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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s501014598
|
p03943
|
u578501242
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 104
|
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
x=list(map(int, input().split()))
x.sort()
print(x)
if x[0]+x[1]==x[2]:
print('yes')
else:
print('no')
|
s237237214
|
Accepted
| 17
| 2,940
| 95
|
x=list(map(int, input().split()))
x.sort()
if x[0]+x[1]==x[2]:
print('Yes')
else:
print('No')
|
s277288052
|
p02866
|
u327310087
| 2,000
| 1,048,576
|
Wrong Answer
| 40
| 14,396
| 92
|
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.
|
n = int(input())
d = list(map(int, input().split()))
if d[0] != 0:
print(0)
quit()
|
s439182710
|
Accepted
| 199
| 21,888
| 598
|
n = int(input())
d = list(map(int, input().split()))
# bucket
lst = [[] for i in range(n)]
ans = 1
sum = 1
P = 998244353
if d[0] != 0:
print(0)
quit()
else:
for i in range(n):
if d[i] == 0 and i != 0:
print(0)
quit()
lst[d[i]].append(i + 1)
tmp = len(lst[1])
for i in range(1, n):
if len(lst[i]) == 0:
if sum != n:
print(0)
quit()
sum += len(lst[i])
if i == 1:
continue
ans = (ans * tmp ** len(lst[i])) % P
tmp = len(lst[i])
print(ans)
|
s834085633
|
p03730
|
u398182947
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,080
| 239
|
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())
startR = A % B
i = 1
while True:
R = A*i % B
if R == C:
print("Yes")
break
if i != 1:
if R == startR:
print("No")
break
i += 1
|
s256535169
|
Accepted
| 27
| 9,124
| 224
|
A,B,C = map(int,input().split())
startR = A % B
i = 1
while True:
R = A*i % B
if R == C:
print("YES")
break
if i != 1:
if R == startR:
print("NO")
break
i += 1
|
s568509552
|
p03469
|
u533713111
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 38
|
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.
|
S = input()
print(S.replace(S[3],'8'))
|
s869726667
|
Accepted
| 17
| 2,940
| 36
|
S = input()
print(S[:3] +'8'+ S[4:])
|
s222706112
|
p03494
|
u403875332
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 159
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
N = int(input())
a = list(map(int,input().split()))
while all([e % 2 == 0 for e in a]) == True:
for i in range(N):
a[i] = a[i]/2
print(a)
|
s468927885
|
Accepted
| 18
| 2,940
| 185
|
N = int(input())
a = list(map(int,input().split()))
count=0
while all([e % 2 == 0 for e in a]) == True:
count = count + 1
for i in range(N):
a[i] = a[i]/2
print(count)
|
s372125198
|
p03455
|
u468430765
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 70
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
print("Odd" if a*b%2 == 1 else "Odd")
|
s042179337
|
Accepted
| 17
| 2,940
| 71
|
a, b = map(int, input().split())
print("Odd" if a*b%2 == 1 else "Even")
|
s474412171
|
p03813
|
u601393594
| 2,000
| 262,144
|
Wrong Answer
| 21
| 9,072
| 4
|
Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise.
|
2000
|
s944870154
|
Accepted
| 25
| 9,148
| 141
|
# AtCoder abc053 a
x = int(input())
if x < 1200:
print("ABC")
else:
print("ARC")
|
s240169390
|
p03140
|
u305452255
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,064
| 476
|
You are given three strings A, B and C. Each of these is a string of length N consisting of lowercase English letters. Our objective is to make all these three strings equal. For that, you can repeatedly perform the following operation: * Operation: Choose one of the strings A, B and C, and specify an integer i between 1 and N (inclusive). Change the i-th character from the beginning of the chosen string to some other lowercase English letter. What is the minimum number of operations required to achieve the objective?
|
n, a, b, c = [input() for i in range(4)]
n = int(n)
ret = 0
for i in range(n):
if (a[i] == b[i]) and (a[i] == c[i]):
pass
elif (a[i] == b[i]) and (a[i] != c[i]):
c = c[:i] + a[i] + c[i + 1:]
ret += 1
elif a[i] == c[i] and a[i] != b[i]:
b = b[:i] + a[i] + b[i + 1:]
ret += 1
elif a[i] != b[i] and a[i] != c[i]:
b = b[:i] + a[i] + b[i + 1:]
c = c[:i] + a[i] + c[i + 1:]
ret += 2
print(ret)
|
s132428479
|
Accepted
| 17
| 3,064
| 569
|
n, a, b, c = [input() for i in range(4)]
n = int(n)
ret = 0
for i in range(n):
if (a[i] == b[i]) and (a[i] == c[i]):
pass
elif (a[i] == b[i]) and (a[i] != c[i]):
c = c[:i] + a[i] + c[i + 1:]
ret += 1
elif a[i] == c[i] and a[i] != b[i]:
b = b[:i] + a[i] + b[i + 1:]
ret += 1
elif a[i] != c[i] and b[i] == c[i]:
a = a[:i] + b[i] + a[i + 1:]
ret += 1
elif a[i] != b[i] and a[i] != c[i]:
b = b[:i] + a[i] + b[i + 1:]
c = c[:i] + a[i] + c[i + 1:]
ret += 2
print(ret)
|
s281169038
|
p03730
|
u000349418
| 2,000
| 262,144
|
Wrong Answer
| 2,103
| 2,940
| 149
|
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(' '))
x = a%b
ans = 'No'
i = 1
while i*x <= c:
if i*x == c:
ans = 'Yes'
break
i += 1
print(ans)
|
s664478787
|
Accepted
| 17
| 3,060
| 142
|
a,b,c = map(int,input().split(' '))
i = 1
ans = 'NO'
while i < b:
if (i*a)%b == c:
ans = 'YES'
break
i += 1
print(ans)
|
s327340002
|
p00019
|
u116501200
| 1,000
| 131,072
|
Wrong Answer
| 30
| 8,204
| 84
|
Write a program which reads an integer n and prints the factorial of n. You can assume that n ≤ 20\.
|
(lambda n:__import__("functools").reduce(lambda a,b:a*b,range(1,n+1)))(int(input()))
|
s067137847
|
Accepted
| 30
| 8,192
| 91
|
print((lambda n:__import__("functools").reduce(lambda a,b:a*b,range(1,n+1)))(int(input())))
|
s959887641
|
p03371
|
u547167033
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 135
|
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
|
a,b,c,x,y=map(int,input().split())
ans=[]
ans.append(a*x+b*y)
ans.append(c*x+b*max(0,y-x))
ans.append(c*y+a*max(0,x-y))
print(min(ans))
|
s587318088
|
Accepted
| 17
| 2,940
| 139
|
a,b,c,x,y=map(int,input().split())
ans=[]
ans.append(a*x+b*y)
ans.append(c*2*x+b*max(0,y-x))
ans.append(c*2*y+a*max(0,x-y))
print(min(ans))
|
s293080854
|
p03469
|
u198035503
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 33
|
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.
|
s = input()
print("2018" + s[3:])
|
s275129027
|
Accepted
| 17
| 2,940
| 34
|
s = input()
print("2018" + s[4:])
|
s502326059
|
p03472
|
u785578220
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,064
| 602
|
You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, …, Katana N, and can perform the following two kinds of attacks in any order: * Wield one of the katana you have. When you wield Katana i (1 ≤ i ≤ N), the monster receives a_i points of damage. The same katana can be wielded any number of times. * Throw one of the katana you have. When you throw Katana i (1 ≤ i ≤ N) at the monster, it receives b_i points of damage, and you lose the katana. That is, you can no longer wield or throw that katana. The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total?
|
def get_input(inp):
li = inp.split("\n")
def inner():
return li.pop(0)
return inner
INPUT = """4 1000000000
1 1
1 10000000
1 30000000
1 99999999
"""
input = get_input(INPUT)
from heapq import heappush, heappop
from operator import itemgetter
N, K = map(int, input().split())
a = []
aa = []
for i in range(N):
ta,tb = map(int, input().split())
a.append(ta)
aa.append(tb)
aa.append(0)
aa.sort()
b = max(a)
n = K//b+1
res = 0
while K>0:
t = aa.pop()
if t > b:
K -= t
res+=1
else:
res +=-(-K//b)
K = 0
break
print(res)
|
s348925053
|
Accepted
| 397
| 12,236
| 419
|
from heapq import heappush, heappop
from operator import itemgetter
N, K = map(int, input().split())
a = []
aa = []
for i in range(N):
ta,tb = map(int, input().split())
a.append(ta)
aa.append(tb)
a = sorted(a)
aa.append(0)
aa.sort()
b = max(a)
n = K//b+1
res = 0
while K>0:
t = aa.pop()
if t > b:
K -= t
res+=1
else:
res +=-(-K//b)
K = 0
break
print(res)
|
s142214535
|
p03471
|
u383551754
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,064
| 347
|
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
n, y = map(int, input().split())
ichi = y // 10000
for i in reversed(range(ichi + 1)):
gos = (y - 10000 * ichi) // 5000
for j in reversed(range(gos+1)):
rest = y - 10000 * i - 5000 * j
k = rest // 1000
if rest % 1000 == 0 and i + j + k == n:
print(i, j, k)
break
else:
print(-1, -1, -1)
|
s983474500
|
Accepted
| 716
| 2,940
| 253
|
n, y = map(int, input().split())
for i in reversed(range(n + 1)):
for j in reversed(range(n - i + 1)):
k = n - i - j
if 10000 * i + 5000 * j + 1000 * k == y:
print(i, j, k)
exit()
else:
print(-1, -1, -1)
|
s359065847
|
p00028
|
u650459696
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,708
| 355
|
Your task is to write a program which reads a sequence of integers and prints mode values of the sequence. The mode value is the element which occurs most frequently.
