for i in range(N)),sum(B) * 2 // N - sum(B) // N) | s667745305 | Accepted | 30 | 7,572 | 244 | A = []
B = []
while True:
try:
x,y = map(int,input().split(','))
A.append(x)
B.append(y)
except EOFError:
break
N = len(A)
print(sum(A [i] * B [i] for i in range(N)))
print(sum(B) * 2 // N - sum(B) // N) |
s418922280 | p02401 | u884012707 | 1,000 | 131,072 | Wrong Answer | 20 | 7,616 | 254 | 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, op, b = input().split()
if op=="+":
print(int(a) + int(b))
elif op=="-":
print(int(a)-int(b))
elif op=="*":
print(int(a)*int(b))
elif op=="/":
print(int(a)/int(b))
else:
break | s938312099 | Accepted | 30 | 7,644 | 255 | while True:
a, op, b = input().split()
if op=="+":
print(int(a) + int(b))
elif op=="-":
print(int(a)-int(b))
elif op=="*":
print(int(a)*int(b))
elif op=="/":
print(int(a)//int(b))
else:
break |
s425864930 | p03712 | u367393577 | 2,000 | 262,144 | Wrong Answer | 22 | 3,316 | 603 | 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. | def twoDimArraywithwall(row,column,n,wallchar): return [[n if j!=0 and j!=(column+1) else wallchar for j in range(column+2)] if i!=0 and i!=(row+1) else [wallchar]*(column+2) for i in range(row+2)]
def maketwodimarraywithwall(matrix,wallchar):
rownum=len(matrix)
colnum=len(matrix[0])
ret=twoDimArraywithwall(rownum,colnum,None,wallchar)
for i in range(rownum):
for j in range(colnum):
ret[i+1][j+1]=matrix[i][j]
return ret
H, W = map(int, input().split())
table = []
for h in range(H):
table.append(list(input()))
print(maketwodimarraywithwall(table,"#")) | s692127071 | Accepted | 21 | 3,188 | 640 | def twoDimArraywithwall(row,column,n,wallchar): return [[n if j!=0 and j!=(column+1) else wallchar for j in range(column+2)] if i!=0 and i!=(row+1) else [wallchar]*(column+2) for i in range(row+2)]
def maketwodimarraywithwall(matrix,wallchar):
rownum=len(matrix)
colnum=len(matrix[0])
ret=twoDimArraywithwall(rownum,colnum,None,wallchar)
for i in range(rownum):
for j in range(colnum):
ret[i+1][j+1]=matrix[i][j]
return ret
H, W = map(int, input().split())
table = []
for h in range(H):
table.append(list(input()))
table=maketwodimarraywithwall(table,"#")
for i in table:
print("".join(i)) |
s122488592 | p03759 | u582084082 | 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 b - a == c - b:
print('Yes')
else:
print('No')
| s745633569 | Accepted | 18 | 2,940 | 94 | a, b, c = map(int, input().split())
if b - a == c - b:
print('YES')
else:
print('NO')
|
s011494895 | p03697 | u956547804 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 111 | You are given two integers A and B as the input. Output the value of A + B. However, if A + B is 10 or greater, output `error` instead. | s=input()
result=[]
ans='yes'
for i in s:
if i in result:
ans='no'
result.append(i)
print(ans)
| s159868860 | Accepted | 17 | 2,940 | 80 | a,b=map(int,input().split())
if a+b>=10:
print('error')
else:
print(a+b) |
s342494231 | p03997 | u904945034 | 2,000 | 262,144 | Wrong Answer | 28 | 8,976 | 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,b,h = [int(input()) for _ in range(3)]
print(((a + b)*h)/2) | s689523350 | Accepted | 26 | 9,028 | 68 | a,b,h = [int(input()) for _ in range(3)]
print(round(((a + b)*h)/2)) |
s947251059 | p03624 | u170183831 | 2,000 | 262,144 | Wrong Answer | 20 | 3,188 | 115 | 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. | diff = set(input()) - set('abcdefghijklmnopqrstuvwxyz')
if diff:
print(sorted(diff)[0])
else:
print('None') | s541191979 | Accepted | 19 | 3,188 | 116 | diff = set('abcdefghijklmnopqrstuvwxyz') - set(input())
if diff:
print(sorted(diff)[0])
else:
print('None')
|
