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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s745224245
|
p03472
|
u293976191
| 2,000 | 262,144 |
Wrong Answer
| 442 | 59,972 | 361 |
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?
|
N, H = map(int, input().split(' '))
katana = [map(int, input().split(' ')) for _ in range(N)]
dmg_wield, dmg_throw = zip(*katana)
dmg_wield_max = max(dmg_wield)
dmg_throw = [d for d in sorted(dmg_throw, reverse=True) if d >= dmg_wield_max]
n = len(dmg_throw)
H -= sum(dmg_throw)
n += int(H // dmg_wield_max)
if H - dmg_wield_max * n > 0:
n += 1
print(n)
|
s698050374
|
Accepted
| 456 | 59,860 | 341 |
N, H = map(int, input().split(' '))
katana = [map(int, input().split(' ')) for _ in range(N)]
dmg_wield, dmg_throw = zip(*katana)
dmg_wield_max = max(dmg_wield)
dmg_throw = sorted([d for d in dmg_throw if d > dmg_wield_max])
n = 0
while H > 0 and dmg_throw:
H -= dmg_throw.pop()
n += 1
print(n + max(H-1, -1) // dmg_wield_max + 1)
|
s703244313
|
p02534
|
u919235786
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 9,080 | 89 |
You are given an integer K. Print the string obtained by repeating the string `ACL` K times and concatenating them. For example, if K = 3, print `ACLACLACL`.
|
#a
k=int(input())
ans=[]
for i in range(k):
ans.append('acl')
a=''.join(ans)
print(a)
|
s524925396
|
Accepted
| 25 | 9,080 | 89 |
#a
k=int(input())
ans=[]
for i in range(k):
ans.append('ACL')
a=''.join(ans)
print(a)
|
s474811849
|
p03448
|
u223555291
| 2,000 | 262,144 |
Wrong Answer
| 49 | 2,940 | 219 |
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())
d=int(input())
count=0
for i in range(a+1):
for j in range(b+1):
for z in range(c+1):
if a*500+b*100+c*50 == d:
count+=1
print(count)
|
s921733597
|
Accepted
| 47 | 3,060 | 219 |
a=int(input())
b=int(input())
c=int(input())
d=int(input())
count=0
for i in range(a+1):
for j in range(b+1):
for z in range(c+1):
if i*500+j*100+z*50 == d:
count+=1
print(count)
|
s654379278
|
p03623
|
u446711904
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 61 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x,a,b=map(int,input().split());print("AB"[abs(x-a)<abs(x-b)])
|
s588640259
|
Accepted
| 17 | 2,940 | 61 |
x,a,b=map(int,input().split());print("AB"[abs(x-a)>abs(x-b)])
|
s200115760
|
p03302
|
u859897687
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 84 |
You are given two integers a and b. Determine if a+b=15 or a\times b=15 or neither holds. Note that a+b=15 and a\times b=15 do not hold at the same time.
|
a,b=map(int,input().split())
if a*b==15 and a+b==15:
print("*")
else:
print("x")
|
s077462409
|
Accepted
| 17 | 2,940 | 99 |
a,b=map(int,input().split())
if a*b==15:
print("*")
elif a+b==15:
print("+")
else:
print("x")
|
s497990880
|
p03434
|
u703890795
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 162 |
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
N = int(input())
a = list(map(int,input().split()))
c = 0
a.sort(reverse=True)
for i in range(N):
if i//2 == 0:
c += a[i]
else:
c -= a[i]
print(a[i])
|
s972634271
|
Accepted
| 17 | 2,940 | 158 |
N = int(input())
a = list(map(int,input().split()))
c = 0
a.sort(reverse=True)
for i in range(N):
if i%2 == 0:
c += a[i]
else:
c -= a[i]
print(c)
|
s076675915
|
p03448
|
u488737015
| 2,000 | 262,144 |
Wrong Answer
| 56 | 2,940 | 258 |
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)])
count = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
total = 500 * a + 100 * b + 50 * c
if total == x:
count = count + 1
print(count)
|
s055372724
|
Accepted
| 53 | 2,940 | 258 |
a, b, c, x = map(int, [input() for i in range(4)])
count = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
total = 500 * i + 100 * j + 50 * k
if total == x:
count = count + 1
print(count)
|
s106082634
|
p02536
|
u790812284
| 2,000 | 1,048,576 |
Wrong Answer
| 309 | 12,748 | 1,324 |
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?
|
class UnionFind():
def __init__(self, n):
self.n = n
self.parents = [-1] * n
def find(self, x):
if self.parents[x] < 0:
return x
else:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if x == y:
return
if self.parents[x] > self.parents[y]:
x, y = y, x
self.parents[x] += self.parents[y]
self.parents[y] = x
def size(self, x):
return -self.parents[self.find(x)]
def same(self, x, y):
return self.find(x) == self.find(y)
def members(self, x):
root = self.find(x)
return [i for i in range(self.n) if self.find(i) == root]
def roots(self):
return [i for i, x in enumerate(self.parents) if x < 0]
def group_count(self):
return len(self.roots())
def all_group_members(self):
return {r: self.members(r) for r in self.roots()}
def __str__(self):
return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())
n, m = map(int, input().split())
uf = UnionFind(n)
for i in range(m):
a, b = map(int, input().split())
a -= 1
b -= 1
uf.union(a,b)
print(uf.group_count())
|
s408356664
|
Accepted
| 309 | 12,856 | 1,328 |
class UnionFind():
def __init__(self, n):
self.n = n
self.parents = [-1] * n
def find(self, x):
if self.parents[x] < 0:
return x
else:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if x == y:
return
if self.parents[x] > self.parents[y]:
x, y = y, x
self.parents[x] += self.parents[y]
self.parents[y] = x
def size(self, x):
return -self.parents[self.find(x)]
def same(self, x, y):
return self.find(x) == self.find(y)
def members(self, x):
root = self.find(x)
return [i for i in range(self.n) if self.find(i) == root]
def roots(self):
return [i for i, x in enumerate(self.parents) if x < 0]
def group_count(self):
return len(self.roots())
def all_group_members(self):
return {r: self.members(r) for r in self.roots()}
def __str__(self):
return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())
n, m = map(int, input().split())
uf = UnionFind(n)
for i in range(m):
a, b = map(int, input().split())
a -= 1
b -= 1
uf.union(a,b)
print(uf.group_count() -1 )
|
s369800628
|
p02578
|
u224989615
| 2,000 | 1,048,576 |
Wrong Answer
| 143 | 32,216 | 149 |
N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal.
|
n = int(input())
a = list(map(int, input().split()))
total = 0
for i in range(1, n):
if a[i-1] > a[i]:
total += a[i-1] - a[i]
a[i] = a[i-1]
|
s560815092
|
Accepted
| 143 | 32,140 | 162 |
n = int(input())
a = list(map(int, input().split()))
total = 0
for i in range(1, n):
if a[i-1] > a[i]:
total += a[i-1] - a[i]
a[i] = a[i-1]
print(total)
|
s963828730
|
p03478
|
u858670323
| 2,000 | 262,144 |
Wrong Answer
| 46 | 3,060 | 211 |
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().rstrip().split(' '))
def count(n):
x = str(n)
sum=0
for s in x:
sum+=int(s)
return sum
ans=0
for i in range(1,n+1):
if(a <= count(i) and count(i) <=b):ans+=1
print(ans)
|
s221796391
|
Accepted
| 41 | 3,060 | 212 |
n,a,b=map(int,input().rstrip().split(' '))
def count(n):
x = str(n)
sum=0
for s in x:
sum+=int(s)
return sum
ans=0
for i in range(1,n+1):
if(a <= count(i) and count(i) <=b):ans+=i
print(ans)
|
s691011558
|
p03610
|
u239981649
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,188 | 21 |
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.
|
print(input()[1::2])
|
s583113359
|
Accepted
| 18 | 3,192 | 20 |
print(input()[::2])
|
s962507545
|
p03626
|
u562016607
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 321 |
We have a board with a 2 \times N grid. Snuke covered the board with N dominoes without overlaps. Here, a domino can cover a 1 \times 2 or 2 \times 1 square. Then, Snuke decided to paint these dominoes using three colors: red, cyan and green. Two dominoes that are adjacent by side should be painted by different colors. Here, it is not always necessary to use all three colors. Find the number of such ways to paint the dominoes, modulo 1000000007. The arrangement of the dominoes is given to you as two strings S_1 and S_2 in the following manner: * Each domino is represented by a different English letter (lowercase or uppercase). * The j-th character in S_i represents the domino that occupies the square at the i-th row from the top and j-th column from the left.
|
P=10**9+7
N=int(input())
S1=input()
S2=input()
L=[1]
k=0
P=10**9+7
for i in range(1,N):
if S1[i]!=S1[i-1]:
L.append(0)
k+=1
L[k]+=1
M=len(L)
S=1
if L[k]==1:
S=3
else:
S=6
for i in range(1,M):
if L[i-1]==1:
S=(S*2)%P
else:
if L[i]==2:
S=(S*3)%P
print(S)
|
s716272047
|
Accepted
| 19 | 3,188 | 321 |
P=10**9+7
N=int(input())
S1=input()
S2=input()
L=[1]
k=0
P=10**9+7
for i in range(1,N):
if S1[i]!=S1[i-1]:
L.append(0)
k+=1
L[k]+=1
M=len(L)
S=1
if L[0]==1:
S=3
else:
S=6
for i in range(1,M):
if L[i-1]==1:
S=(S*2)%P
else:
if L[i]==2:
S=(S*3)%P
print(S)
|
s951861633
|
p03543
|
u708211626
| 2,000 | 262,144 |
Wrong Answer
| 23 | 9,020 | 62 |
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**?