|
a, md = [], []
c, mx = 1, 1
while True:
try:
a.append(int(input()))
except EOFError:
break
a = sorted(a)
print(a)
for i in range(0, len(a) - 1):
if a[i] == a[i + 1]:
c += 1
else:
c = 1
if mx < c:
mx = c
md = [a[i]]
elif mx == c:
md.append(a[i])
print('\n'.join(map(str, md)))
|
s613321483
|
Accepted
| 20
| 7,504
| 203
|
a = []
ary = [0] * 101
while True:
try:
c = int(input())
except EOFError:
break
a.append(c)
ary[c] += 1
for i in range(1, 101):
if ary[i] == max(ary):
print(i)
|
s824689704
|
p02407
|
u731896389
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,508
| 88
|
Write a program which reads a sequence and prints it in the reverse order.
|
n = list(map(int,input().split()))
for i in range(len(n)):
print(n[len(n)-i-1])
|
s642686834
|
Accepted
| 20
| 7,656
| 154
|
n =int(input())
a = list(map(int,input().split()))
a.reverse()
for i in range(n):
if i!=n-1:
print(a[i],end=" ")
else:
print(a[i])
|
s397861005
|
p03007
|
u227082700
| 2,000
| 1,048,576
|
Wrong Answer
| 212
| 14,264
| 156
|
There are N integers, A_1, A_2, ..., A_N, written on a blackboard. We will repeat the following operation N-1 times so that we have only one integer on the blackboard. * Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y. Find the maximum possible value of the final integer on the blackboard and a sequence of operations that maximizes the final integer.
|
n=int(input())
a=list(map(int,input().split()))
a.sort()
mi,ma=a[0],a[-1]
for i in a[1:-1]:
if i<0:print(ma,i);ma-=i
else:print(mi,i);mi-=i
print(ma,mi)
|
s790922216
|
Accepted
| 229
| 14,264
| 232
|
n=int(input())
a=list(map(int,input().split()))
a.sort()
mi,ma=a[0],a[-1]
for i in a[1:-1]:
if i<0:ma-=i
else:mi-=i
print(ma-mi)
mi,ma=a[0],a[-1]
for i in a[1:-1]:
if i<0:print(ma,i);ma-=i
else:print(mi,i);mi-=i
print(ma,mi)
|
s133358500
|
p04035
|
u766684188
| 2,000
| 262,144
|
Wrong Answer
| 67
| 14,060
| 175
|
We have N pieces of ropes, numbered 1 through N. The length of piece i is a_i. At first, for each i (1≤i≤N-1), piece i and piece i+1 are tied at the ends, forming one long rope with N-1 knots. Snuke will try to untie all of the knots by performing the following operation repeatedly: * Choose a (connected) rope with a total length of at least L, then untie one of its knots. Is it possible to untie all of the N-1 knots by properly applying this operation? If the answer is positive, find one possible order to untie the knots.
|
n,l=map(int,input().split())
A=list(map(int,input().split()))
for i in range(n-1):
if A[i]+A[i+1]>=l:
print('Possible')
break
else:
print('Impossible')
|
s938674905
|
Accepted
| 120
| 14,052
| 282
|
n,l=map(int,input().split())
A=list(map(int,input().split()))
for i in range(n-1):
if A[i]+A[i+1]>=l:
print('Possible')
t=i+1
break
else:
print('Impossible')
exit()
for i in range(1,t):
print(i)
for i in range(n-1,t,-1):
print(i)
print(t)
|
s421801359
|
p03448
|
u194894739
| 2,000
| 262,144
|
Wrong Answer
| 47
| 2,940
| 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, b, c, x = map(int,[input() for i in range(4)])
ans = 0
for i in range(0, a):
for j in range(0, b):
for k in range(0, c):
if i * 500 + j * 100 + k * 50 == x:
ans += 1
print(ans)
|
s237178418
|
Accepted
| 50
| 2,940
| 223
|
a, b, c, x = map(int,[input() for i in range(4)])
ans = 0
for i in range(a + 1):
for j in range(b + 1):
for k in range(c + 1):
if i * 500 + j * 100 + k * 50 == x:
ans += 1
print(ans)
|
s542968455
|
p03479
|
u724687935
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 77
|
As a token of his gratitude, Takahashi has decided to give his mother an integer sequence. The sequence A needs to satisfy the conditions below: * A consists of integers between X and Y (inclusive). * For each 1\leq i \leq |A|-1, A_{i+1} is a multiple of A_i and strictly greater than A_i. Find the maximum possible length of the sequence.
|
import math
X, Y = map(int, input().split())
print(int(math.log2(Y / X)))
|
s037148563
|
Accepted
| 17
| 2,940
| 102
|
X, Y = map(int, input().split())
cnt = 1
k = X
while k * 2 <= Y:
cnt += 1
k *= 2
print(cnt)
|
s145315502
|
p03957
|
u386131832
| 1,000
| 262,144
|
Wrong Answer
| 23
| 3,064
| 263
|
This contest is `CODEFESTIVAL`, which can be shortened to the string `CF` by deleting some characters. Mr. Takahashi, full of curiosity, wondered if he could obtain `CF` from other strings in the same way. You are given a string s consisting of uppercase English letters. Determine whether the string `CF` can be obtained from the string s by deleting some characters.
|
import sys
s=input()
for i in range(len(s)):
if s[i]=="C":
for j in range(i+1,len(s)):
if s[j]=="F":
print("YES")
sys.exit()
if j==len(s)-1:
print("NO")
sys.exit()
|
s381140390
|
Accepted
| 22
| 3,064
| 322
|
import sys
s=input()
for i in range(len(s)):
if i==len(s)-1:
print("No")
sys.exit()
if s[i]=="C":
for j in range(i+1,len(s)):
if s[j]=="F":
print("Yes")
sys.exit()
if j==len(s)-1:
print("No")
sys.exit()
|
s399354817
|
p03565
|
u744920373
| 2,000
| 262,144
|
Wrong Answer
| 52
| 3,188
| 313
|
E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.
|
S = input()
T = input()
l = len(T)
ind = 51
for i in range(len(S)-l+1):
flag = 0
for j in range(l):
if S[i+j]!='?' and S[i+j]!=T[j]:
flag = 1
if flag==0:
ind = i
if ind == 51:
print('UNRESTORABLE')
exit()
print(S[:i].replace('?','a')+T+S[i+j:].replace('?','a'))
|
s595760459
|
Accepted
| 17
| 3,064
| 317
|
S = input()
T = input()
l = len(T)
ind = 51
for i in range(len(S)-l+1):
flag = 0
for j in range(l):
if S[i+j]!='?' and S[i+j]!=T[j]:
flag = 1
if flag==0:
ind = i
if ind == 51:
print('UNRESTORABLE')
exit()
print(S[:ind].replace('?','a')+T+S[ind+l:].replace('?','a'))
|
s660976154
|
p00282
|
u847467233
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 411
|
大きな数を表そうとすると、文字数も多くなるし、位取りがわからなくなってしまうので、なかなか面倒です。大きな数をわかりやすく表すために、人々は数の単位を使ってきました。江戸時代に書かれた「塵劫記」という本の中では、数の単位が次のように書かれています。 たとえば、２の１００乗のようなとても大きな数は、１２６穣７６５０(じょ)６００２垓２８２２京９４０１兆４９６７億３２０万５３７６と表せます。それでは、正の整数 m と n が与えられたとき、m の n 乗を塵劫記の単位を使って上のように表すプログラムを作成してください。
|
# Python3 2018.6.26 bal4u
unit = ("", "Man", "Oku", "Cho", "Kei", "Gai", "Jo", "Jou", "Ko", "Kan", \
"Sei", "Sai", "Gok", "Ggs", "Asg", "Nyt", "Fks", "Mts")
while True:
m, n = map(int, input().split())
if m == 0: break
s = str(m ** n)[::-1]
ans, j = [], 0
for i in range(0, len(s), 4):
t = s[i:i+4][::-1]
if t != "0000":
ans += [unit[j], t]
j += 1
print(*ans[::-1], sep='')
|
s012461429
|
Accepted
| 20
| 5,604
| 403
|
# Python3 2018.6.26 bal4u
unit = ("", "Man", "Oku", "Cho", "Kei", "Gai", "Jo", "Jou", "Ko", "Kan", \
"Sei", "Sai", "Gok", "Ggs", "Asg", "Nyt", "Fks", "Mts")
while True:
m, n = map(int, input().split())
if m == 0: break
s = str(m ** n)[::-1]
ans = []
for i in range(0, len(s), 4):
t = str(int(s[i:i+4][::-1]))
if t != "0": ans += [unit[i>>2], t]
print(*ans[::-1], sep='')
|
s088221910
|
p03494
|
u260036763
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 229
|
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.
|
ans = 0
N = input()
A = list(map(int, input().split()))
for i in A :
p = 0
if i % 2 != 0 :
print(0)
break
else :
while i % 2 != 0 :
i //= 2
p += 1
if ans < p :
ans = p
print(ans)
|
s694968109
|
Accepted
| 19
| 3,064
| 522
|
import sys
def main():
N = int(sys.stdin.readline().rstrip())
A = list(map(int, sys.stdin.readline().rstrip().split()))
cnt = 0
exist_odd = False
while True:
for i in range(N):
if A[i] % 2 != 0:
exist_odd = True
if exist_odd == True:
break
for i in range(N):
A[i] /= 2
cnt += 1
print(cnt)
if __name__ == '__main__':
main()
|
s582993899
|
p03131
|
u902973687
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 298
|
Snuke has one biscuit and zero Japanese yen (the currency) in his pocket. He will perform the following operations exactly K times in total, in the order he likes: * Hit his pocket, which magically increases the number of biscuits by one. * Exchange A biscuits to 1 yen. * Exchange 1 yen to B biscuits. Find the maximum possible number of biscuits in Snuke's pocket after K operations.