s701987104 | p03575 | u316386814 | 2,000 | 262,144 | Wrong Answer | 103 | 3,436 | 916 | You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges. The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i. An edge whose removal disconnects the graph is called a _bridge_. Find the number of the edges that are bridges among the M edges. | from collections import deque
n, m = list(map(int, input().split()))
li = []
for _ in range(m):
li.append(list(map(int, input().split())))
# construct indirectional graphs
edges = [[] for _ in range(n)]
for i, j in li:
i -= 1
j -= 1
edges[i].append(j)
edges[j].append(i)
import copy
edges2 = copy.deepcopy(edges)
S = deque()
S.append([0])
while len(S) > 0:
path = S.pop()
# if full path,
if len(path) == n:
# used path list
used = [[] for _ in range(n)]
for i, j in zip(path[:-1], path[1:]):
used[i].append(j)
for i in range(len(edges2)):
for j in edges2[i][::-1]:
if j not in used[i]:
edges2[i].remove(j)
continue
for v in edges[path[-1]]:
if v not in path:
S.append(path + [v])
ans = sum(len(e) for e in edges2)
print(ans) | s565022261 | Accepted | 24 | 3,316 | 663 | from collections import deque
n, m = list(map(int, input().split()))
li = []
for _ in range(m):
li.append(list(map(int, input().split())))
ans = 0
for k in range(len(li)):
li0 = li[:k] + li[k + 1:]
# construct indirectional graphs
edges = [[] for _ in range(n)]
for i, j in li0:
i -= 1
j -= 1
edges[i].append(j)
edges[j].append(i)
S = deque()
S.append(0)
visited = set()
while len(S) > 0:
u = S.pop()
visited.add(u)
for v in edges[u]:
if v not in visited:
S.append(v)
if len(visited) < n:
ans += 1
print(ans) |
s690132614 | p03556 | u140251125 | 2,000 | 262,144 | Wrong Answer | 28 | 2,940 | 77 | 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. | # input
N = int(input())
i = 1
while i ** 2 <= N:
i += 1
print(i ** 2) | s280822688 | Accepted | 29 | 2,940 | 83 | # input
N = int(input())
i = 1
while i ** 2 <= N:
i += 1
print((i - 1) ** 2) |
s864157252 | p02842 | u805011545 | 2,000 | 1,048,576 | Wrong Answer | 26 | 9,164 | 98 | Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact. | N = int(input())
X = -(-N//1.08)
n = X*108//100
if(int(N)==int(n)):
print(X)
else:
print(':(') | s910036954 | Accepted | 31 | 9,144 | 103 | N = int(input())
X = -(-N//1.08)
n = X*108//100
if(int(N)==int(n)):
print(int(X))
else:
print(':(') |
s694998735 | p03836 | u280978334 | 2,000 | 262,144 | Wrong Answer | 19 | 3,064 | 408 | Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him. | sx,sy,tx,ty = map(int,input().split())
ans = ""
X = tx - sx
Y = ty - sy
for i in range(Y):
ans += "U"
for i in range(X):
ans += "R"
for i in range(Y):
ans += "D"
for i in range(X):
ans += "L"
ans += "L"
for i in range(Y):
ans += "U"
for i in range(X+1):
ans += "R"
ans += "D"
ans += "R"
for i in range(Y+1):
ans += "D"
for i in range(X+1):
ans += "L"
ans += "U"
print(ans) | s524881029 | Accepted | 20 | 3,064 | 411 | sx,sy,tx,ty = map(int,input().split())
ans = ""
X = tx - sx
Y = ty - sy
for i in range(Y):
ans += "U"
for i in range(X):
ans += "R"
for i in range(Y):
ans += "D"
for i in range(X):
ans += "L"
ans += "L"
for i in range(Y+1):
ans += "U"
for i in range(X+1):
ans += "R"
ans += "D"
ans += "R"
for i in range(Y+1):
ans += "D"
for i in range(X+1):
ans += "L"
ans += "U"
print(ans) |
s600395710 | p03610 | u636822224 | 2,000 | 262,144 | Wrong Answer | 34 | 3,188 | 74 | You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1. | s=input()
t=s[0]
for i in range(int(len(s)/2)-1):
t+=s[2*(i+1)]
print(t) | s665834310 | Accepted | 30 | 3,188 | 124 | s=input()
t=s[0]
if len(s)/2==0:
n=int(len(s)/2)-1
else:
n=int((len(s)-1)/2)
for i in range(n):
t+=s[2*(i+1)]
print(t) |
s031041980 | p03044 | u797016134 | 2,000 | 1,048,576 | Wrong Answer | 378 | 4,720 | 179 | We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem. | n = int(input())