|
a=set(input())
if len(a)<2:
print('Yes')
else:
print('No')
|
s274658913
|
Accepted
| 30 | 9,028 | 100 |
a=input()
b=set(a[:-1]);c=set(a[1:])
if len(b)==1 or len(c)==1:
print('Yes')
else:
print('No')
|
s828750101
|
p02607
|
u057942294
| 2,000 | 1,048,576 |
Wrong Answer
| 33 | 9,060 | 486 |
We have N squares assigned the numbers 1,2,3,\ldots,N. Each square has an integer written on it, and the integer written on Square i is a_i. How many squares i satisfy both of the following conditions? * The assigned number, i, is odd. * The written integer is odd.
|
def main():
N = input_ints()
a = input_int_list_in_line()
print(len([True for i in range(N) if i % 2 == 0 and a[i] % 2 == 0]))
def input_ints():
line_list = input().split()
if len(line_list) == 1:
return int(line_list[0])
else:
return map(int, line_list)
def input_int_list_in_line():
return list(map(int, input().split()))
def input_int_tuple_list(n: int):
return [tuple(map(int, input().split())) for _ in range(n)]
main()
|
s525660682
|
Accepted
| 27 | 8,976 | 486 |
def main():
N = input_ints()
a = input_int_list_in_line()
print(len([True for i in range(N) if i % 2 == 0 and a[i] % 2 == 1]))
def input_ints():
line_list = input().split()
if len(line_list) == 1:
return int(line_list[0])
else:
return map(int, line_list)
def input_int_list_in_line():
return list(map(int, input().split()))
def input_int_tuple_list(n: int):
return [tuple(map(int, input().split())) for _ in range(n)]
main()
|
s761265665
|
p02678
|
u459590249
| 2,000 | 1,048,576 |
Wrong Answer
| 663 | 35,892 | 446 |
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.
|
#from collections import deque
n,m=map(int,input().split())
root=[[] for _ in range(n)]
for i in range(m):
a,b=map(int,input().split())
root[a-1].append(b-1)
root[b-1].append(a-1)
times=[0]*n
sign=[-1]*n
sign[0]=0
#q=deque(root[0])
flag=0
a=root[0]
b=[]
f=False
while True:
for i in a:
if sign[i]==-1:
sign[i]=flag
b.extend(root[i])
f=True
flag=i
a=b
b=[]
if f==False:
break
f=False
for i in range(n-1):
print(sign[i+1]+1)
|
s646634758
|
Accepted
| 812 | 35,724 | 589 |
from collections import deque
n,m=map(int,input().split())
root=[[] for _ in range(n)]
for i in range(m):
a,b=map(int,input().split())
root[a-1].append(b-1)
root[b-1].append(a-1)
times=[0]*n
sign=[-1]*n
sign[0]=0
color=[False]*n
que=deque([0])
while que:
i=que.popleft()
if color[i]==False:
for edge in root[i]:
if color[edge]==False and sign[edge]==-1:
sign[edge]=i
que.append(edge)
color[i]=True
print('Yes')
for i in range(n-1):
print(sign[i+1]+1)
|
s854871099
|
p03486
|
u027675217
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,060 | 193 |
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
s=input()
t=input()
s_l,t_l=[],[]
for i in s:
s_l.append(i)
for i in t:
t_l.append(i)
s_l.sort(reverse=True)
t_l.sort()
s="".join(s_l)
t="".join(t_l)
if s<t:
print("Yes")
else:
print("No")
|
s913322999
|
Accepted
| 17 | 3,060 | 193 |
s=input()
t=input()
s_l,t_l=[],[]
for i in s:
s_l.append(i)
for i in t:
t_l.append(i)
s_l.sort()
t_l.sort(reverse=True)
s="".join(s_l)
t="".join(t_l)
if s<t:
print("Yes")
else:
print("No")
|
s573028989
|
p00740
|
u135085051
| 1,000 | 131,072 |
Wrong Answer
| 1,980 | 5,600 | 701 |
One of the oddest traditions of the town of Gameston may be that even the town mayor of the next term is chosen according to the result of a game. When the expiration of the term of the mayor approaches, at least three candidates, including the mayor of the time, play a game of pebbles, and the winner will be the next mayor. The rule of the game of pebbles is as follows. In what follows, _n_ is the number of participating candidates. Requisites A round table, a bowl, and plenty of pebbles. Start of the Game A number of pebbles are put into the bowl; the number is decided by the Administration Commission using some secret stochastic process. All the candidates, numbered from 0 to _n_ -1 sit around the round table, in a counterclockwise order. Initially, the bowl is handed to the serving mayor at the time, who is numbered 0. Game Steps When a candidate is handed the bowl and if any pebbles are in it, one pebble is taken out of the bowl and is kept, together with those already at hand, if any. If no pebbles are left in the bowl, the candidate puts all the kept pebbles, if any, into the bowl. Then, in either case, the bowl is handed to the next candidate to the right. This step is repeated until the winner is decided. End of the Game When a candidate takes the last pebble in the bowl, and no other candidates keep any pebbles, the game ends and that candidate with all the pebbles is the winner. A math teacher of Gameston High, through his analysis, concluded that this game will always end within a finite number of steps, although the number of required steps can be very large.
|
n, p = map(int,input().split())
def solve(n, p):
tmp = p
i = 0
person = [0] * n
while(1):
if tmp != 0:
person[i] += 1
tmp -= 1
if person[i] == p:
return i
else:
tmp = person[i]
person[i] = 0
i += 1
if i == n:
i = 0
if p == 0:
for i in range(n):
if person[i] > 0:
return i + 1
ans = []
while(1):
n, p = map(int,input().split())
if n == 0 and p == 0:
for i in ans:
print(i)
break
else:
ans.append(solve(n, p))
|
s197619347
|
Accepted
| 3,530 | 5,596 | 604 |
ans = []
while True:
n, p = map(int, input().split())
if n == 0 and p == 0:
break
stone = p
person = [0] * n
i = 0
while True:
i = i % n
if stone > 1:
person[i] += 1
stone -= 1
i += 1
elif stone == 1:
stone -= 1
person[i] += 1
if person[i] == p:
ans.append(i)
break
else:
i += 1
else:
stone += person[i]
person[i] = 0
i += 1
for i in ans:
print(i)
|
s069600338
|
p03853
|
u516447519
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,188 | 133 |
There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, print a image with a height of 2H pixels and a width of W pixels where the pixel at the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division is rounded down).
|
H,W = [int(i) for i in input().split()]
C = [list(input()) for i in range(H)]
for num in C:
print(str(num))
print(str(num))
|
s545356173
|
Accepted
| 18 | 3,060 | 112 |
H,W = [int(i) for i in input().split()]
for i in range(H):
s = input()
print(str(s))
print(str(s))
|
s415925069
|
p03543
|
u170324846
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 99 |
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()
if int(N[1:4]) % 111 == 0 or int(N[2:5]) % 111 == 0:
print("Yes")
else:
print("No")
|
s466224608
|
Accepted
| 17 | 2,940 | 100 |
N = input()
if int(N[0:3]) % 111 == 0 or int(N[1:4]) % 111 == 0:
print("Yes")
else:
print("No")
|
s271258719
|
p00705
|
u798803522
| 1,000 | 131,072 |
Wrong Answer
| 240 | 7,596 | 780 |
The ICPC committee would like to have its meeting as soon as possible to address every little issue of the next contest. However, members of the committee are so busy maniacally developing (possibly useless) programs that it is very difficult to arrange their schedules for the meeting. So, in order to settle the meeting date, the chairperson requested every member to send back a list of convenient dates by E-mail. Your mission is to help the chairperson, who is now dedicated to other issues of the contest, by writing a program that chooses the best date from the submitted lists. Your program should find the date convenient for the most members. If there is more than one such day, the earliest is the best.
|
cond = [int(n) for n in input().split(" ")]
while True:
answer = [101,0]
case = []
sets = set()
for c in range(cond[0]):
case.append([int(n) for n in input().split(" ")[1:]])
sets = sets | {n for n in case[-1]}
#print(sets,case)
else:
for num in sets:
disp = [num,0]
for c in case:
if num in c:
disp[1] += 1
else:
#print(disp,answer,cond)
if (disp[1] >= cond[1] and disp[1] > answer[1] ) or (disp[1] == answer[1] and disp[0] < answer[0]):
answer = disp
else:
print(answer[0])
cond = [int(n) for n in input().split(" ")]
if cond == [0,0]:
break
|
s456287606
|
Accepted
| 240 | 7,716 | 860 |
cond = [int(n) for n in input().split(" ")]
while True:
answer = [101,0]
case = []
sets = set()
for c in range(cond[0]):
case.append([int(n) for n in input().split(" ")[1:]])
sets = sets | {n for n in case[-1]}
#print(sets,case)
else:
for num in sets:
disp = [num,0]
for c in case:
if num in c:
disp[1] += 1
else:
#print(disp,answer,cond)
if (disp[1] >= cond[1] and disp[1] > answer[1] ) or (disp[1] == answer[1] and disp[0] < answer[0]):
answer = disp
else:
if answer[0] == 101:
print(0)
else:
print(answer[0])
cond = [int(n) for n in input().split(" ")]
if cond == [0,0]:
break
|
s018270949
|
p03544
|
u088488125
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,088 | 154 |
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())
l_b=2
l=1
if n=="1":
print("2")
elif n=="2":
print("1")
else:
for i in range(3,n+1):
l_bb=l_b
l_b=l
l=l_b+l_bb
print(l)
|
s370154535
|
Accepted
| 28 | 9,176 | 150 |
n=int(input())
l_b=2
l=1
if n==0:
print("2")
elif n==1:
print("1")
else:
for i in range(2,n+1):
l_bb=l_b
l_b=l
l=l_b+l_bb
print(l)
|
s297917306
|
p03759
|
u999893056
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 76 |
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 = list(map(int, input().split()))
print("Yes" if b-a == c-b else "No")
|
s317016931
|
Accepted
| 17 | 2,940 | 78 |
a, b, c = map(int, input().split())
print("YES" if b - a == c - b else "NO")
|
s630956807
|
p02261
|
u007270338
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,352 | 683 |
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
#coding:utf-8
from copy import deepcopy
n = int(input())
A = list(input().split())
B = deepcopy(A)
def BubbleSort(A,N):
for i in range(N):
for j in range(N-1,i,-1):
if A[j][1] < A[j-1][1]:
A[j], A[j-1] = A[j-1], A[j]
def SelectionSort(A,N):
for i in range(N):
minj = i
for j in range(i,N):
if A[j][1] < A[minj][1]:
minj = j
A[i], A[minj] = A[minj], A[i]
BubbleSort(A,n)
SelectionSort(B,n)
A = " ".join([data for data in A])
B = " ".join([data for data in B])
print(A)
print("Stable")
print(B)
if A == B:
print("Stable")
else:
print("No stable")
|
s011224374
|
Accepted
| 30 | 6,352 | 684 |
#coding:utf-8
from copy import deepcopy
n = int(input())
A = list(input().split())
B = deepcopy(A)
def BubbleSort(A,N):
for i in range(N):
for j in range(N-1,i,-1):
if A[j][1] < A[j-1][1]:
A[j], A[j-1] = A[j-1], A[j]
def SelectionSort(A,N):
for i in range(N):
minj = i
for j in range(i,N):
if A[j][1] < A[minj][1]:
minj = j
A[i], A[minj] = A[minj], A[i]
BubbleSort(A,n)
SelectionSort(B,n)
A = " ".join([data for data in A])
B = " ".join([data for data in B])
print(A)
print("Stable")
print(B)
if A == B:
print("Stable")
else:
print("Not stable")
|
s796813013
|
p02612
|
u331105860
| 2,000 | 1,048,576 |
Wrong Answer
| 124 | 26,884 | 189 |
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.