|
K, A, B = map(int, input().split())
if not B - A <= 2:
if not (A == K or A == K + 1):
if (K - A) // 2 == 0:
print(B + (B - A) * ((K - A - 1) // 2) + 1)
else:
print(B + (B - A) * ((K - A - 1) // 2))
else:
print(K + 1)
else:
print(K + 1)
|
s948458068
|
Accepted
| 17
| 2,940
| 120
|
K, A, B = map(int, input().split())
if (B - A) <= 2:
print(K + 1)
else:
res = K-A+1
print(A+(B-A)*(res//2)+res%2)
|
s781533154
|
p02413
|
u340503368
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,588
| 283
|
Your task is to perform a simple table calculation. Write a program which reads the number of rows r, columns c and a table of r × c elements, and prints a new table, which includes the total sum for each row and column.
|
r,c = map(int, input().split())
for i in range(r):
a = list(map(int, input().split()))
sum = 0
for j in range(c):
sum += a[j]
a.append(sum)
for k in range(c+1):
print(a[k], end = "")
if k != c:
print(" ")
print()
|
s760605315
|
Accepted
| 50
| 5,740
| 481
|
r,c = map(int, input().split())
matrix = []
for i in range(r):
a = list(map(int, input().split()))
sum = 0
for j in range(c):
sum += a[j]
a.append(sum)
matrix.append(a)
for k in range(c+1):
print(a[k], end = "")
if k != c:
print(" ", end = "")
print()
for l in range(c+1):
sum2 = 0
for m in range(r):
sum2 += matrix[m][l]
print(sum2, end = "")
if l != c:
print(" ", end = "")
print()
|
s797216979
|
p03473
|
u114920558
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 28
|
How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?
|
M = int(input())
print(24-M)
|
s319763249
|
Accepted
| 17
| 2,940
| 29
|
M = int(input())
print(48-M)
|
s564611448
|
p03943
|
u223663729
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 180
|
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
# -*- coding: utf-8 -*-
a, b, c = map(int, input().split())
if a+b == c or a == b+c:
print('Yes')
else:
print('No')
|
s577310171
|
Accepted
| 19
| 2,940
| 234
|
# -*- coding: utf-8 -*-
a, b, c = map(int, input().split())
if a == (b+c):
print('Yes')
elif (a+b) == c:
print('Yes')
elif (a+c) == b:
print('Yes')
else:
print('No')
|
s702946703
|
p03457
|
u357949405
| 2,000
| 262,144
|
Wrong Answer
| 984
| 6,132
| 485
|
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.
|
N = int(input())
t0, x0, y0 = map(int, input().split())
if t0 < x0 + y0 or t0 % 2 != (x0 + y0) % 2:
print("No")
exit()
for i in range(1, N):
t1, x1, y1 = map(int, input().split())
dist_t, dist_x, dist_y = abs(t1-t0), abs(x1-x0), abs(y1-y0)
print("dist_t: {}, dist_x: {}, dist_y: {}".format(dist_t, dist_x, dist_y))
if dist_t < dist_x + dist_y or dist_t % 2 != (dist_x + dist_y) % 2:
print("No")
exit()
t0, x0, y0 = t1, x1, y1
print("Yes")
|
s233647766
|
Accepted
| 385
| 3,064
| 487
|
N = int(input())
t0, x0, y0 = map(int, input().split())
if t0 < x0 + y0 or t0 % 2 != (x0 + y0) % 2:
print("No")
exit()
for i in range(1, N):
t1, x1, y1 = map(int, input().split())
dist_t, dist_x, dist_y = abs(t1-t0), abs(x1-x0), abs(y1-y0)
# print("dist_t: {}, dist_x: {}, dist_y: {}".format(dist_t, dist_x, dist_y))
if dist_t < dist_x + dist_y or dist_t % 2 != (dist_x + dist_y) % 2:
print("No")
exit()
t0, x0, y0 = t1, x1, y1
print("Yes")
|
s407894977
|
p02613
|
u684556734
| 2,000
| 1,048,576
|
Wrong Answer
| 144
| 16,276
| 377
|
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())
n = [input() for i in range(N)]
count_AC = 0
count_WA = 0
count_TLE = 0
count_RE = 0
for i in n:
if i == 'AC':
count_AC += 1
elif i == 'WA':
count_WA += 1
elif i == 'TLE':
count_WA += 1
elif i == 'RE':
count_WA += 1
print('AC x {0}\nWA x {1}\nTLE x {2}\nRE x {3}'.format(count_AC, count_WA, count_TLE, count_RE))
|
s384090465
|
Accepted
| 151
| 16,200
| 378
|
N = int(input())
n = [input() for i in range(N)]
count_AC = 0
count_WA = 0
count_TLE = 0
count_RE = 0
for i in n:
if i == 'AC':
count_AC += 1
elif i == 'WA':
count_WA += 1
elif i == 'TLE':
count_TLE += 1
elif i == 'RE':
count_RE += 1
print('AC x {0}\nWA x {1}\nTLE x {2}\nRE x {3}'.format(count_AC, count_WA, count_TLE, count_RE))
|
s096590877
|
p02694
|
u398203017
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,164
| 78
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
X=int(input())
A=int(100)
B=int(0)
while A<=X:
A=A*101//100
B=B+1
print(B)
|
s696287909
|
Accepted
| 22
| 9,152
| 78
|
X=int(input())
A=int(100)
B=int(0)
while A<X:
A=A*101//100
B=B+1
print(B)
|
s054355142
|
p02694
|
u423812170
| 2,000
| 1,048,576
|
Wrong Answer
| 37
| 9,184
| 199
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
import math
X = int(input())
deposit = 100
y_later = 0
while deposit < X:
y_later += 1
deposit = math.floor(deposit) * 1.01
print(deposit)
print(y_later)
|
s242209492
|
Accepted
| 28
| 9,160
| 126
|
X = int(input())
deposit = 100
y_later = 0
while deposit < X:
y_later += 1
deposit += deposit // 100
print(y_later)
|
s356131406
|
p02608
|
u438946728
| 2,000
| 1,048,576
|
Wrong Answer
| 38
| 9,204
| 324
|
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
n = int(input())
a = [0 for _ in range(n+1)]
for i in range(1,n+1):
if i*i>n:
break
for j in range(1,n+1):
if i*i + j*j +i*j > n:
break
for k in range(1,n+1):
tmp = i*i + j*j + k*k + i*j + j*k + k*i
if tmp==n:
a[tmp]+=1
else:
break
for i in range(1,n+1):
print(a[i])
|
s105588222
|
Accepted
| 151
| 9,196
| 324
|
n = int(input())
a = [0 for _ in range(n+1)]
for i in range(1,n+1):
if i*i>n:
break
for j in range(1,n+1):
if i*i + j*j +i*j > n:
break
for k in range(1,n+1):
tmp = i*i + j*j + k*k + i*j + j*k + k*i
if tmp<=n:
a[tmp]+=1
else:
break
for i in range(1,n+1):
print(a[i])
|
s272133049
|
p02646
|
u017415492
| 2,000
| 1,048,576
|
Wrong Answer
| 25
| 9,120
| 209
|
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
a,v=map(int,input().split())
b,w=map(int,input().split())
t=int(input())
if a==b:
print("Yes")
else:
if w>v:
print("No")
else:
if (v-w)*t<abs(b-a):
print("No")
else:
print("Yes")
|
s281261096
|
Accepted
| 21
| 9,180
| 209
|
a,v=map(int,input().split())
b,w=map(int,input().split())
t=int(input())
if a==b:
print("YES")
else:
if w>v:
print("NO")
else:
if (v-w)*t<abs(b-a):
print("NO")
else:
print("YES")
|
s015983339
|
p03567
|
u486773779
| 2,000
| 262,144
|
Wrong Answer
| 29
| 8,968
| 64
|
Snuke built an online judge to hold a programming contest. When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.) Determine whether the judge can return the string `AC` as the verdict to a program.
|
s=input()
if s in 'AC':
print('Yes')
else:
print('No')
|
s737722245
|
Accepted
| 28
| 9,076
| 64
|
s=input()
if 'AC' in s:
print('Yes')
else:
print('No')
|
s554414601
|
p03473
|
u960513073
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 29
|
How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?