ans = [0]*n
for i in range(n-1):
u, v, w = map(int, input().split())
if w % 2 == 0:
ans[u-1] = 1
ans[v-1] = 1
for i in ans:
print(i) | s661516394 | Accepted | 702 | 79,408 | 500 | import sys
sys.setrecursionlimit(10**7)
n = int(input())
link = [[] for i in range(n)]
for i in range(n-1):
u, v, w = map(int, input().split())
u -= 1
v -= 1
link[u].append((v,w))
link[v].append((u,w))
ans = [-1]*n
def dfs(v, now):
ans[v] = now
for next_v ,w in link[v]:
if ans[next_v] != -1:
continue
if w % 2:
dfs(next_v, 1-now)
else:
dfs(next_v, now)
dfs(0,0)
for i in ans:
print(i) |
s535344378 | p03487 | u171065106 | 2,000 | 262,144 | Wrong Answer | 163 | 22,436 | 241 | You are given a sequence of positive integers of length N, a = (a_1, a_2, ..., a_N). Your objective is to remove some of the elements in a so that a will be a **good sequence**. Here, an sequence b is a **good sequence** when the following condition holds true: * For each element x in b, the value x occurs exactly x times in b. For example, (3, 3, 3), (4, 2, 4, 1, 4, 2, 4) and () (an empty sequence) are good sequences, while (3, 3, 3, 3) and (2, 4, 1, 4, 2) are not. Find the minimum number of elements that needs to be removed so that a will be a good sequence. | from collections import Counter
n = int(input())
a = list(map(int, input().split()))
count = 0
for i, j in Counter(a).items():
print(i, j, j-i, j)
if j-i < 0:
count += j
else:
count += min(j-i, j)
print(count) | s736193821 | Accepted | 75 | 22,236 | 217 | from collections import Counter
n = int(input())
a = list(map(int, input().split()))
count = 0
for i, j in Counter(a).items():
if j-i < 0:
count += j
else:
count += min(j-i, j)
print(count) |
s636772674 | p02412 | u535719732 | 1,000 | 131,072 | Wrong Answer | 20 | 5,592 | 277 | Write a program which identifies the number of combinations of three integers which satisfy the following conditions: * You should select three distinct integers from 1 to n. * A total sum of the three integers is x. For example, there are two combinations for n = 5 and x = 9. * 1 + 3 + 5 = 9 * 2 + 3 + 4 = 9 | while True:
num = list(map(int,input().split()))
if(num[0] == 0 and num[1] == 0): break
c = 0
for i in range(1,num[0] - 2):
for j in range(i+1,num[0] - 1):
for k in range(j+1,num[0]):
if(i+j+k == num[1]): c
print(c)
| s307325820 | Accepted | 60 | 5,596 | 326 | while True:
num = list(map(int,input().split()))
if(num[0] == 0 and num[1] == 0): break
c = 0
for i in range(1,num[0]+1):
for j in range(i+1,num[0]+1):
if j <= i:
continue
k = num[1]-(i+j)
if k > j and k <= num[0]:
c+= 1
print(c)
|
s896537547 | p03696 | u698176039 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 384 | You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_. Here, a correct bracket sequence is defined as follows: * `()` is a correct bracket sequence. * If X is a correct bracket sequence, the concatenation of `(`, X and `)` in this order is also a correct bracket sequence. * If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence. * Every correct bracket sequence can be derived from the rules above. Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one. | N = int(input())
S = list(input())
ans = (S)
cnt = 0
idx = 0
for s in S:
if s == ')':
cnt += 1
if cnt > 0:
ans = ans[:idx] + ['('] + ans[idx:]
idx += 1
cnt = 0
else :
cnt -= 1
idx += 1
if cnt<0:
for i in range(-cnt):
ans = ans[:idx] + [')'] + ans[idx:]
idx += 1
print(''.join(ans))
| s415174267 | Accepted | 17 | 3,064 | 302 | N = int(input())
S = list(input())
ans = (S)
cnt = 0
for i,s in enumerate(S):
if cnt < 0:
ans = ['('] + ans
cnt = 0
if s == '(':
cnt += 1
else:
cnt -= 1
if cnt> 0:
ans = ans + [')'] * cnt
elif cnt < 0:
ans = ['('] * (-cnt) + ans