|
import sys
import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N = int(readline())
print((N-1000) % 1000)
|
s977123320
|
Accepted
| 116 | 27,004 | 230 |
import sys
import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N = int(readline())
if N % 1000 == 0:
print(0)
else:
print(1000 - N % 1000)
|
s547658160
|
p03796
|
u396266329
| 2,000 | 262,144 |
Wrong Answer
| 27 | 2,940 | 76 |
Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.
|
N = int(input())
res = 1
for i in range(N):
res *= i
print(res%1000000007)
|
s910920430
|
Accepted
| 45 | 2,940 | 87 |
N = int(input())
res = 1
for i in range(N):
res *= i+1
res %= 1000000007
print(res)
|
s550638171
|
p03359
|
u763534217
| 2,000 | 262,144 |
Wrong Answer
| 26 | 9,132 | 68 |
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
a,b=map(int, input().split())
if a<b:
print(a)
else:
print(a-1)
|
s467632200
|
Accepted
| 27 | 9,044 | 77 |
a, b = map(int, input().split())
if a <= b:
print(a)
else:
print(a-1)
|
s893298069
|
p04029
|
u582243208
| 2,000 | 262,144 |
Wrong Answer
| 23 | 3,064 | 29 |
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
n=int(input())
print(n*(n+1))
|
s923418088
|
Accepted
| 23 | 3,064 | 32 |
n=int(input())
print(n*(n+1)//2)
|
s623644450
|
p03730
|
u421499233
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 183 |
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())
flag = False
for i in range(a):
tmp = b*i + c
if tmp%a == 0:
flag = True
break
if flag:
print("TES")
else:
print("NO")
|
s083803718
|
Accepted
| 20 | 3,060 | 183 |
a,b,c = map(int,input().split())
flag = False
for i in range(a):
tmp = b*i + c
if tmp%a == 0:
flag = True
break
if flag:
print("YES")
else:
print("NO")
|
s165101166
|
p02606
|
u205758185
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,096 | 117 |
How many multiples of d are there among the integers between L and R (inclusive)?
|
L, R, d = input().split()
L = int()
R = int()
d = int()
for i in range(L, R):
if i % d == 0:
print(i)
|
s718756925
|
Accepted
| 24 | 9,092 | 131 |
L, R, d = map(int,input().split())
ans = 0
for i in range(L, R+1):
if i % d == 0:
ans = ans + 1
print(ans)
|
s114185758
|
p03408
|
u252964975
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 245 |
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.
|
N = int(input())
S = []
for i in range(N):
Si = str(input())
S.append(Si)
M = int(input())
T = []
for i in range(M):
Ti = str(input())
T.append(Ti)
total = 0
for i in range(N):
if not S[i] in T:
total = total + 1
print(total)
|
s054537432
|
Accepted
| 18 | 3,064 | 324 |
N = int(input())
S = []
for i in range(N):
Si = str(input())
S.append(Si)
M = int(input())
T = []
for i in range(M):
Ti = str(input())
T.append(Ti)
max_point = 0
for i in range(N):
point = len([1 for s in S if s==S[i]]) - len([1 for t in T if t==S[i]])
max_point = max([max_point,point])
print(max_point)
|
s948934692
|
p02796
|
u178235755
| 2,000 | 1,048,576 |
Wrong Answer
| 606 | 51,576 | 323 |
In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep.
|
n = int(input())
arms = []
for i in range(n):
x, l = [int(i) for i in input().split(' ')]
arms.append({"st": x-l, "ed": x+l})
arms.sort(key=lambda x: x["ed"])
print(arms)
n_arm = 1
tot_end = arms[0]["ed"]
for i in range(1, n):
if tot_end <= arms[i]["st"]:
n_arm += 1
tot_end = arms[i]["ed"]
print(n_arm)
|
s815823243
|
Accepted
| 510 | 41,696 | 311 |
n = int(input())
arms = []
for i in range(n):
x, l = [int(i) for i in input().split(' ')]
arms.append({"st": x-l, "ed": x+l})
arms.sort(key=lambda x: x["ed"])
n_arm = 1
tot_end = arms[0]["ed"]
for i in range(1, n):
if tot_end <= arms[i]["st"]:
n_arm += 1
tot_end = arms[i]["ed"]
print(n_arm)
|
s730031938
|
p03150
|
u131443777
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 3,188 | 424 |
A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.
|
import re
str_in_org = input()
KEYENCE = 'keyence'
flag = False
for n in range(1,7):
str_in = str_in_org
print (n)
str_div1 = KEYENCE[:n]
str_div2 = KEYENCE[n:]
print(str_div1)
if( str_div1 in str_in ):
str_in = re.split( str_div1, str_in, 2 )[1]
print (str_in)
if( str_div2 in str_in ):
flag = True
if flag == True :
print ('YES')
else:
print ('NO')
|
s251310345
|
Accepted
| 19 | 3,188 | 474 |
import re
str_in_org = input()
KEYENCE = 'keyence'
flag = False
for n in range(1,7):
str_in = str_in_org
len_in = len(str_in)
len_tail = len_in - (7 - n)
str_div1 = str_in[:n]
str_div2 = str_in[len_tail:]
key_div1 = KEYENCE[:n]
key_div2 = KEYENCE[n:]
if( str_div1 == key_div1 and str_div2 == key_div2 ):
if( str_div1 + str_div2 == KEYENCE ):
flag = True
if flag == True :
print ('YES')
else:
print ('NO')
|
s107121395
|
p03795
|
u943657163
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 81 |
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())
p, d =1, 10**9+7
for i in range(1,n+1):
p = (p*i) % d
print(p)
|
s507023373
|
Accepted
| 17 | 2,940 | 49 |
n = int(input())
print(800 * n - 200 * (n // 15))
|
s988836542
|
p03129
|
u902973687
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 113 |
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
N, K = map(int, input().split())
if sum([N - k for k in range(3, N+1)]) >= K:
print("YES")
else:
print("NO")
|
s344064371
|
Accepted
| 18 | 2,940 | 243 |
N, K = map(int, input().split())
if N % 2 == 1:
if len([i for i in range(N+1) if i % 2 == 1]) >= K:
print("YES")
else:
print("NO")
else:
if len([i for i in range(N) if i % 2 == 1]) >= K:
print("YES")
else:
print("NO")
|
s838081387
|
p03131
|
u803848678
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,060 | 167 |
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 = list(map(int, input().split()))
if a-1 >= k or a >= b-2:
print(k+1)
exit()
ans = a
k -= a-1
print(ans, k)
ans += (b-a)*(k//2)
ans += k%2
print(ans)
|
s431461297
|
Accepted
| 18 | 3,060 | 152 |
k, a, b = list(map(int, input().split()))
if a-1 >= k or a >= b-2:
print(k+1)
exit()
ans = a
k -= a-1
ans += (b-a)*(k//2)
ans += k%2
print(ans)
|
s055642970
|
p03485
|
u429443682
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 102 |
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a, b = map(int, input().split())
if a % b == 0:
print(int(a / b))
else:
print(int(a / b + 1))
|
s970694287
|
Accepted
| 17 | 2,940 | 120 |
a, b = map(int, input().split())
if (a + b) % 2 == 0:
print(int((a + b) / 2))
else:
print(int((a + b) / 2) + 1)
|
s472749224
|
p02865
|
u858670323
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 76 |
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
N = int(input())
if N % 2 == 0:
print(N/2 -1)
else:
print((N+1)/2-1)
|
s930941616
|
Accepted
| 20 | 3,060 | 86 |
N = int(input())
if N % 2 == 0:
print(int(N/2 -1))
else:
print(int((N+1)/2-1))
|
s518269977
|
p03623
|
u626228246
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 62 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x,a,b = map(int,input().split())
print(min(abs(x-a),abs(x-b)))
|
s838914789
|
Accepted
| 17 | 2,940 | 76 |
x,a,b = map(int,input().split())
print("A" if abs(x-a) < abs(x-b) else "B")
|
s438441302
|
p03372
|
u532966492
| 2,000 | 262,144 |
Wrong Answer
| 536 | 41,616 | 895 |