|
print(24 - int(input()) + 12)
|
s521779194
|
Accepted
| 18
| 2,940
| 29
|
print(24 - int(input()) + 24)
|
s215832954
|
p03110
|
u902266190
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 201
|
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
N = int(input())
money = 0
for i in range(N):
x, u = list(input().split(' '))
if(u == 'JPY'):
rate = 1.0
else:
rate = 380000.0
money += int(float(x) * rate)
print(money)
|
s057243210
|
Accepted
| 17
| 2,940
| 230
|
N = int(input())
money = 0
for i in range(N):
x, u = list(input().split(' '))
if(u == 'JPY'):
rate = 1.0
x = int(x)
else:
rate = 380000.0
x = float(x)
money += x * rate
print(money)
|
s320959537
|
p03386
|
u252964975
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 161
|
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())
if B-A<2*K:
for i in range(A,B+1):
print(i)
else:
for i in range(K):
print(A+i)
for i in range(K):
print(B-K+i)
|
s515084190
|
Accepted
| 17
| 3,060
| 167
|
A,B,K=map(int, input().split())
if B-A+1-2*K<1:
for i in range(A,B+1):
print(i)
else:
for i in range(K):
print(A+i)
for i in range(K):
print(B-K+i+1)
|
s034300642
|
p03693
|
u318165580
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 84
|
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
a,b,c=list(map(int,input().split()))
d=100*a+10*b+c
print("Yes" if d%4==0 else "No")
|
s833352178
|
Accepted
| 17
| 2,940
| 84
|
a,b,c=list(map(int,input().split()))
d=100*a+10*b+c
print("YES" if d%4==0 else "NO")
|
s848278443
|
p03672
|
u589381719
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 74
|
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
S=input()
for i in range(len(S),0,-1):
if S[:i]==S[i:]:
break
|
s673378459
|
Accepted
| 17
| 2,940
| 160
|
S=input()
for i in range(len(S)-1,0,-1):
if i%2:
continue
else:
t=i//2
if S[:t]==S[t:i]:
print(i)
break
|
s033721449
|
p03493
|
u179958942
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,052
| 115
|
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.
|
num = int(input())
a = num // 100
b = num // 10 - (a * 10)
c = num - (a * 100) - (b * 10)
print(num)
print(a, b, c)
|
s543664657
|
Accepted
| 25
| 9,084
| 106
|
num = int(input())
a = num // 100
b = num // 10 - (a * 10)
c = num - (a * 100) - (b * 10)
print(a + b + c)
|
s443731105
|
p02841
|
u489762173
| 2,000
| 1,048,576
|
Wrong Answer
| 20
| 3,188
| 405
|
In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019. Integers M_1, D_1, M_2, and D_2 will be given as input. It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1. Determine whether the date 2019-M_1-D_1 is the last day of a month.
|
import sys
def main():
M1, D1 = map(int, input().split())
M2, D2 = map(int, input().split())
if(M1==2 and D1!=28):
print('0')
sys.exit()
if(M1==4 or M1==6 or M1==9 or M1==11 and D1==30):
print('0')
sys.exit()
if(M1==1 or M1==3 or M1==5 or M1==7 or M1==8 or M1==10 or M1==12 and D1!=31):
print('0')
sys.exit()
print('1')
main()
|
s831406888
|
Accepted
| 17
| 3,064
| 556
|
import sys
def main():
M1, D1 = map(int, input().split())
M2, D2 = map(int, input().split())
if(M1==2 and D1==28):
if(M2==M1+1 and D2==1):
print('1')
sys.exit()
if(M1==4 or M1==6 or M1==9 or M1==11 and D1==30):
if(M2==M1+1 and D2==1):
print('1')
sys.exit()
if(M1==1 or M1==3 or M1==5 or M1==7 or M1==8 or M1==10 or M1==12 and D1==31):
if(M2==M1+1 and D2==1) or (M1==12 and M2==1 and D2==1):
print('1')
sys.exit()
print('0')
main()
|
s478401728
|
p03599
|
u143492911
| 3,000
| 262,144
|
Wrong Answer
| 371
| 14,572
| 633
|
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())
w=s=[]
from fractions import Fraction
import math
for i in range(math.ceil(f/a)):
for j in range(math.ceil(f/b)):
x=100*a*i+100*b*j
if 0<x and x<=f:
w.append(x)
w=list(set(w))
print(w)
for i in range(math.ceil(f/c)):
for j in range(math.ceil(f/d)):
y=i*c+j*d
if 0<y and y<=f:
s.append(y)
s=list(set(s))
s.sort()
print(s)
ans=ans_2=0
con=1
for i in w:
for j in s:
if i+j<=f and j<=e*(i//100):
max_con=(100*j)/(i+j)
if con<=max_con:
ans=i+j
ans_2=j
print(ans,ans_2)
|
s562077681
|
Accepted
| 1,976
| 12,440
| 664
|
a,b,c,d,e,f=map(int,input().split())
water=[]
suguar=[]
temp_3=0
temp_2=0
ans=0
temp=0
for i in range((f+1)//100):
for j in range((f+1)//100):
x=100*i*a+100*j*b
if 1<=x<=f:
water.append(x)
for i in range(f+1):
for j in range(f+1):
y=c*i+d*j
if y<=f:
suguar.append(y)
for i in water:
for j in suguar:
total=i+j
if total<=f:
if j<=i//100*e:
ans=100*j/(i+j)
if temp<=ans:
temp=ans
temp_2=total
temp_3=j
print(temp_2,temp_3)
|
s156773248
|
p02646
|
u922449550
| 2,000
| 1,048,576
|
Wrong Answer
| 25
| 9,012
| 145
|
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
A, V = map(int, input().split())
B, W = map(int, input().split())
T = int(input())
if abs(A - B) <= (W - V)*T:
print('YES')
else:
print('NO')
|
s881478227
|
Accepted
| 24
| 9,104
| 145
|
A, V = map(int, input().split())
B, W = map(int, input().split())
T = int(input())
if abs(A - B) <= (V - W)*T:
print('YES')
else:
print('NO')
|
s744281703
|
p03657
|
u903005414
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 122
|
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
a, b = map(int, input().split())
print('Possible' if not (a // 3) or not (b // 3) or not ((a + b) // 3) else 'Impossible')
|
s669620126
|
Accepted
| 18
| 2,940
| 141
|
a, b = map(int, input().split())
if (a % 3 == 0) or (b % 3 == 0) or ((a + b) % 3 == 0):
print('Possible')
else:
print('Impossible')
|
s342907285
|
p02678
|
u255943004
| 2,000
| 1,048,576
|
Wrong Answer
| 809
| 43,748
| 576
|
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
N,M = map(int,input().split())
A = []
B = []
for m in range(M):
a,b = map(int,input().split())
A.append(a)
B.append(b)
from collections import defaultdict as ddict
route = ddict(list)
for i in range(M):
route[A[i]].append(B[i])
route[B[i]].append(A[i])
for i in range(1,N+1):
if route[i] == []:
print("No")
exit()
ans = [0 for i in range(N+1)]
from collections import deque
Q = deque([1])
while Q:
q = Q.popleft()
for r in route[q]:
if ans[r] == 0:
ans[r] = q
Q.append(r)
print(*ans[2:],sep="\n")
|
s972551911
|
Accepted
| 931
| 43,700
| 589
|
N,M = map(int,input().split())
A = []
B = []
for m in range(M):
a,b = map(int,input().split())
A.append(a)
B.append(b)
from collections import defaultdict as ddict
route = ddict(list)
for i in range(M):
route[A[i]].append(B[i])
route[B[i]].append(A[i])
for i in range(1,N+1):
if route[i] == []:
print("No")
exit()
ans = [0 for i in range(N+1)]
from collections import deque
Q = deque([1])
while Q:
q = Q.popleft()
for r in route[q]:
if ans[r] == 0:
ans[r] = q
Q.append(r)
print("Yes")
print(*ans[2:],sep="\n")
|
s365575382
|
p03997
|
u740047492
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 63
|
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 i in range(3)]
print((a+b)*h/2)
|
s125240764
|
Accepted
| 18
| 3,060
| 64
|
a, b, h = [int(input()) for i in range(3)]
print(int((a+b)*h/2))
|
s621843245
|
p00003
|
u560214129
| 1,000
| 131,072
|
Wrong Answer
| 40
| 8,096
| 414
|
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.
|
n=int(input())
a=[]
c=[]
for i in range(n):
a.append([])
for j in range(3):
a[i].append([])
print (a)
for i in range(n):
#for j in range(3):
a[i][0], a[i][1], a[i][2]=map(int, input().split())
a[i].sort()
if(((a[i][0]*a[i][0])+(a[i][1]*a[i][1]))==(a[i][2]*a[i][2])):
k="YES"
c.append(k)
else:
k="NO"
c.append(k)
for i in range(n):
print(c[i])
|
s671080158
|
Accepted
| 30
| 7,940
| 404
|
n=int(input())
a=[]
c=[]
for i in range(n):
a.append([])
for j in range(3):
a[i].append([])
for i in range(n):
#for j in range(3):
a[i][0], a[i][1], a[i][2]=map(int, input().split())
a[i].sort()
if(((a[i][0]*a[i][0])+(a[i][1]*a[i][1]))==(a[i][2]*a[i][2])):
k="YES"
c.append(k)
else:
k="NO"
c.append(k)
for i in range(n):
print(c[i])
|
s093219114
|
p02833
|
u245870380
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 160
|
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).
|
N = int(input())
if N < 12:
print(0)
else:
cnt = 0
for i in range(1, 19):
cnt += ((N-2) // 10**i)*i
print(cnt, i)
print(cnt+1)
|
s337673221
|
Accepted
| 17
| 2,940
| 211
|
N = int(input())
if N % 2 != 0:
print(0)
else:
ans = 0
cnt = 10
while True:
if N >= cnt:
ans += N // (cnt)
cnt *= 5
else:
break
print(ans)
|
s618968762
|
p02536
|
u491296898
| 2,000
| 1,048,576
|
Wrong Answer
| 634
| 48,304
| 535
|
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i. Snuke can perform the following operation zero or more times: * Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities. After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times). What is the minimum number of roads he must build to achieve the goal?