print(''.join(ans))
|
s137525658 | p03729 | u623349537 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 100 | You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`. | A, B, C = input().split()
if A[-1] == B[0] and B[-1] == C[0]:
print("Yes")
else:
print("No") | s987975783 | Accepted | 17 | 2,940 | 100 | A, B, C = input().split()
if A[-1] == B[0] and B[-1] == C[0]:
print("YES")
else:
print("NO") |
s464957382 | p03861 | u583507988 | 2,000 | 262,144 | Wrong Answer | 2,104 | 3,060 | 165 | 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())
ans = 0
for i in range(b):
if i % x == 0:
ans += 1
for j in range(a):
if j % x == 0:
ans -= 1
print(ans) | s824610016 | Accepted | 31 | 9,092 | 91 | a, b, x = map(int, input().split())
c=b//x
if a%x==0:
d=a//x-1
else:
d=a//x
print(c-d) |
s215524212 | p03377 | u044964932 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 235 | There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals. | def main():
a, b, x = map(int, input().split())
if x < a:
ans = "No"
else:
if (x - a) > b:
ans = "No"
else:
ans = "Yes"
print(ans)
if __name__ == "__main__":
main()
| s427371502 | Accepted | 17 | 3,064 | 235 | def main():
a, b, x = map(int, input().split())
if x < a:
ans = "NO"
else:
if (x - a) > b:
ans = "NO"
else:
ans = "YES"
print(ans)
if __name__ == "__main__":
main()
|
s306517685 | p03369 | u782516204 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 73 | 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. | l = input()
ans = 0
for i in l:
if i=="o":
ans+=1
print(ans) | s179436008 | Accepted | 17 | 2,940 | 85 | l = input()
ans = 0
for i in l:
if i=="o":
ans+=1
print(ans * 100 + 700) |
s731892519 | p03505 | u344627992 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 172 | _ButCoder Inc._ runs a programming competition site called _ButCoder_. In this site, a user is given an integer value called rating that represents his/her skill, which changes each time he/she participates in a contest. The initial value of a new user's rating is 0, and a user whose rating reaches K or higher is called _Kaiden_ ("total transmission"). Note that a user's rating may become negative. Hikuhashi is a new user in ButCoder. It is estimated that, his rating increases by A in each of his odd-numbered contests (first, third, fifth, ...), and decreases by B in each of his even-numbered contests (second, fourth, sixth, ...). According to this estimate, after how many contests will he become Kaiden for the first time, or will he never become Kaiden? | K, A, B = map(int, input().split())
if A <= B :
print ('-1')
elif K < B:
print ('1')
else :
delta = A - B
ans = (K - A) // delta * 2 + 1
print ('%ld' % ans)
| s990639677 | Accepted | 19 | 3,060 | 186 | K, A, B = map(int, input().split())
if K <= A :
print ('1')
elif A <= B :
print ('-1')
else :
delta = A - B
ans = (K - A + delta - 1) // delta * 2 + 1
print ('%ld' % ans)
|
s244864625 | p00462 | u901080241 | 1,000 | 131,072 | Wrong Answer | 20 | 7,552 | 307 | JOI ピザでは,市の中心部を通る全長 _d_ メートルの環状線の沿線上でピザの宅配販 売を行っている. JOI ピザは,環状線上に _n_ 個の店舗 _S_ 1, ... , _S n_ を持つ.本店は _S_ 1 である. _S_ 1 から _S i_ まで,時計回りに環状線を移動したときの道のりを _d i_ メートルとおく. _d_ 2, ... , _d n_ は 1 以上 _d_ \- 1 以下の整数である. _d_ 2 , ... , _d n_ は全て異なる. ピザの注文を受けると, ピザが冷めないように, 宅配先までの移動距離がもっとも短い店舗でピザを焼き宅配する. 宅配先の位置は 0 以上 _d_ \- 1 以下の整数 _k_ で表される.これは, 本店 _S_ 1 から宅配先までに時計回りで環状線を移動したときの道のりが _k_ メートルであることを意味する. ピザの宅配は環状線に沿って行われ, それ以外の道を通ることは許されない. ただし, 環状線上は時計回りに移動しても反時計回りに移動してもよい. 例えば,店舗の位置と宅配先の位置が下図のようになっている場合 (この例は「入出力の例」の例 1 と対応している), 宅配先 1 にもっとも近い店舗は _S_ 2 なので, 店舗 _S_ 2 から宅配する.このとき, 店舗からの移動距離は 1 である.また, 宅配先 2 にもっとも近い店舗は _S_ 1 (本店) なので, 店舗 _S_ 1 (本店) から宅配する.このとき,店舗からの移動距離は 2 である. 環状線の全長 _d_ , JOI ピザの店舗の個数 _n_ , 注文の個数 _m_ , 本店以外の位置を表す _n_ \- 1 個の整数 _d_ 2, ... , _d n_ , 宅配先の場所を表す整数 _k_ 1, ... , _k m_ が与えられたとき, 各注文に対する宅配時の移動距離 (すなわち,最寄店舗から宅配先までの道のり) の全注文にわたる総和を求めるプログラムを作成せよ. | import bisect