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter. Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories. Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter. Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
|
def main():
from itertools import accumulate as ac
n,c=map(int,input().split())
xv=[list(map(int,input().split())) for _ in [0]*n]
rmemo=[0]*n
lmemo=[0]*n
rmemo[0]=xv[0][1]-xv[0][0]
lmemo[n-1]=xv[n-1][1]-(c-xv[n-1][0])
for i in range(1,n):
rmemo[i]=rmemo[i-1]+xv[i][1]-xv[i][0]+xv[i-1][0]
for i in range(n-2,-1,-1):
lmemo[i]=lmemo[i+1]+xv[i][1]+xv[i][0]-xv[i+1][0]
temp=rmemo[0]
rmemo2=[temp]*(n+1)
rmemo2[-1]=0
temp=lmemo[-1]
lmemo2=[temp]*(n+1)
lmemo2[-1]=0
for i in range(1,n):
rmemo2[i]=max(rmemo[i],rmemo2[i-1])
for i in range(n-2,-1,-1):
lmemo2[i]=max(lmemo[i],lmemo2[i+1])
#print(rmemo2)
#print(lmemo)
ans=max(rmemo+lmemo+[0])
for i in range(n):
ans=max(ans,rmemo[i]+lmemo2[i+1]-xv[i][0])
ans=max(ans,lmemo[i]+rmemo2[i+1]+xv[i][0]-c)
print(ans)
main()
|
s353602430
|
Accepted
| 518 | 42,384 | 858 |
def main():
from itertools import accumulate as ac
n,c=map(int,input().split())
xv=[list(map(int,input().split())) for _ in [0]*n]
rmemo=[0]*n
lmemo=[0]*n
rmemo[0]=xv[0][1]-xv[0][0]
lmemo[n-1]=xv[n-1][1]-(c-xv[n-1][0])
for i in range(1,n):
rmemo[i]=rmemo[i-1]+xv[i][1]-xv[i][0]+xv[i-1][0]
for i in range(n-2,-1,-1):
lmemo[i]=lmemo[i+1]+xv[i][1]+xv[i][0]-xv[i+1][0]
temp=rmemo[0]
rmemo2=[temp]*(n)
temp=lmemo[-1]
lmemo2=[temp]*(n+1)
lmemo2[-1]=0
for i in range(1,n):
rmemo2[i]=max(rmemo[i],rmemo2[i-1])
for i in range(n-2,-1,-1):
lmemo2[i]=max(lmemo[i],lmemo2[i+1])
rmemo2=[0]+rmemo2
ans=max(rmemo+lmemo+[0])
for i in range(n):
ans=max(ans,rmemo[i]+lmemo2[i+1]-xv[i][0])
ans=max(ans,lmemo[i]+rmemo2[i]+xv[i][0]-c)
print(ans)
main()
|
s976093191
|
p03713
|
u786020649
| 2,000 | 262,144 |
Wrong Answer
| 254 | 9,232 | 336 |
There is a bar of chocolate with a height of H blocks and a width of W blocks. Snuke is dividing this bar into exactly three pieces. He can only cut the bar along borders of blocks, and the shape of each piece must be a rectangle. Snuke is trying to divide the bar as evenly as possible. More specifically, he is trying to minimize S_{max} \- S_{min}, where S_{max} is the area (the number of blocks contained) of the largest piece, and S_{min} is the area of the smallest piece. Find the minimum possible value of S_{max} - S_{min}.
|
h,w=map(int,input().split())
a1=min(h,w)
a2=10**6
a3=10**6
for i in range(1,h):
mx=max(w*i,(w//2)*(h-i),((w+1)//2)*(h-i))
mn=min(w*i,(w//2)*(h-i),((w+1)//2)*(h-i))
a2=min(a2,mx-mn)
for j in range(1,w):
mx=max(h*i,(h//2)*(w-i),((h+1)//2)*(w-i))
mn=min(h*i,(h//2)*(w-i),((h+1)//2)*(w-i))
a3=min(a3,mx-mn)
print(min(a1,a2,a3))
|
s905222243
|
Accepted
| 259 | 9,236 | 354 |
h,w=map(int,input().split())
a1=min(h*(w%3!=0),w*(h%3!=0))
a2=10**6
a3=10**6
for i in range(1,h):
mx=max(w*i,(w//2)*(h-i),((w+1)//2)*(h-i))
mn=min(w*i,(w//2)*(h-i),((w+1)//2)*(h-i))
a2=min(a2,mx-mn)
for i in range(1,w):
mx=max(h*i,(h//2)*(w-i),((h+1)//2)*(w-i))
mn=min(h*i,(h//2)*(w-i),((h+1)//2)*(w-i))
a3=min(a3,mx-mn)
print(min(a1,a2,a3))
|
s757674844
|
p02612
|
u589361760
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 9,144 | 46 |
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())
price = n % 1000
print(price)
|
s461433169
|
Accepted
| 33 | 9,156 | 87 |
n = int(input())
price = n % 1000
if price == 0:
print(0)
else:
print(1000 - price)
|
s988003801
|
p03564
|
u814986259
| 2,000 | 262,144 |
Wrong Answer
| 21 | 2,940 | 184 |
Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value: * Operation A: The displayed value is doubled. * Operation B: The displayed value increases by K. Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations.
|
N=int(input())
K=int(input())
ans=2**N
tmp=1
for i in range(2**N - 1):
for j in range(N):
if i>>j %2==1:
tmp*=2
else:
tmp+=K
ans=min(ans,tmp)
print(ans)
|
s716805251
|
Accepted
| 20 | 2,940 | 184 |
N=int(input())
K=int(input())
ans=2**N
for i in range(2**N):
tmp=1
for j in range(N):
if (i>>j) %2==1:
tmp*=2
else:
tmp+=K
ans=min(ans,tmp)
print(ans)
|
s128100649
|
p02612
|
u193264896
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,216 | 246 |
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.
|
import sys
read = sys.stdin.read
readline = sys.stdin.buffer.readline
sys.setrecursionlimit(10 ** 8)
INF = float('inf')
MOD = 10 ** 9 + 7
def main():
N = int(readline())
ans = 1000-N
print(ans)
if __name__ == '__main__':
main()
|
s515984454
|
Accepted
| 28 | 9,220 | 301 |
import sys
read = sys.stdin.read
readline = sys.stdin.buffer.readline
sys.setrecursionlimit(10 ** 8)
INF = float('inf')
MOD = 10 ** 9 + 7
def main():
N = int(readline())
ans = 1000-N%1000
if ans ==1000:
print(0)
else:
print(ans)
if __name__ == '__main__':
main()
|
s067737458
|
p03698
|
u049182844
| 2,000 | 262,144 |
Wrong Answer
| 25 | 8,984 | 107 |
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
def main():
s = input()
ans = 'no'
if len(set(s)) == len(s):
ans = 'yes'
print(ans)
|
s096086127
|
Accepted
| 29 | 9,020 | 149 |
def main():
s = input()
ans = 'no'
if len(set(s)) == len(s):
ans = 'yes'
print(ans)
if __name__ == '__main__':
main()
|
s072319015
|
p02409
|
u007858722
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,616 | 408 |
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.
|
roomsOfBuilding = [[[0 for i in range(10)] for j in range(3)] for k in range(4)]
n = int(input())
for l in range(n):
b, f, r, v = map(int, input().split())
roomsOfBuilding[b - 1][f - 1][r - 1] += v
for building in roomsOfBuilding:
for rooms in building:
for room in rooms:
print(room, end=' ')
print('')
print('#' * 20)
|
s054836764
|
Accepted
| 20 | 5,624 | 470 |
roomsOfBuilding = [[[0 for i in range(10)] for j in range(3)] for k in range(4)]
n = int(input())
for l in range(n):
b, f, r, v = map(int, input().split())
roomsOfBuilding[b - 1][f - 1][r - 1] += v
counter = 0
for building in roomsOfBuilding:
counter += 1
for rooms in building:
for room in rooms:
print(' ' + str(room), end='')
print('')
if counter < 4:
print('#' * 20)
|
s273131324
|
p04043
|
u949414593
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 60 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
print("YNeos"[sorted(map(int,input().split()))!=[5,5,7]::2])
|
s082951527
|
Accepted
| 17 | 2,940 | 60 |
print("YNEOS"[sorted(map(int,input().split()))!=[5,5,7]::2])
|
s614276295
|
p03545
|
u338824669
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 477 |
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
ABCD=list(map(int,list(input())))
for i in range(2**3):
op=[0]*3
for j in range(3):
op[j]=i%2
i//=2
val=ABCD[0]
for j in range(3):
if op[j]==1:
val+=ABCD[j+1]
else:
val-=ABCD[j+1]
if val==7:
ans=str(ABCD[0])
for j in range(3):
if op[j]==1:
ans+="+"
else:
ans+="-"
ans+=str(ABCD[j+1])
print(ans)
exit()
|
s701595536
|
Accepted
| 17 | 3,064 | 482 |
ABCD=list(map(int,list(input())))
for i in range(2**3):
op=[0]*3
for j in range(3):
op[j]=i%2
i//=2
val=ABCD[0]
for j in range(3):
if op[j]==1:
val+=ABCD[j+1]
else:
val-=ABCD[j+1]
if val==7:
ans=str(ABCD[0])
for j in range(3):
if op[j]==1:
ans+="+"
else:
ans+="-"
ans+=str(ABCD[j+1])
print(ans+"=7")
exit()
|
s202813242
|
p03852
|
u559722042
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 169 |
Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.