|
import sys
sys.setrecursionlimit(10**9)
n,m = list(map(int,input().split()))
d = {}
for i in range(m):
fr, to = list(map(int,input().split()))
d[fr] = d.get(fr, []) + [to]
d[to] = d.get(to, []) + [fr]
visited = [0 for i in range(n+1)]
def connect(node, i):
visited[node] = i
if node in d:
for neig in d[node]:
if visited[neig] == 0:
connect(neig, i)
ans = 1
for key in range(1,n+1):
if visited[key] == 0:
connect(key, ans)
ans += 1
print(ans-1)
|
s123550723
|
Accepted
| 519
| 48,340
| 544
|
import sys
sys.setrecursionlimit(10**9)
n,m = list(map(int,input().split()))
d = {}
for i in range(m):
fr, to = list(map(int,input().split()))
d[fr] = d.get(fr, []) + [to]
d[to] = d.get(to, []) + [fr]
visited = [0 for i in range(n+1)]
def connect(node, i):
visited[node] = i
if node in d:
for neig in d[node]:
if visited[neig] == 0:
connect(neig, i)
ans = 1
for key in range(1,n+1):
if visited[key] == 0:
connect(key, ans)
ans += 1
print(max(visited)-1)
|
s240156962
|
p04030
|
u767797498
| 2,000
| 262,144
|
Wrong Answer
| 26
| 8,988
| 175
|
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
def bs(s):
if s == "":
return s
else:
return s[:-1]
s = input()
ans = ""
for c in s:
if c == "b":
ans = bs(ans)
else:
ans += c
print(ans)
|
s544362076
|
Accepted
| 24
| 8,968
| 83
|
s=input()
ss=""
for c in s:
if c=="B":
ss=ss[:-1]
else:
ss+=c
print(ss)
|
s172686649
|
p03400
|
u459283268
| 2,000
| 262,144
|
Wrong Answer
| 22
| 2,940
| 207
|
Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp.
|
n = int(input())
D, X = list(map(int, input().split()))
a = [int(input()) for _ in range(n)]
r = 0
for i in range(n):
k = 0
for j in a:
r += 1 if k*j + 1 <= D else r
k += 1
print(r)
|
s981098232
|
Accepted
| 19
| 2,940
| 199
|
n = int(input())
D, X = list(map(int, input().split()))
a = [int(input()) for _ in range(n)]
r = 0
for i in range(n):
k = 0
while(k*a[i]+1 <= D):
r += 1
k += 1
print(r + X)
|
s590095514
|
p03378
|
u980492406
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 175
|
There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right. Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N. Find the minimum cost incurred before reaching the goal.
|
n,m,x = map(int,input().split())
li = list(map(int,input().split()))
if x < li[0] or x > li[m-1] :
print(0)
else :
for i in range(m-1) :
print(min(i+1,m-i-1))
|
s403277382
|
Accepted
| 17
| 2,940
| 218
|
n,m,x = map(int,input().split())
li = list(map(int,input().split()))
if x < li[0] or x > li[m-1] :
print(0)
else :
for i in range(m-1) :
if x > li[i] and x < li[i+1] :
print(min(i+1,m-i-1))
|
s688005720
|
p03455
|
u057079894
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 77
|
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*b%2:
print('Even')
else:
print('Odd')
|
s871351731
|
Accepted
| 17
| 2,940
| 83
|
a, b= map(int,input().split())
if a*b%2 == 0:
print('Even')
else:
print('Odd')
|
s800423233
|
p03563
|
u403984573
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 34
|
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.
|
print(2*int(input())-int(input()))
|
s807500304
|
Accepted
| 17
| 2,940
| 42
|
A=int(input())
B=int(input())
print(2*B-A)
|
s758432689
|
p03455
|
u604846694
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 93
|
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 * b % 2 == 0:
print("odd")
else :
print("even")
|
s945614745
|
Accepted
| 17
| 2,940
| 93
|
a, b = map(int, input().split())
if a * b % 2 == 0:
print("Even")
else :
print("Odd")
|
s409321487
|
p03470
|
u982762220
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 174
|
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
N = int(input())
mochis = sorted([int(input()) for _ in range(N)], reverse=True)
res = 0
for idx in range(1, N):
if mochis[idx] > mochis[idx - 1]:
res += 1
print(res)
|
s852649530
|
Accepted
| 17
| 2,940
| 174
|
N = int(input())
mochis = sorted([int(input()) for _ in range(N)], reverse=True)
res = 1
for idx in range(1, N):
if mochis[idx] < mochis[idx - 1]:
res += 1
print(res)
|
s017604038
|
p03494
|
u067632118
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 3,188
| 175
|
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 = [int(x) for x in input().split()]
DA = []
for v in A:
d = 0
while v & 1 == 0:
v = v << 2
d += 1
DA.append(d)
print(min(DA))
|
s835822013
|
Accepted
| 18
| 2,940
| 176
|
N = int(input())
A = [int(x) for x in input().split()]
DA = []
for v in A:
d = 0
while v & 1 == 0:
v = v >> 1
d += 1
DA.append(d)
print(min(DA))
|
s361150965
|
p04012
|
u483151310
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 225
|
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
w = input()
dct_str = dict.fromkeys(w, 0)
for alp in w:
dct_str[alp] += 1
flag = True
for alp in dct_str.keys():
if dct_str[alp] % 2 != 0:
flag = False
if flag == True:
print("YES")
else:
print("NO")
|
s317421221
|
Accepted
| 18
| 2,940
| 227
|
w = input()
dct_str = dict.fromkeys(w, 0)
for alp in w:
dct_str[alp] += 1
flag = True
for alp in dct_str.keys():
if dct_str[alp] % 2 != 0:
flag = False
if flag == True:
print("Yes")
else:
print("No")
|
s507377505
|
p03399
|
u694422786
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 126
|
You planned a trip using trains and buses. The train fare will be A yen (the currency of Japan) if you buy ordinary tickets along the way, and B yen if you buy an unlimited ticket. Similarly, the bus fare will be C yen if you buy ordinary tickets along the way, and D yen if you buy an unlimited ticket. Find the minimum total fare when the optimal choices are made for trains and buses.
|
a = []
for i in range(4):
a.append(int(input()))
print(a)
train = min(a[0],a[1])
bus = min(a[2],a[3])
print(train + bus)
|
s095347184
|
Accepted
| 17
| 3,060
| 118
|
a = []
for i in range(4):
a.append(int(input()))
train = min(a[0],a[1])
bus = min(a[2],a[3])
print(train + bus)
|
s216770480
|
p03578
|
u089142196
| 2,000
| 262,144
|
Wrong Answer
| 319
| 57,136
| 300
|
Rng is preparing a problem set for a qualification round of CODEFESTIVAL. He has N candidates of problems. The difficulty of the i-th candidate is D_i. There must be M problems in the problem set, and the difficulty of the i-th problem must be T_i. Here, one candidate of a problem cannot be used as multiple problems. Determine whether Rng can complete the problem set without creating new candidates of problems.
|
N=int(input())
D=list(map(int,input().split()))
M=int(input())
T=list(map(int,input().split()))
from collections import Counter
a=Counter(D)
b=Counter(T)
for keys in b:
if keys not in a:
print("No")
break
else:
if b[keys]>a[keys]:
print("No")
break
else:
print("Yes")
|
s896998853
|
Accepted
| 324
| 56,800
| 300
|
N=int(input())
D=list(map(int,input().split()))
M=int(input())
T=list(map(int,input().split()))
from collections import Counter
a=Counter(D)
b=Counter(T)
for keys in b:
if keys not in a:
print("NO")
break
else:
if b[keys]>a[keys]:
print("NO")
break
else:
print("YES")
|
s158773763
|
p02690
|
u927373043
| 2,000
| 1,048,576
|
Wrong Answer
| 2,205
| 9,152
| 262
|
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
|
X = int(input())
b = 0
flag = True
while(flag):
for i in range(1,X+1):
if(X % i == 0):
print(i)
if(pow(b+i,5)-pow(b,5) == X):
print("%d %d"%(b+i,b))
flag = False
break
b -= 1
if(-pow(b,5) > X):
b = int(X**(1/5))
|
s613213974
|
Accepted
| 23
| 9,152
| 308
|
X = int(input())
a = []
b = 0
flag = True
num = 120
if(X <= 120):
num = X
for i in range(1,num+1):
if(X%i== 0):
a.append(i)
while(flag):
for i in range(len(a)):
if(pow(b+a[i],5)-pow(b,5) == X):
print("%d %d"%(b+a[i],b))
flag = False
break
b -= 1
if(b < -119):
b = 119
|
s333648048
|
p02407
|
u105694406
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,636
| 110
|
Write a program which reads a sequence and prints it in the reverse order.
|
a = list(map(int, input().split()))
a.reverse()
sum = ""
for i in a:
sum = sum + str(i) + " "
print(sum[:-1])
|
s367817310
|
Accepted
| 30
| 7,588
| 118
|
input()
a = list(map(int, input().split()))
a.reverse()
sum = ""
for i in a:
sum = sum + str(i) + " "
print(sum[:-1])
|
s691754988
|
p03386
|
u210817970
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 348
|
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.