d = int(input())
n = int(input())
m = int(input())
shops = [0]*n
for i in range(1,n):
shops[i] = int(input())
shops.sort()
shops.append(d)
d = 0
for i in range(m):
cus = int(input())
req = bisect.bisect(shops,cus)
d += min(abs(cus-shops[req-1]), abs(cus-shops[req]))
print(d) | s004591373 | Accepted | 390 | 11,840 | 394 | import bisect
while True:
d = int(input())
if d==0: break
n = int(input())
m = int(input())
shops = [0]*n
for i in range(1,n):
shops[i] = int(input())
shops.sort()
shops.append(d)
d = 0
for i in range(m):
cus = int(input())
req = bisect.bisect(shops,cus)
d += min(abs(cus-shops[req-1]), abs(cus-shops[req]))
print(d) |
s053286348 | p02795 | u899782392 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 132 | We have a grid with H rows and W columns, where all the squares are initially white. You will perform some number of painting operations on the grid. In one operation, you can do one of the following two actions: * Choose one row, then paint all the squares in that row black. * Choose one column, then paint all the squares in that column black. At least how many operations do you need in order to have N or more black squares in the grid? It is guaranteed that, under the conditions in Constraints, having N or more black squares is always possible by performing some number of operations. | import sys
H = int(sys.stdin.readline())
W = int(sys.stdin.readline())
N = int(sys.stdin.readline())
print((N - 1) / max(H, W) + 1) | s293662347 | Accepted | 17 | 2,940 | 134 | import sys
H = int(sys.stdin.readline())
W = int(sys.stdin.readline())
N = int(sys.stdin.readline())
print((N - 1) // max(H, W) + 1) |
s107888733 | p03478 | u635391905 | 2,000 | 262,144 | Wrong Answer | 36 | 3,064 | 691 | Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive). | # -*- coding: utf-8 -*-
import sys
from math import *
if __name__ =="__main__":
str_in = input('input with blocks:')
num = [int(n) for n in str_in.split()]
N,A,B=num
SUM=0
if N<0 or N>(10**4) or A<0 or B<0 or A>36 or B>36 or A>B:
print("error input!")
else:
for i in range(1,(N+1)):
#print(i)
ii0 = int((i / 1) % 10 )
ii1 = int((i / 10) % 10)
ii2 = int((i / 100) % 10)
ii3 = int((i / 1000) % 10)
ii4 = int((i / 10000) % 10)
#print(ii4,ii3,ii2,ii1,ii0)
SUMi=ii4+ii3+ii2+ii1+ii0
if(SUMi>=A and SUMi<=B):
SUM+=i
print(SUM) | s371401615 | Accepted | 36 | 3,064 | 674 | #2
# -*- coding: utf-8 -*-
import sys
from math import *
if __name__ =="__main__":
str_in = input()
num = [int(n) for n in str_in.split()]
N,A,B=num
SUM=0
if N<0 or N>(10**4) or A<0 or B<0 or A>36 or B>36 or A>B:
print("error input!")
else:
for i in range(1,(N+1)):
#print(i)
ii0 = int((i / 1) % 10 )
ii1 = int((i / 10) % 10)
ii2 = int((i / 100) % 10)
ii3 = int((i / 1000) % 10)
ii4 = int((i / 10000) % 10)
#print(ii4,ii3,ii2,ii1,ii0)
SUMi=ii4+ii3+ii2+ii1+ii0
if(SUMi>=A and SUMi<=B):
SUM+=i
print(SUM) |
s086710086 | p02742 | u619084943 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 291 | We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move: | def main():
h,w = map(int,input().split())
if w==1 or h==1:
print(1)
return
if w%2==0:
print(h*w/2)
else:
if h%2==0:
print((w//2+1)*(h/2) + (w//2)*(h/2))
else:
print((w//2+1)*(h//2+1) + (w//2)*(h//2))
main() | s637479254 | Accepted | 17 | 3,060 | 306 | def main():
h,w = map(int,input().split())
if w==1 or h==1:
print(1)
return
if w%2==0:
print(int(h*w/2))
else:
if h%2==0:
print(int((w//2+1)*(h/2) + (w//2)*(h/2)))
else:
print(int((w//2+1)*(h//2+1) + (w//2)*(h//2)))
main() |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.