|
if __name__ == "__main__":
c = input()
if c == "a" or c == "i" or c == "u" or c == "e" or c == "o":
print("voewl")
else:
print("consonant")
|
s646157888
|
Accepted
| 17 | 2,940 | 169 |
if __name__ == "__main__":
c = input()
if c == "a" or c == "i" or c == "u" or c == "e" or c == "o":
print("vowel")
else:
print("consonant")
|
s585201697
|
p02831
|
u265118937
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 9,116 | 62 |
Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests.
|
import math
a,b=map(int,input().split())
print(math.gcd(a, b))
|
s611036745
|
Accepted
| 27 | 9,152 | 108 |
import math
a,b=map(int,input().split())
def lcm(x, y):
return (x * y) // math.gcd(x, y)
print(lcm(a,b))
|
s800725971
|
p03729
|
u386089355
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 121 |
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[len(a) - 1] == b[0]) and (b[len(b) - 1] == c[0]):
print("Yes")
else:
print("No")
|
s866092335
|
Accepted
| 17 | 2,940 | 101 |
a, b, c = input().split()
if a[-1] == b[0] and b[-1] == c[0]:
print("YES")
else:
print("NO")
|
s548632563
|
p03861
|
u953379577
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,156 | 50 |
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,c = map(int,input().split())
print(a//c-b//c)
|
s824802729
|
Accepted
| 31 | 9,156 | 97 |
a,b,c = map(int,input().split())
if a%c == 0:
print(b//c-a//c+1)
else:
print(b//c-a//c)
|
s028951481
|
p02927
|
u127499732
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 160 |
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())
x=int(d//10)
a,b,c=list(range(1,10)),list(range(1,x+1)),list(range(1,m+1))
d=sum([i*b for i in a],[])
e=len(set(c)&set(d))
print(e)
|
s494981533
|
Accepted
| 18 | 3,060 | 408 |
def main():
import sys
import bisect
from itertools import accumulate
m, d = map(int, input().split())
if d < 22:
print(0)
return
count = 0
for i in range(22, d + 1):
d1 = int(str(i)[1])
d10 = int(str(i)[0])
x = d1 * d10
if 1 <= x <= m and d1 >= 2:
count += 1
print(count)
if __name__ == '__main__':
main()
|
s448958338
|
p03050
|
u136395536
| 2,000 | 1,048,576 |
Wrong Answer
| 174 | 2,940 | 182 |
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())
total = 0
for i in range(1,int(math.sqrt(N))):
amari = i
if N % amari == 0:
total += N//amari - 1
#print(amari)
print(total)
|
s590414018
|
Accepted
| 174 | 3,060 | 287 |
import math
N = int(input())
total = 0
if N != 1:
for i in range(1,int(math.sqrt(N)+1)):
amari = i
if N % amari == 0:
k = N // amari - 1
if amari == N % k:
total += N//amari - 1
#print(amari)
print(total)
|
s505660631
|
p02663
|
u357751375
| 2,000 | 1,048,576 |
Wrong Answer
| 21 | 9,012 | 316 |
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?
|
t = list(input())
a = 0
b = 0
for i in range(len(t)):
if i == 0 and t[i] == '?':
t[i] = 'D'
if 0 < i < len(t) - 1 and t[i] == '?':
if t[i-1] == 'P':
t[i] = 'D'
else:
t[i] = 'P'
if i == len(t) -1 and t[i] == '?':
t[i] = 'D'
print(''.join(t))
|
s652236470
|
Accepted
| 22 | 9,172 | 99 |
H1,M1,H2,M2,K = map(int,input().split())
A = (H1 * 60) + M1
B = (H2 * 60) + M2
C = B - A
print(C-K)
|
s091514483
|
p03150
|
u297574184
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,068 | 163 |
A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.
|
Ss = input()
for i in range(8):
print(i, Ss[:i] + Ss[-7+i:])
if Ss[:i] + Ss[-7+i:] == 'keyence':
print('YES')
break
else:
print('NO')
|
s837839606
|
Accepted
| 17 | 2,940 | 137 |
Ss = input()
for i in range(8):
if Ss[:i] + Ss[len(Ss)-7+i:] == 'keyence':
print('YES')
break
else:
print('NO')
|
s315522468
|
p02694
|
u297080498
| 2,000 | 1,048,576 |
Wrong Answer
| 24 | 9,168 | 146 |
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())
r = 0.01
i = 0
s = 100
while True:
i += 1
s = int(s * (1 + r))
if s > x:
print(i)
break
|
s046220471
|
Accepted
| 22 | 9,108 | 163 |
import math
x = int(input())
r = 0.01
i = 0
s = 100
while True:
i += 1
s = math.floor(s * (1 + r))
if s >= x:
print(i)
break
|
s315846348
|
p03172
|
u858742833
| 2,000 | 1,048,576 |
Wrong Answer
| 1,661 | 37,860 | 432 |
There are N children, numbered 1, 2, \ldots, N. They have decided to share K candies among themselves. Here, for each i (1 \leq i \leq N), Child i must receive between 0 and a_i candies (inclusive). Also, no candies should be left over. Find the number of ways for them to share candies, modulo 10^9 + 7. Here, two ways are said to be different when there exists a child who receives a different number of candies.
|
def main():
N, K = map(int, input().split())
A = list(map(int, input().split()))
u = [1] + [0] * K
v = [1] + [0] * K
s = 0
for i in A:
u, v = v, u
s += i
t = 0
for k in range(0, i):
t += u[k]
v[k] = t
for k in range(i, min(s, K) + 1):
t += u[i]
v[k] = t
t -= u[k - i]
print(v[K] % (10 ** 9 + 7))
main()
|
s265106111
|
Accepted
| 1,650 | 38,504 | 434 |
def main():
N, K = map(int, input().split())
A = map(int, input().split())
u = [1] + [0] * K
v = [1] + [0] * K
s = 1
for a in A:
u, v = v, u
s += a
c = 0
for i, t in enumerate(u[:a + 1]):
c += t
v[i] = c
for i, t, tt in zip(range(a + 1, K + 1), u[a + 1:s], u):
c += t - tt
v[i] = c
print(v[K] % (10 ** 9 + 7))
main()
|
s806050599
|
p03943
|
u337851472
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 99 |
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a, b, c = map(int,input().split())
print("YES" if (a+b) == c or (b+c) == a or (c+a) == b else "NO")
|
s886348695
|
Accepted
| 17 | 2,940 | 99 |
a, b, c = map(int,input().split())
print("Yes" if (a+b) == c or (b+c) == a or (c+a) == b else "No")
|
s860412690
|
p03371
|
u306950978
| 2,000 | 262,144 |
Wrong Answer
| 162 | 3,064 | 211 |
"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())
n = max(x,y)
kar = a*x + b*y
for i in range(1,n+1):
if kar > a*(max(0,x-i)) + b*(max(0,y-i)) +c*i:
kar = a*(max(0,x-i)) + b*(max(0,y-i)) +2*c*i
print(kar)
|
s703808509
|
Accepted
| 156 | 3,064 | 213 |
a , b , c , x , y = map(int,input().split())
n = max(x,y)
kar = a*x + b*y
for i in range(1,n+1):
if kar > a*(max(0,x-i)) + b*(max(0,y-i)) +2*c*i:
kar = a*(max(0,x-i)) + b*(max(0,y-i)) +2*c*i
print(kar)
|
s662011836
|
p03448
|
u747005359
| 2,000 | 262,144 |
Wrong Answer
| 49 | 3,060 | 227 |
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)])
cnt = 0
for a in range(a+1):
for b in range(b+1):
for c in range(c+1):
if 500*a + 100*b + 50*c == x:
cnt += 1
print(cnt)
|
s449162526
|
Accepted
| 52 | 3,060 | 211 |
a, b, c, x = map(int, [input() for i in range(4)])
cnt = 0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if 500*i + 100*j + 50*k == x:
cnt += 1
print(cnt)
|
s185270838
|
p03854
|
u137722467
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,188 | 374 |
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
s = input()
while(1):
if s.endswith('dream'):
s = s.rstrip('dream')
elif s.endswith('dreamer'):
s = s.rstrip('dreamer')
elif s.endswith('eraser'):
s = s.rstrip('eraser')
elif s.endswith('erase'):
s = s.rstrip('erase')
else:
print("NO")
quit()
if s == '':
break
print("YES")
|
s618370961
|
Accepted
| 67 | 3,188 | 306 |
s = input()
while(1):
if s.endswith('dream'):
s = s[:-5]
elif s.endswith('dreamer'):
s = s[:-7]
elif s.endswith('eraser'):
s = s[:-6]
elif s.endswith('erase'):
s = s[:-5]
else:
print("NO")
quit()
if s == '':
break
print("YES")
|
s007458711
|
p03997
|
u900140292
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 61 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a=int(input())
b=int(input())
h=int(input())/2
print((a+b)*h)
|
s275753708
|
Accepted
| 17 | 2,940 | 63 |
a=int(input())
b=int(input())
h=int(input())//2
print((a+b)*h)
|
s196419734
|
p03547
|
u962942039
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 119 |
In programming, hexadecimal notation is often used. In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively. In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`. When X and Y are seen as hexadecimal numbers, which is larger?
|
x, y = input().split()
x = int(x, 16)
y = int(y, 16)
if x > y:
print('<')
elif x < y:
print('>')
else:
print('=')
|
s091734701
|
Accepted
| 17 | 3,060 | 119 |
x, y = input().split()
x = int(x, 16)
y = int(y, 16)
if x > y:
print('>')
elif x < y:
print('<')
else:
print('=')
|
s963627881
|
p00006
|
u114860785
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,548 | 19 |
Write a program which reverses a given string str.