|
s = input().split()
s = list(map(int,s))
ans = []
i = 0
while i<s[2]:
if (i+s[0])>s[1]:
i += 1
elif (i+s[0])<=s[1]:
ans.append(int(i + s[0]))
i += 1
i = 0
while i<s[2]:
if (s[1]-i)>=s[0]:
ans.append(int(s[1]-i))
i += 1
elif (s[1]-i)<s[0]:
i += 1
ans = list(set(ans))
for a in ans:
print(a)
|
s609373125
|
Accepted
| 17
| 3,064
| 359
|
s = input().split()
s = list(map(int,s))
ans = []
i = 0
while i<s[2]:
if (i+s[0])>s[1]:
i += 1
elif (i+s[0])<=s[1]:
ans.append(int(i + s[0]))
i += 1
i = 0
while i<s[2]:
if (s[1]-i)>=s[0]:
ans.append(int(s[1]-i))
i += 1
elif (s[1]-i)<s[0]:
i += 1
ans = list(set(ans))
ans.sort()
for a in ans:
print(a)
|
s066611922
|
p03909
|
u506127000
| 2,000
| 262,144
|
Wrong Answer
| 140
| 4,024
| 245
|
There is a grid with H rows and W columns. The square at the i-th row and j-th column contains a string S_{i,j} of length 5. The rows are labeled with the numbers from 1 through H, and the columns are labeled with the uppercase English letters from `A` through the W-th letter of the alphabet. Exactly one of the squares in the grid contains the string `snuke`. Find this square and report its location. For example, the square at the 6-th row and 8-th column should be reported as `H6`.
|
from string import ascii_uppercase
H, W = [int(i) for i in input().split()]
S = [input().split() for _ in range(H)]
for h in range(H):
for w in range(W):
if S[h][w] == "snuke":
print("{}{}".format(ascii_uppercase[w], h))
|
s323055567
|
Accepted
| 62
| 4,024
| 247
|
from string import ascii_uppercase
H, W = [int(i) for i in input().split()]
S = [input().split() for _ in range(H)]
for h in range(H):
for w in range(W):
if S[h][w] == "snuke":
print("{}{}".format(ascii_uppercase[w], h+1))
|
s454492445
|
p03471
|
u999503965
| 2,000
| 262,144
|
Wrong Answer
| 341
| 8,944
| 226
|
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
n,y=map(int,input().split())
a,b,c=-1,-1,-1
for i in range(n+1):
for j in range(n+1):
if i+j>0:
continue
h=n-i-j
if 10000*i+5000*j+1000*h==y:
a,b,c=i,j,h
break
print("{} {} {}".format(a,b,c))
|
s693138229
|
Accepted
| 1,108
| 9,108
| 206
|
n,y=map(int,input().split())
a,b,c=-1,-1,-1
for i in range(n+1):
for j in range(n+1):
h=n-i-j
if 10000*i+5000*j+1000*h==y and h>=0:
a,b,c=i,j,h
break
print("{} {} {}".format(a,b,c))
|
s715336720
|
p02396
|
u498462680
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,540
| 17
|
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.
|
Indata = input()
|
s933660597
|
Accepted
| 140
| 5,604
| 152
|
i=0
while True:
Indata = int(input())
if Indata == 0:
break
else:
i=i+1
print("Case " +str(i) + ": " + str(Indata))
|
s071039525
|
p02612
|
u131411061
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,168
| 48
|
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())
tmp = N//1000
print(N-tmp*1000)
|
s082824767
|
Accepted
| 31
| 9,152
| 105
|
N = int(input())
if N%1000==0:
print(0)
else:
tmp = N//1000
res = tmp+1
print(res*1000-N)
|
s322624083
|
p03448
|
u038027079
| 2,000
| 262,144
|
Wrong Answer
| 148
| 4,080
| 272
|
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()) for _ in range(4)]
count = 0
for i in range(A[0] + 1):
for j in range(A[1] + 1):
for k in range(A[2] + 1):
sum = i * 500 + j * 100 + k * 50
if sum == A[3]:
count += 1
print(sum)
print(count)
|
s103471025
|
Accepted
| 55
| 3,060
| 249
|
A = [int(input()) for _ in range(4)]
count = 0
for i in range(A[0] + 1):
for j in range(A[1] + 1):
for k in range(A[2] + 1):
sum = i * 500 + j * 100 + k * 50
if sum == A[3]:
count += 1
print(count)
|
s176079997
|
p02795
|
u417589709
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 73
|
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.
|
h = int(input())
w = int(input())
n = int(input())
t = 0
print(max(h,w))
|
s971866381
|
Accepted
| 17
| 2,940
| 83
|
h = int(input())
w = int(input())
n = int(input())
print(int((n-1) / max(h,w))+1)
|
s330798692
|
p03598
|
u543954314
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 134
|
There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i). Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N. In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B. Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i). Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N. When activated, each type of robot will operate as follows. * When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. * When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.
|
n = int(input())
k = int(input())
x = list(map(int, input().split()))
ans = 0
for i in range(n):
ans += min(x[i], k-x[i])
print(ans)
|
s733230763
|
Accepted
| 18
| 2,940
| 136
|
n = int(input())
k = int(input())
x = list(map(int, input().split()))
ans = 0
for i in range(n):
ans += min(x[i], k-x[i])
print(ans*2)
|
s929609686
|
p03409
|
u404710939
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,064
| 390
|
On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (a_i, b_i), and the coordinates of the i-th blue point are (c_i, d_i). A red point and a blue point can form a _friendly pair_ when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point. At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs.
|
N = int(input())
ab = [[int(i) for i in input().split(" ")] for _ in range(N)]
cd = [[int(i) for i in input().split(" ")] for _ in range(N)]
ab_sort = sorted(ab, key=lambda x:(-x[0], -x[1]))
cd_sort = sorted(cd, key=lambda x:(-x[0], -x[1]))
cnt = 0
for c, d in cd_sort:
for a, b in ab_sort:
if a < c and b < d:
ab_sort.remove([a, b])
cnt += 1
print(cnt)
|
s907436121
|
Accepted
| 18
| 3,064
| 390
|
N = int(input())
ab = [[int(i) for i in input().split(" ")] for _ in range(N)]
cd = [[int(i) for i in input().split(" ")] for _ in range(N)]
ab_sort = sorted(ab, key=lambda x:-x[1])
cd_sort = sorted(cd, key=lambda x:x[0])
cnt = 0
for c, d in cd_sort:
for a, b in ab_sort:
if a < c and b < d:
ab_sort.remove([a, b])
cnt += 1
break
print(cnt)
|
s771540514
|
p03970
|
u396391104
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 111
|
CODE FESTIVAL 2016 is going to be held. For the occasion, Mr. Takahashi decided to make a signboard. He intended to write `CODEFESTIVAL2016` on it, but he mistakenly wrote a different string S. Fortunately, the string he wrote was the correct length. So Mr. Takahashi decided to perform an operation that replaces a certain character with another in the minimum number of iterations, changing the string to `CODEFESTIVAL2016`. Find the minimum number of iterations for the rewrite operation.
|
S = input()
T = "CODEFESTIVAL2016"
ans = 0
for i in range(len(S)):
if S[i] == T[i]:
ans += 1
print(ans)
|
s798073754
|
Accepted
| 17
| 2,940
| 111
|
S = input()
T = "CODEFESTIVAL2016"
ans = 0
for i in range(len(S)):
if S[i] != T[i]:
ans += 1
print(ans)
|
s254967090
|
p03658
|
u419963262
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 126
|
Snuke has N sticks. The length of the i-th stick is l_i. Snuke is making a snake toy by joining K of the sticks together. The length of the toy is represented by the sum of the individual sticks that compose it. Find the maximum possible length of the toy.
|
n,k=map(int,input().split())
a=list(map(int,input().split()))
b=sorted(a)
ans=0
for i in range(k):
ans+=a[(-1)*i]
print(ans)
|
s256327395
|
Accepted
| 19
| 2,940
| 131
|
n,k=map(int,input().split())
a=list(map(int,input().split()))
b=sorted(a)
ans=0
for i in range(1,k+1):
ans+=b[(-1)*i]
print(ans)
|
s660197813
|
p03386
|
u562016607
| 2,000
| 262,144
|
Wrong Answer
| 2,151
| 797,592
| 166
|
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=[]
for i in range(A,max([A+K,B+1])):
S.append(i)
for i in range(min([B-K+1,A]),B+1):
S.append(i)
S.sort()
for i in S:
print(i)
|
s205772618
|
Accepted
| 17
| 3,060
| 173
|
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([B-K+1,A]),B+1):
S.add(i)
T=list(S)
T.sort()
for i in T:
print(i)
|
s703690049
|
p03555
|
u898967808
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 94
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
c1=input()
c2=input()
if c1==c2[::-1] and c2==c1[::-1]:
print('Yes')
else:
print('No')
|
s313158266
|
Accepted
| 17
| 2,940
| 113
|
c1=input()
c2=input()
if c1[0]==c2[2] and c1[1] == c2[1] and c1[2] == c2[0]:
print('YES')
else:
print('NO')
|
s940282698
|
p03544
|
u131625544
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 166
|
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)
|
n = int(input())
luc = [2, 1]
if n < 3:
print(luc[n - 1])
else:
for _ in range(n - 3):
tmp = sum(luc)
luc = luc[1:] + [tmp]
print(luc[1])
|
s359446200
|
Accepted
| 17
| 3,064
| 162
|
n = int(input())
luc = [2, 1]
if n < 2:
print(luc[n])
else:
for _ in range(n - 1):
tmp = sum(luc)
luc = luc[1:] + [tmp]
print(luc[1])
|
s190754776
|
p03160
|
u364061715
| 2,000
| 1,048,576
|
Wrong Answer
| 155
| 13,924
| 337
|
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
N = int(input())
costlist = list(map(int,input().split()))
DP = [0 for i in range(N)]
DP[1]=abs(costlist[1]-costlist[0])
for i in range(N-2):
probmin1 = abs(costlist[i+2]-costlist[i+1]) + costlist[i+1]
probmin2 = abs(costlist[i+2]-costlist[i]) + costlist[i]
plusmin = min(probmin1,probmin2)
DP[i+2]=plusmin
print(DP[-1])
|
s790281660
|
Accepted
| 141
| 13,924
| 325
|
N = int(input())
costlist = list(map(int,input().split()))
DP = [0 for i in range(N)]
DP[1]=abs(costlist[1]-costlist[0])
for i in range(N-2):
probmin1 = abs(costlist[i+2]-costlist[i+1]) + DP[i+1]
probmin2 = abs(costlist[i+2]-costlist[i]) + DP[i]
plusmin = min(probmin1,probmin2)
DP[i+2]=plusmin
print(DP[-1])
|
s149415827
|
p03408
|
u622570247
| 2,000
| 262,144
|
Wrong Answer
| 31
| 9,412
| 285
|
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
from collections import Counter
n = int(input())
s = [None] * n
for i in range(n):
s[i] = input().rstrip()
m = int(input())
t = [None] * m
for i in range(m):
t[i] = input().rstrip()
ctr = Counter(s) - Counter(t)
if ctr:
print(ctr.most_common(1)[0][0])
else:
print(0)
|
s663848275
|
Accepted
| 32
| 9,408
| 285
|
from collections import Counter
n = int(input())
s = [None] * n
for i in range(n):
s[i] = input().rstrip()
m = int(input())
t = [None] * m
for i in range(m):
t[i] = input().rstrip()
ctr = Counter(s) - Counter(t)
if ctr:
print(ctr.most_common(1)[0][1])
else:
print(0)
|
s698967176
|
p02419
|
u324811972
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,556
| 181
|
Write a program which reads a word W and a text T, and prints the number of word W which appears in text T T consists of string Ti separated by space characters and newlines. Count the number of Ti which equals to W. The word and text are case insensitive.
|
target = input()
text = input()
text2 = text.replace('\n',' ')
sentenceList = text2.split()
count = 0
for word in sentenceList:
if(word.lower() == target):
count+=1
print(count)
|
s922396626
|
Accepted
| 20
| 5,552
| 285
|
target = input()
total_text = ""
while True:
text = input()
if(text=="END_OF_TEXT"):
break
total_text+=" "
total_text+=text
#text2 = text.replace('\n',' ')
sentenceList = total_text.split()
count = 0
for word in sentenceList:
if(word.lower() == target):
count+=1
print(count)
|
s681956776
|
p02410
|
u757827098
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,668
| 309
|
Write a program which reads a $ n \times m$ matrix $A$ and a $m \times 1$ vector $b$, and prints their product $Ab$. A column vector with m elements is represented by the following equation. \\[ b = \left( \begin{array}{c} b_1 \\\ b_2 \\\ : \\\ b_m \\\ \end{array} \right) \\] A $n \times m$ matrix with $m$ column vectors, each of which consists of $n$ elements, is represented by the following equation. \\[ A = \left( \begin{array}{cccc} a_{11} & a_{12} & ... & a_{1m} \\\ a_{21} & a_{22} & ... & a_{2m} \\\ : & : & : & : \\\ a_{n1} & a_{n2} & ... & a_{nm} \\\ \end{array} \right) \\] $i$-th element of a $m \times 1$ column vector $b$ is represented by $b_i$ ($i = 1, 2, ..., m$), and the element in $i$-th row and $j$-th column of a matrix $A$ is represented by $a_{ij}$ ($i = 1, 2, ..., n,$ $j = 1, 2, ..., m$). The product of a $n \times m$ matrix $A$ and a $m \times 1$ column vector $b$ is a $n \times 1$ column vector $c$, and $c_i$ is obtained by the following formula: \\[ c_i = \sum_{j=1}^m a_{ij}b_j = a_{i1}b_1 + a_{i2}b_2 + ... + a_{im}b_m \\]
|
n,m =[int(r) for r in input().split()]
matrix = []
for ni in range(n):
matrix.append([int(a)for a in input().split()])
vector = []
for mi in range(m):
vector.append(int(input()))
for i in range(len(matrix)):
sum = 0
for j in range(len(matrix[i])):
sum += matrix[i][j] * vector[j]
|
s515347797
|
Accepted
| 30
| 7,988
| 306
|
n,m =[int(r) for r in input().split()]
matrix = []
for ni in range(n):
matrix.append([int(a)for a in input().split()])
vector = []
for mi in range(m):
vector.append(int(input()))
for ni in range(n):
sum = 0
for mi in range(m):
sum += matrix[ni][mi] * vector[mi]
print(sum)
|
s070563010
|
p03068
|
u250102437
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 425
|
You are given a string S of length N consisting of lowercase English letters, and an integer K. Print the string obtained by replacing every character in S that differs from the K-th character of S, with `*`.
|
# -*- coding: utf-8 -*-
from sys import stdin
############################################
# read data for n sequences.
n = stdin.readline()
N = int(n)
n = stdin.readline()
S = str(n)
n = stdin.readline()
K = int(n)
#data = [int(stdin.readline().rstrip()) for _ in range(n)]
key = S[K]
out=[]
for i,l in enumerate(S):
if l != key:
out.append("*")
else:
out.append(l)
out1=''
for o in out:
out1+=o
print(out1)
|
s055559991
|
Accepted
| 17
| 3,064
| 467
|
# -*- coding: utf-8 -*-
from sys import stdin
############################################
# read data for n sequences.
n = stdin.readline()
N = int(n)
n = stdin.readline()
S = str(n)
n = stdin.readline()
K = int(n)
#data = [int(stdin.readline().rstrip()) for _ in range(n)]
key = S[K-1]
out=[]
for i,l in enumerate(S):
if i < N:
if l != key:
out.append("*")
elif l =="\n":
x=1
else:
out.append(l)
out1=''
for o in out:
out1+=o
print(out1)
|
s655324624
|
p03079
|
u114648678
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 105
|
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 and B+C>=A and C+A>=B:
print('YES')
else :
print('NO')
|
s633917944
|
Accepted
| 18
| 2,940
| 91
|
A,B,C=map(int,input().split())
if A==B and B==C:
print('Yes')
else :
print('No')
|
s121524113
|
p02690
|
u735891571
| 2,000
| 1,048,576
|
Wrong Answer
| 32
| 9,184
| 281
|
Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X.
|
X = int(input())
lis = []
for i in range(-200, 201):
lis.append(i**5)
flag = 0
for a, i in enumerate(lis):
for b, j in enumerate(lis):
if (i + j == X) or (j - i == X):
flag = 1
break
if flag == 1:
break
print(a-200, b-200)
|
s644411459
|
Accepted
| 37
| 9,196
| 293
|
X = int(input())
lis = []
for i in range(-200, 201):
lis.append(i**5)
flag = 0
for a, i in enumerate(lis):
for b, j in enumerate(lis):
if (i + j == X) or (j - i == X):
flag = 1
break
if flag == 1:
break
print((a-200)*(-1), (b-200)*(-1))
|
s188458620
|
p02865
|
u541475502
| 2,000
| 1,048,576
|
Wrong Answer
| 20
| 3,316
| 27
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
a = int(input())
print(a-1)
|
s384327586
|
Accepted
| 17
| 2,940
| 72
|
a = int(input())
if a % 2 == 0:
print(a//2-1)
else:
print(a//2)
|
s108576715
|
p03846
|
u941753895
| 2,000
| 262,144
|
Wrong Answer
| 101
| 13,812
| 752
|
There are N people, conveniently numbered 1 through N. They were standing in a row yesterday, but now they are unsure of the order in which they were standing. However, each person remembered the following fact: the absolute difference of the number of the people who were standing to the left of that person, and the number of the people who were standing to the right of that person. According to their reports, the difference above for person i is A_i. Based on these reports, find the number of the possible orders in which they were standing. Since it can be extremely large, print the answer modulo 10^9+7. Note that the reports may be incorrect and thus there may be no consistent order. In such a case, print 0.
|
a=1000000007
n=int(input())
if n==5:
exit()
l=list(map(int,input().split()))
l.sort()
if n%2==1:
if l[0]!=0:
print(0)
exit()
else:
l.remove(0)
sh=2
f=0
for i in l:
if f==0:
if i!=sh:
print(0)
exit()
else:
f=1
else:
if i!=sh:
print(0)
exit()
else:
sh+=2
f=0
sh=(n-1)//2
b=1
for i in range(sh):
b*=2
b%=a
print(b)
else:
sh=1
f=0
for i in l:
if f==0:
if i!=sh:
print(0)
exit()
else:
f=1
else:
if i!=sh:
print(0)
exit()
else:
sh+=2
f=0
sh=n//2
b=1
for i in range(sh):
b*=2
b%=a
print(b)
|
s786978794
|
Accepted
| 104
| 16,300
| 549
|
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
def LI(): return list(map(int,input().split()))
def I(): return int(input())
def LS(): return input().split()
def S(): return input()
def main():
n=I()
l=LI()
l.sort()
if n%2==0:
for i in range(n//2):
if l[i]!=i//2*2+1:
return 0
else:
if l[0]!=0:
return 0
for i in range(1,n):
if -(-i//2)*2!=l[i]:
return 0
return pow(2,n//2)%mod
print(main())
|
s024013374
|
p03386
|
u606033239
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 2,940
| 112
|
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 = list(i for i in range(a,b+1) if i<=a+k-1 or i>=b+1-k)
s.sort()
print(s)
|
s579568070
|
Accepted
| 19
| 3,060
| 99
|
a,b,k=map(int,input().split())
s=range(a,b+1)
for i in sorted(set(s[:k])|set(s[-k:])):
print(i)
|
s377844458
|
p03369
|
u496744988
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 131
|
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
s=input()
price=0
if s[0] == 'o':
price += 700
if s[1] == 'o':
price += 100
if s[2] == 'o':
price += 100
print(price)
|
s645034901
|
Accepted
| 17
| 2,940
| 133
|
s=input()
price=700
if s[0] == 'o':
price += 100
if s[1] == 'o':
price += 100
if s[2] == 'o':
price += 100
print(price)
|
s031530029
|
p03493
|
u221888299
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 229
|
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.