|
print(input()[-1])
|
s703888402
|
Accepted
| 20 | 5,548 | 21 |
print(input()[::-1])
|
s325622089
|
p00063
|
u150984829
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,536 | 54 |
半角アルファベット文字列からなる、1 行あたり 100 文字以内のデータがあります。いくつかの行は対称（左端から読んでも右端から読んでも同じ）です。このデータを読み込んで、その中の対称な文字列の数を出力するプログラムを作成してください。なお、１文字だけからなる行は対称であるとします。
|
import sys
print(e[:-1]==e[-2::-1]for e in sys.stdin)
|
s735478237
|
Accepted
| 20 | 5,552 | 59 |
import sys
print(sum(e[:-1]==e[-2::-1]for e in sys.stdin))
|
s094930587
|
p02613
|
u960201620
| 2,000 | 1,048,576 |
Wrong Answer
| 76 | 16,272 | 397 |
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.
|
from sys import stdin
n = int(stdin.readline().rstrip())
list = []
i = 0
for i in range(n):
list.append(stdin.readline().rstrip())
AC = WA = TLE = RE = 0
for k in list:
if k == 'AC':
AC = AC + 1
if k == 'WA':
WA = WA + 1
if k == 'TLE':
TLE = TLE + 1
if k == 'RE':
RE = RE + 1
print(f'AC × {AC}')
print(f'WA × {WA}')
print(f'TLE × {TLE}')
print(f'RE × {RE}')
|
s042164689
|
Accepted
| 76 | 16,240 | 398 |
from sys import stdin
n = int(stdin.readline().rstrip())
list = []
i = 0
for i in range(n):
list.append(stdin.readline().rstrip())
AC = WA = TLE = RE = 0
for k in list:
if k == 'AC':
AC = AC + 1
if k == 'WA':
WA = WA + 1
if k == 'TLE':
TLE = TLE + 1
if k == 'RE':
RE = RE + 1
print(f'AC x {AC}')
print(f'WA x {WA}')
print(f'TLE x {TLE}')
print(f'RE x {RE}')
|
s267279188
|
p03760
|
u242518667
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 201 |
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
|
pw=[input() for i in range(2)]
length=len(pw[0])
full_pw=[]
i=0
while i < length:
full_pw.append(pw[0][i])
if i !=len(pw[0])-1:
full_pw.append(pw[1][i])
i+=1
print(*full_pw,sep='')
|
s689924806
|
Accepted
| 17 | 3,060 | 201 |
pw=[input() for i in range(2)]
length=len(pw[0])
full_pw=[]
i=0
while i < length:
full_pw.append(pw[0][i])
if i <=len(pw[1])-1:
full_pw.append(pw[1][i])
i+=1
print(*full_pw,sep='')
|
s156211437
|
p03854
|
u009219947
| 2,000 | 262,144 |
Wrong Answer
| 43 | 3,188 | 275 |
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
strings = input()
word_list = ["dream", "erase", "dreamer", "eraser"]
rev_list = ["maerd", "esare", "remaerd", "resare"]
match = ""
for s in reversed(strings):
match += s
if match in rev_list:
match = ""
if len(match) == 0:
print("Yes")
else:
print("No")
|
s898362750
|
Accepted
| 43 | 3,188 | 223 |
strings = input()
rev_list = ["maerd", "esare", "remaerd", "resare"]
match = ""
for s in reversed(strings):
match += s
if match in rev_list:
match = ""
if len(match) == 0:
print("YES")
else:
print("NO")
|
s689335288
|
p03795
|
u736729525
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 42 |
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
N = int(input())
print(N*800 - (N // 15))
|
s675051875
|
Accepted
| 17 | 2,940 | 46 |
N = int(input())
print(N*800 - (N // 15)*200)
|
s962859454
|
p04043
|
u134520518
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 146 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
l = list(map(int,input().split()))
s = [5,7,5]
for a in l:
if a in s:
s.remove(a)
if s == []:
print('Yes')
else:
print('No')
|
s871847965
|
Accepted
| 17 | 2,940 | 146 |
l = list(map(int,input().split()))
s = [5,7,5]
for a in l:
if a in s:
s.remove(a)
if s == []:
print('YES')
else:
print('NO')
|
s612185866
|
p03371
|
u441320782
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 218 |
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
|
A,B,C,X,Y=map(int,input().split())
M=[A,B,C]
ans=[]
ans.append(A*X+B*Y)
ans.append(C*max(X,Y)*2)
if A>B:
ans.append(C*min(X,Y)*2+A*(X-min(X,Y)))
if B>A:
ans.append(C*min(X,Y)*2+B*(Y-min(X,Y)))
print(min(ans))
|
s593839015
|
Accepted
| 101 | 3,064 | 176 |
A,B,C,X,Y=map(int,input().split())
ans=5000*(10**5)*2+1
for i in range(max(X,Y)+1):
total=0
total=2*C*i+A*max(X-i,0)+B*max(Y-i,0)
if ans>=total:
ans=total
print(ans)
|
s420950523
|
p03679
|
u374146618
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 107 |
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
x, a, b = [int(_) for _ in input().split()]
if (a+b)>x:
print("dangerous")
else:
print("delicious")
|
s945531594
|
Accepted
| 17 | 2,940 | 152 |
x, a, b = map(int, input().split())
if b <= a:
ans = "delicious"
elif b > a and b <= a + x:
ans = "safe"
else:
ans = "dangerous"
print(ans)
|
s008441133
|
p02936
|
u610473220
| 2,000 | 1,048,576 |
Wrong Answer
| 2,108 | 82,552 | 478 |
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
N, Q = map(int, input().split())
ab = []
for i in range(N-1):
ab.append(list(map(int, input().split())))
l = [[] for i in range(N)]
for pi in ab:
l[pi[0]-1].append(pi[1]-1)
cou = [0] * N
for i in range(Q):
p, x = map(int, input().split())
p -= 1
cou[p] += x
for j in l[p]:
cou[j] += x
for k in l[j]:
cou[k] += x
for h in l[k]:
cou[h] += x
for i in range(N):
print(str(cou[i]), end = " ")
|
s832673887
|
Accepted
| 1,441 | 271,380 | 728 |
import sys
sys.setrecursionlimit(10**6)
input = sys.stdin.readline
def g(ab: list, ans: list, check: list, i: int = 0):
if len(ab[i]) == 1 and i != 0:
return
check[i] = False
for j in ab[i]:
if check[j]:
ans[j] += ans[i]
g(ab, ans, check, j)
def main():
N, Q = map(int, input().split())
ab = [[] for _ in range(N)]
for _ in range(N-1):
a, b = map(int, input().split())
ab[a-1].append(b-1)
ab[b-1].append(a-1)
ans = [0] * N
check = [True] * N
for _ in range(Q):
p, x = map(int, input().split())
ans[p-1] += x
g(ab, ans, check)
print(' '.join(list(map(str, ans))))
if __name__ == '__main__': main()
|
s203931953
|
p03050
|
u510097799
| 2,000 | 1,048,576 |
Wrong Answer
| 149 | 3,060 | 146 |
Snuke received a positive integer N from Takahashi. A positive integer m is called a _favorite number_ when the following condition is satisfied: * The quotient and remainder of N divided by m are equal, that is, \lfloor \frac{N}{m} \rfloor = N \bmod m holds. Find all favorite numbers and print the sum of those.
|
n = int(input())
l = int(-(-(n ** (1/2)) // 1))
print(l)
total = 0
for i in range(1, l):
if n % i == 0:
total += (n - i) // i
print(int(total))
|
s853073979
|
Accepted
| 142 | 3,064 | 156 |
n = int(input())
l = int(-(-(n ** (1/2)) // 1))
total = 0
for i in range(1, l):
if n % i == 0 and n != i * (i+1):
total += (n - i) // i
print(int(total))
|
s057322023
|
p04045
|
u940652437
| 2,000 | 262,144 |
Wrong Answer
| 71 | 9,312 | 383 |
Iroha is very particular about numbers. There are K digits that she dislikes: D_1, D_2, ..., D_K. She is shopping, and now paying at the cashier. Her total is N yen (the currency of Japan), thus she has to hand at least N yen to the cashier (and possibly receive the change). However, as mentioned before, she is very particular about numbers. When she hands money to the cashier, the decimal notation of the amount must not contain any digits that she dislikes. Under this condition, she will hand the minimum amount of money. Find the amount of money that she will hand to the cashier.
|
N,K = map(int,input().split())
num_list=[]
kari = 0
N -= 1
num_list.append(input().split())
num_list_3 = sorted(num_list)
while(kari != 1):
count = 0
N += 1
for i in range(K - 1):
a = num_list_3[0][i]
s = str(N).count(a)
print(s)
if(s != 0):
count+= 1
print(count)
if (count == 0):
kari = 1
print(N)
|
s664422588
|
Accepted
| 412 | 9,132 | 290 |
n,k=map(int,input().split())
d = list(map(int,input().split()))
#print(d)
for i in range(n,10**5):
count=0
for j in range(k):
u = str(d[j])
if(str(i).count(u)==0):
count+=1
#print("count:"+str(count))
if(count==k):
print(i)
break
|
s760669999
|
p02394
|
u295538678
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,596 | 121 |
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
|
w,h,x,y,r =[int(i) for i in input().split()]
if(w < x+r ):
print('YES')
elif(h < y+r):
print('YES')
else:
print('NO')
|
s333548042
|
Accepted
| 20 | 7,712 | 139 |
w,h,x,y,r =[int(i) for i in input().split()]
if((x+r <= w) and (y+r <= h) and (0 <= x-r) and (0 <= y-r)):
print('Yes')
else:
print('No')
|
s506398384
|
p03493
|
u363118893
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 35 |
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 = str(input())
print(S.find("1"))
|
s433360179
|
Accepted
| 17 | 2,940 | 38 |
S = str(input())
print(S.count("1"))
|
s588302472
|
p03478
|
u578146839
| 2,000 | 262,144 |
Wrong Answer
| 39 | 3,628 | 159 |
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())
ans = 0
for i in range(1, n + 1):
s=sum(map(int, str(i)))
print(s)
if a <= s <= b:
ans += i
print(ans)
|
s412016113
|
Accepted
| 30 | 2,940 | 146 |
n, a, b = map(int, input().split())
ans = 0
for i in range(1, n + 1):
s=sum(map(int, str(i)))
if a <= s <= b:
ans += i
print(ans)
|
s849286403
|
p03377
|
u407730443
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 96 |
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.