|
print('数字を連続で入力')
s1,s2,s3 = map(int, input())
while s1<0 or s1>1 or s2<0 or s2>1 or s3<0 or s3>1:
print('数字を連続で再入力')
s1,s2,s3 = map(int, input())
ans = s1+s2+s3
print("{}".format(ans))
|
s722553288
|
Accepted
| 17
| 2,940
| 31
|
n = input()
print(n.count('1'))
|
s001594685
|
p03795
|
u982591663
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 45
|
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
N = int(input())
print(800*N + 200*(N//15))
|
s566036847
|
Accepted
| 17
| 2,940
| 45
|
N = int(input())
print(800*N - 200*(N//15))
|
s899331886
|
p03698
|
u445404615
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 110
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
s = input()
for i in range(len(s)):
if s.count(s[i]) != 1:
print('No')
exit()
print('Yes')
|
s048115512
|
Accepted
| 17
| 2,940
| 110
|
s = input()
for i in range(len(s)):
if s.count(s[i]) != 1:
print('no')
exit()
print('yes')
|
s214162820
|
p03605
|
u844789719
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 77
|
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
N = input()
if N[0] == 9 or N[1] == 9:
print('Yes')
else:
print('No')
|
s771563712
|
Accepted
| 17
| 2,940
| 81
|
N = input()
if N[0] == '9' or N[1] == '9':
print('Yes')
else:
print('No')
|
s608645557
|
p02409
|
u682153677
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,628
| 1,047
|
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
# -*- coding: utf-8 -*-
a = {}
b = {}
c = {}
d = {}
for i in range(1, 30):
a[i] = 0
b[i] = 0
c[i] = 0
d[i] = 0
n = int(input())
for j in range(n):
m = list(map(int, input().split()))
if m[0] == 1:
a[m[2] + 10 * (m[1] - 1)] = a[m[2] + 10 * (m[1] - 1)] + m[3]
elif m[0] == 2:
b[m[2] + 10 * (m[1] - 1)] = b[m[2] + 10 * (m[1] - 1)] + m[3]
elif m[0] == 3:
c[m[2] + 10 * (m[1] - 1)] = b[m[2] + 10 * (m[1] - 1)] + m[3]
elif m[0] == 4:
d[m[2] + 10 * (m[1] - 1)] = d[m[2] + 10 * (m[1] - 1)] + m[3]
for i in range(1, 3):
for j in range(1, 10):
print(' {0}'.format(a[i * j]), end='')
print('####################')
for i in range(1, 3):
for j in range(1, 10):
print(' {0}'.format(b[i * j]), end='')
print('####################')
for i in range(1, 3):
for j in range(1, 10):
print(' {0}'.format(c[i * j]), end='')
print('####################')
for i in range(1, 3):
for j in range(1, 10):
print(' {0}'.format(d[i * j]), end='')
|
s503111558
|
Accepted
| 20
| 5,640
| 1,139
|
# -*- coding: utf-8 -*-
a = {}
b = {}
c = {}
d = {}
for i in range(1, 31):
a[i] = 0
b[i] = 0
c[i] = 0
d[i] = 0
n = int(input())
for j in range(n):
m = list(map(int, input().split()))
if m[0] == 1:
a[m[2] + 10 * (m[1] - 1)] = a[m[2] + 10 * (m[1] - 1)] + m[3]
elif m[0] == 2:
b[m[2] + 10 * (m[1] - 1)] = b[m[2] + 10 * (m[1] - 1)] + m[3]
elif m[0] == 3:
c[m[2] + 10 * (m[1] - 1)] = c[m[2] + 10 * (m[1] - 1)] + m[3]
elif m[0] == 4:
d[m[2] + 10 * (m[1] - 1)] = d[m[2] + 10 * (m[1] - 1)] + m[3]
for i in range(1, 4):
for j in range(1, 11):
print(' {0}'.format(a[j + 10 * (i - 1)]), end='')
print()
print('####################')
for i in range(1, 4):
for j in range(1, 11):
print(' {0}'.format(b[j + 10 * (i - 1)]), end='')
print()
print('####################')
for i in range(1, 4):
for j in range(1, 11):
print(' {0}'.format(c[j + 10 * (i - 1)]), end='')
print()
print('####################')
for i in range(1, 4):
for j in range(1, 11):
print(' {0}'.format(d[j + 10 * (i - 1)]), end='')
print()
|
s451866700
|
p02694
|
u861471387
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,260
| 118
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
import math
X = int(input())
m = 100
ans = 0
while m<X:
m=math.floor(m*1.01)
ans+=1
print(m)
print(ans)
|
s682066078
|
Accepted
| 20
| 9,168
| 105
|
import math
X = int(input())
m = 100
ans = 0
while m<X:
m=math.floor(m*1.01)
ans+=1
print(ans)
|
s194864077
|
p02694
|
u706908631
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,156
| 127
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
N = int(input())
X = 100
count = 0
while True:
if X >= N:
print(count)
break
else:
X = X + (X * 0.01)
|
s013048766
|
Accepted
| 23
| 9,168
| 142
|
N = int(input())
X = 100
count = 0
while True:
if X >= N:
print(count)
break
else:
X = int(X + (X * 0.01))
count += 1
|
s173439186
|
p03711
|
u706828591
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 243
|
Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.
|
x, y = map(int,input().split())
#ls = list(map(int,input().split()))
#r = int(input())
#s = input()
ls = [4,6,9,11]
if (ls.count(x) == 1 and ls.count(y) == 1) or (x == 2 and y == 2):
print("Yes")
else:
print("No")
|
s892766986
|
Accepted
| 17
| 3,060
| 308
|
x, y = map(int,input().split())
#ls = list(map(int,input().split()))
#r = int(input())
#s = input()
ls = [4,6,9,11]
if (ls.count(x) == 1 and ls.count(y) == 1) or (x == 2 and y == 2) or (ls.count(x) != 1 and x != 2 and y != 2 and ls.count(y) != 1):
print("Yes")
else:
print("No")
|
s380998996
|
p03681
|
u622045059
| 2,000
| 262,144
|
Wrong Answer
| 2,107
| 83,316
| 289
|
Snuke has N dogs and M monkeys. He wants them to line up in a row. As a Japanese saying goes, these dogs and monkeys are on bad terms. _("ken'en no naka", literally "the relationship of dogs and monkeys", means a relationship of mutual hatred.)_ Snuke is trying to reconsile them, by arranging the animals so that there are neither two adjacent dogs nor two adjacent monkeys. How many such arrangements there are? Find the count modulo 10^9+7 (since animals cannot understand numbers larger than that). Here, dogs and monkeys are both distinguishable. Also, two arrangements that result from reversing each other are distinguished.
|
import sys
sys.setrecursionlimit(1000000)
N,M = map(int, input().split())
print(sys.getrecursionlimit())
if abs(N-M) > 1:
print(0)
exit()
def frac(n):
if n == 1: return 1
return n * frac(n-1)
if N-M == 0:
print(frac(N)*frac(M)*2)
else:
print(frac(N)*frac(M))
|
s732989693
|
Accepted
| 491
| 5,184
| 306
|
from math import factorial
N,M = map(int, input().split())
mod = 10 ** 9 + 7
if abs(N-M) > 1:
print(0)
exit()
n_frac = factorial(N)
m_frac = n_frac
if N-M == 0:
print((n_frac*m_frac)*2%mod)
else:
if N > M:
m_frac //= N
else:
m_frac *= M
print(n_frac*m_frac%mod)
|
s476664930
|
p03433
|
u951980350
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 161
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
N = int(input())
A = int(input())
if A >= N:
print("yes")
else:
n = N % 500
if n != 0 and A >= n:
print("yes")
else:
print("no")
|
s503200843
|
Accepted
| 18
| 2,940
| 173
|
N = int(input())
A = int(input())
if A >= N:
print("Yes")
else:
n = N % 500
if n == 0 or (n != 0 and A >= n):
print("Yes")
else:
print("No")
|
s216508789
|
p03400
|
u280512618
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 107
|
Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp.
|
n=int(input())
d,x=map(int,input().split())
s=d*(1+d)//2
print(x+n*d+sum(int(input()) for _ in range(n))*s)
|
s470955275
|
Accepted
| 20
| 2,940
| 97
|
n=int(input())
d,x=map(int,input().split())
print(x+n+sum((d-1)//int(input()) for _ in range(n)))
|
s201415269
|
p03699
|
u488178971
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 247
|
You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade?
|
N = int(input())
S = []
for i in range(N):
S.append(int(input()))
sorted(S)
if sum(S)%10!=0:
print(sum(S))
else:
while len(S)>0:
S.pop(0)
if sum(S)%10!=0 or len(S)==0:
print(sum(S))
exit()
|
s958115346
|
Accepted
| 18
| 3,064
| 261
|
N = int(input())
S = []
for i in range(N):
S.append(int(input()))
nb10 = ([s for s in S if s%10 != 0])
b10 = ([s for s in S if s%10 == 0])
if sum(S)%10!=0:
print(sum(S))
elif sum(S) ==sum(b10):
print(0)
else:
print(sum(S)-sorted(nb10)[0])
|
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