|
a, b, x = [int(m) for m in input().split(" ")]
print("Yes" if 0 <= x-a and x <= a+b else "No" )
|
s800465338
|
Accepted
| 17 | 2,940 | 95 |
a, b, x = [int(m) for m in input().split(" ")]
print("YES" if 0 <= x-a and x <= a+b else "NO")
|
s050360941
|
p02977
|
u832039789
| 2,000 | 1,048,576 |
Wrong Answer
| 269 | 16,952 | 1,379 |
You are given an integer N. Determine if there exists a tree with 2N vertices numbered 1 to 2N satisfying the following condition, and show one such tree if the answer is yes. * Assume that, for each integer i between 1 and N (inclusive), Vertex i and N+i have the weight i. Then, for each integer i between 1 and N, the bitwise XOR of the weights of the vertices on the path between Vertex i and N+i (including themselves) is i.
|
n = int(input())
if n == 12:
print('Yes')
l = [12,1,2,3,5,6,7,9,10,11]
p = l[:]
for i in p:
l.append(i + n)
m = [4, 8, 12, 16, 20]
for i,j in zip(l, l[1:]):
print(i, j)
for i,j in zip(m, m[1:]):
print(i, j)
exit()
if n == 24:
print('Yes')
l = [24,2,4,6,7,8,9,10,11,13,14,15,17,18,19,21,22]
p = l[:]
for i in p:
l.append(i + n)
m = [1,3,5,12,16,20,23]
p = m[:]
m.append(24)
for i in p:
m.append(i + n)
for i,j in zip(l, l[1:]):
print(i, j)
for i,j in zip(m, m[1:]):
print(i, j)
exit()
if n == 1 or n % 2 == 0:
print('No')
exit()
l = [False] * (n + 1)
p = [n]
b = bin(n)[2:][::-1]
pw = 0
for i in b:
if i == '1':
p.append(2 ** int(pw))
l[2 ** pw] = True
pw += 1
q = []
xr = 0
for i in range(1, n):
if not l[i]:
xr ^= i
q.append(i)
# print(xr)
# print(p)
# print(q)
res1 = []
for i in p:
res1.append(i)
for i in p:
res1.append(i + n)
res2 = []
if xr == 0:
res2 = [n]
for i in q:
res2.append(i)
for i in q:
res2.append(i + n)
else:
res2 = []
for i in q:
res2.append(i)
res2.append(xr)
for i in q:
res2.append(i + n)
print('Yes')
for i,j in zip(res1, res1[1:]):
print(i, j)
for i,j in zip(res2, res2[1:]):
print(i, j)
|
s945567829
|
Accepted
| 266 | 39,200 | 760 |
n = int(input())
nn = n
f = lambda x, y: x + y * nn
g = lambda x: x if x % 2 == 1 else x + nn
b = bin(n)
if b.count('0') == len(b) - 2:
print('No')
exit(0)
ret = [
[f(1, 0), f(2, 0)],
[f(2, 0), f(3, 0)],
[f(3, 0), f(1, 1)],
[f(1, 1), f(2, 1)],
[f(2, 1), f(3, 1)],
]
if n % 2 == 0:
p = 2 ** (len(b) - 3)
q = n ^ 1 ^ p
ret.append([g(p), f(n, 0)])
ret.append([g(q), f(n, 1)])
n -= 1
while n > 3:
ret.append([f(n - 1, 0), f(n, 0)])
ret.append([f(n, 0), f(1, 1)])
ret.append([f(1, 1), f(n - 1, 1)])
ret.append([f(n - 1, 1), f(n, 1)])
n -= 2
print('Yes')
for x, y in ret:
print(x, y)
|
s149174730
|
p03457
|
u059436995
| 2,000 | 262,144 |
Wrong Answer
| 940 | 3,444 | 292 |
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())
t,x,y =(0, 0, 0)
for _ in range(n):
nt, nx, ny =map(int,input().split())
dis=abs(nx-x)+abs(ny-y)
tim = nt - t
print(dis,tim)
if tim >= dis and (tim - dis) % 2 ==0:
t,x,y=nt, nx, ny
else:
print('No')
break
else:
print('Yes')
|
s537843112
|
Accepted
| 192 | 10,284 | 364 |
from sys import stdin
input = stdin.readline
lines = stdin.readlines
N = int(input())
txy = ((map(int, line.split())) for line in lines())
t,x,y =(0, 0, 0)
for nt, nx, ny in txy:
dis=abs(nx-x)+abs(ny-y)
tim = nt - t
if tim >= dis and (tim - dis) % 2 ==0:
t,x,y=nt, nx, ny
else:
print('No')
break
else:
print('Yes')
|
s867515027
|
p03852
|
u103178403
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 179 |
Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.
|
s=input()
t=['dream','dreamer','erase','eraser']
t=sorted(t,reverse=True)
a=0
for i in range(4):
for j in range(4):
if t[i]+t[j]==s:
a+=1
print("YES"if a>0 else "NO")
|
s043695409
|
Accepted
| 17 | 2,940 | 76 |
s=input()
a=['a','e','i','u','o']
print('vowel' if s in a else 'consonant')
|
s298357130
|
p03251
|
u864013199
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 267 |
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())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
if X>=Y:
print("War")
exit()
for Z in range(X,Y):
if max(x)<=Z<min(y):
continue
else:
print("War")
exit()
print("No War")
|
s371728697
|
Accepted
| 18 | 3,060 | 240 |
N,M,X,Y = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
if X>=Y:
print("War")
exit()
for Z in range(X,Y):
if max(x)<=Z<min(y):
print("No War")
exit()
print("War")
|
s641592170
|
p00456
|
u352394527
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 125 |
先日，オンラインでのプログラミングコンテストが行われた． W大学とK大学のコンピュータクラブは以前からライバル関係にあり，このコンテストを利用して両者の優劣を決めようということになった． 今回，この2つの大学からはともに10人ずつがこのコンテストに参加した．長い議論の末，参加した10人のうち，得点の高い方から3人の得点を合計し，大学の得点とすることにした． W大学およびK大学の参加者の得点のデータが与えられる．このとき，おのおのの大学の得点を計算するプログラムを作成せよ．
|
def func():
lst = []
for i in range(10):
lst.append(int(input()))
lst.sort()
print(sum(lst[-3:]))
func()
func()
|
s745400558
|
Accepted
| 20 | 5,596 | 132 |
def func():
lst = []
for i in range(10):
lst.append(int(input()))
lst.sort()
return sum(lst[-3:])
print(func(),func())
|
s546572710
|
p02694
|
u624613992
| 2,000 | 1,048,576 |
Wrong Answer
| 22 | 9,124 | 132 |
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
goal = int(input())
cnt = 0
money = 100
while goal >= money:
money = math.floor(money * 1.01)
cnt +=1
print(cnt)
|
s554097630
|
Accepted
| 25 | 9,168 | 131 |
import math
goal = int(input())
cnt = 0
money = 100
while goal > money:
money = math.floor(money * 1.01)
cnt +=1
print(cnt)
|
s302108313
|
p03836
|
u816732092
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,064 | 330 |
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())
print('R'*(tx-sx),end = "")
print('U'*(ty-sy),end = "")
print('L'*(tx-sx),end = "")
print('D'*(ty-sx+1),end = "")
print('R'*(tx-sx+1),end = "")
print('U'*(ty-sx+1),end = "")
print('L',end = "")
print('U',end = "")
print('R'*(tx-sx+1),end = "")
print('D'*(ty-sx+1),end = "")
print('L')
|
s000832451
|
Accepted
| 18 | 3,064 | 338 |
sx , sy , tx, ty = map(int , input().split())
lx = abs(tx - sx)
ly = abs(ty - sy)
print('U'*ly,end = "")
print('R'*lx,end = "")
print('D'*ly,end = "")
print('L'*(lx+1),end = "")
print('U'*(ly+1),end = "")
print('R'*(lx+1),end = "")
print('D',end = "")
print('R',end = "")
print('D'*(ly+1),end = "")
print('L'*(lx+1),end = "")
print('U')
|
s665062071
|
p03623
|
u744898490
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 201 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
#from time import time
#t = time()
nvm = input().split(' ')
if int(nvm[1]) - int(nvm[0]) > int(nvm[1]) - int(nvm[2]):
print('A')
else:
print('B')
#print(mn[0] * mm[1])
#tt= time()
#print(tt-t)
|
s646374100
|
Accepted
| 17 | 2,940 | 216 |
#from time import time
#t = time()
nvm = input().split(' ')
if abs(int(nvm[0]) - int(nvm[1])) < abs(int(nvm[0]) - int(nvm[2])):
print('A')
else:
print('B')
#print(mn[0] * mm[1])
#tt= time()
#print(tt-t)
|
s383313182
|
p02678
|
u946404093
| 2,000 | 1,048,576 |
Wrong Answer
| 780 | 47,344 | 571 |
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())
room = {}
startRoom = {1:1}
for i in range(M):
A, B = map(int,input().split())
room.setdefault(A, []).append(B)
room.setdefault(B, []).append(A)
#print(room)
targetRoom = [1]
while len(targetRoom) != 0:
nextTarget = []
for i in targetRoom:
roomList = room[i]
for j in roomList:
if j not in startRoom:
startRoom[j] = i
nextTarget.append(j)
targetRoom = nextTarget
#print(startRoom)
if len(startRoom) != N:
print("no")
else:
print("yes")
for i in range(2, N+1):
print(startRoom[i])
|
s272345767
|
Accepted
| 807 | 47,444 | 540 |
N, M = map(int,input().split())
room = {}
startRoom = {1:1}
for i in range(M):
A, B = map(int,input().split())
room.setdefault(A, []).append(B)
room.setdefault(B, []).append(A)
targetRoom = [1]
while len(targetRoom) != 0:
nextTarget = []
for i in targetRoom:
roomList = room[i]
for j in roomList:
if j not in startRoom:
startRoom[j] = i
nextTarget.append(j)
targetRoom = nextTarget
if len(startRoom) != N:
print("No")
else:
print("Yes")
for i in range(2, N+1):
print(startRoom[i])
|
s033396414
|
p04012
|
u199459731
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 164 |
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.
|
c_list = list(input())
flag = True
for i in range(len(c_list)):
if(c_list.count(c_list[i])%2 == 1):
print("NO")
flag = False
break
if(flag):
print("YES")
|
s262636221
|
Accepted
| 17 | 2,940 | 165 |
c_list = list(input())
flag = True
for i in range(len(c_list)):
if(c_list.count(c_list[i])%2 == 1):
print("No")
flag = False
break
if(flag):
print("Yes")
|
s884532910
|
p03759
|
u117193815
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 79 |
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("TES")
else:
print("NO")
|
s327871304
|
Accepted
| 17 | 2,940 | 80 |
a,b,c=map(int, input().split())
if b-a==c-b:
print("YES")
else:
print("NO")
|
s527184130
|
p03740
|
u368796742
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 14 |
Alice and Brown loves games. Today, they will play the following game. In this game, there are two piles initially consisting of X and Y stones, respectively. Alice and Bob alternately perform the following operation, starting from Alice: * Take 2i stones from one of the piles. Then, throw away i of them, and put the remaining i in the other pile. Here, the integer i (1≤i) can be freely chosen as long as there is a sufficient number of stones in the pile. The player who becomes unable to perform the operation, loses the game. Given X and Y, determine the winner of the game, assuming that both players play optimally.
|
print("Alice")
|
s520065233
|
Accepted
| 17 | 2,940 | 114 |
x,y = map(int,input().split())
if y > x:
x,y = y,x
if x - y <= 1:
print("Brown")
else:
print("Alice")
|
s315921859
|
p03494
|
u982471399
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 253 |
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.
|
A=list(map(int,input().split()))
flag=0
count=0
ans=[]
for a in A:
if a%2==0:
count=a/2
else:
count=(a-1)/2
flag=1
ans.append(count)
if flag==1:
print(int(min(ans)))
else:
print(0)
|
s484628323
|
Accepted
| 19 | 3,060 | 289 |
N=int(input())
A=list(map(int,input().split()))
flag=0
ans=[]
for a in A:
if a%2==0:#odd
count=0
while a%2==0:
a=a/2
count=count+1
ans.append(count)
else:#even
flag=1
if flag==0:
print(int(min(ans)))
else:
print(0)
|
s186065016
|
p03447
|
u320325426
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 80 |
You went shopping to buy cakes and donuts with X yen (the currency of Japan). First, you bought one cake for A yen at a cake shop. Then, you bought as many donuts as possible for B yen each, at a donut shop. How much do you have left after shopping?
|
x, a, b = [int(input()) for i in range(3)]
x -= a
while x > 0:
x -= b
print(x)
|
s324396036
|
Accepted
| 17 | 2,940 | 61 |
x, a, b = [int(input()) for i in range(3)]
print((x - a) % b)
|
s043184097
|
p03192
|
u619670102
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 83 |
You are given an integer N that has exactly four digits in base ten. How many times does `2` occur in the base-ten representation of N?
|
#coding:utf-8
s=input()
res =0
for i in s:
if i==2:
res+=1
print(res)
|
s356577480
|
Accepted
| 17 | 2,940 | 84 |
#coding:utf-8
s = input()
res = 0
for i in s:
if i =='2':
res +=1
print(res)
|
s206490297
|
p03448
|
u644210195
| 2,000 | 262,144 |
Wrong Answer
| 222 | 4,456 | 324 |
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(0, 4)])
count = 0
for a_i in range(0, a + 1):
if a_i * 500 > x:
break
for b_i in range(0, b + 1):
for c_i in range(0, c + 1):
print(a_i, b_i, c_i)
if a_i * 500 + b_i * 100 + c_i * 50 == x:
count += 1
print(count)
|
s241784857
|
Accepted
| 42 | 3,060 | 292 |
a, b, c, x = map(int, [input() for i in range(0, 4)])
count = 0
for a_i in range(0, a + 1):
if a_i * 500 > x:
break
for b_i in range(0, b + 1):
for c_i in range(0, c + 1):
if a_i * 500 + b_i * 100 + c_i * 50 == x:
count += 1
print(count)
|
s071964053
|
p03685
|
u075012704
| 2,000 | 262,144 |
Wrong Answer
| 406 | 12,028 | 862 |
Snuke is playing a puzzle game. In this game, you are given a rectangular board of dimensions R × C, filled with numbers. Each integer i from 1 through N is written twice, at the coordinates (x_{i,1},y_{i,1}) and (x_{i,2},y_{i,2}). The objective is to draw a curve connecting the pair of points where the same integer is written, for every integer from 1 through N. Here, the curves may not go outside the board or cross each other. Determine whether this is possible.
|
from collections import defaultdict
R, C, N = map(int, input().split())
Numbers = []
for i in range(N):
x1, y1, x2, y2 = map(int, input().split())
if (x1 in [0, C] and x2 in [0, C]) or (y1 in [0, R] and y2 in [0, R]):
Numbers.append([i+1, x1, y1])
Numbers.append([i+1, x2, y2])
Numbers.sort(key=lambda n: n[1])
D = defaultdict(set)
for i in range(1, len(Numbers), 2):
D[Numbers[i][1]].add(Numbers[i][0])
for i in range(0, len(Numbers), 2):
if Numbers[i][0] not in D[Numbers[i + 1][1]]:
print("NO")
exit()
Numbers.sort(key=lambda n: n[2])
D = defaultdict(set)
for i in range(1, len(Numbers), 2):
D[Numbers[i][2]].add(Numbers[i][0])
for i in range(0, len(Numbers), 2):
print("NO")
exit()
print("YES")
|
s745219423
|
Accepted
| 728 | 36,616 | 1,047 |
R, C, N = map(int, input().split())
UP, RIGHT, DOWN, LEFT = [], [], [], []
for i in range(N):
x1, y1, x2, y2 = map(int, input().split())
if 0 < x1 < R and 0 < y1 < C:
continue
if 0 < x2 < R and 0 < y2 < C:
continue
if x1 == 0: UP.append([i, y1])
elif x1 == R: DOWN.append([i, y1])
elif y1 == 0: LEFT.append([i, x1])
elif y1 == C: RIGHT.append([i, x1])
if x2 == 0: UP.append([i, y2])
elif x2 == R: DOWN.append([i, y2])
elif y2 == 0: LEFT.append([i, x2])
elif y2 == C: RIGHT.append([i, x2])
UP.sort(key=lambda x: x[1])
RIGHT.sort(key=lambda x: x[1])
DOWN.sort(key=lambda x: x[1], reverse=True)
LEFT.sort(key=lambda x: x[1], reverse=True)
Numbers = UP + RIGHT + DOWN + LEFT
stack = []
for n, z in Numbers:
if stack and stack[-1] == n:
stack.pop()
else:
stack.append(n)
print("NO" if stack else "YES")
|
s889400360
|
p03854
|
u236592202
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,188 | 156 |
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
S=input()
S=S.replace('dreamr','')
S=S.replace('eraser','')
S=S.replace('dream','')
S=S.replace('erase','')
if S=='':
print('Yes')
else:
print('No')
|
s382493950
|
Accepted
| 18 | 3,188 | 164 |
S=input()
S = S.replace('eraser','')
S = S.replace('erase','')
S = S.replace('dreamer','')
S = S.replace('dream','')
if S=='':
print('YES')
else:
print('NO')
|
s791276823
|
p03998
|
u016622494
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 420 |
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
A = list(reversed(input()))
B = list(reversed(input()))
C = list(reversed(input()))
P = A.pop()
while True:
if P == "a":
if len(A) == 0 :
print("A")
exit()
P = A.pop()
if P == "b":
if len(B) == 0:
print("B")
exit()
P = B.pop()
if P == "c":
if len(C) == 0:
print("C")
exit()
P == C.pop()
|
s550272754
|
Accepted
| 17 | 3,064 | 342 |
sa = list(reversed(input()))
sb = list(reversed(input()))
sc = list(reversed(input()))
s0 = sa.pop()
while True:
if s0 == 'a':
if not sa:
print('A')
break
s0 = sa.pop()
elif s0 == 'b':
if not sb:
print('B')
break
s0 = sb.pop()
else:
if not sc:
print('C')
break
s0 = sc.pop()
|
s349969250
|
p02396
|
u984892564
| 1,000 | 131,072 |
Wrong Answer
| 130 | 5,596 | 96 |
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.
|
for i in range(10000):
x = input('Case {0}:'.format(i+1))
if int(x) == 0:
break
|
s802306499
|
Accepted
| 160 | 5,596 | 114 |
for i in range(10000):
x = input()
if int(x) == 0:
break
print('Case ', i+1, ': ', x, sep='')
|
s881637851
|
p03623
|
u374082254
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 101 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x, a, b = map(int, input().split())
if abs(a - x) > abs(b - x):
print("A")
else:
print("B")
|
s231097192
|
Accepted
| 17 | 2,940 | 101 |
x, a, b = map(int, input().split())
if abs(a - x) > abs(b - x):
print("B")
else:
print("A")
|
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