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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s395593260
|
p03377
|
u347397127
| 2,000
| 262,144
|
Wrong Answer
| 23
| 9,020
| 83
|
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 = map(int,input().split())
if a<x and b>x:
print("Yes")
else:
print("No")
|
s069276569
|
Accepted
| 28
| 9,132
| 88
|
a,b,x = map(int,input().split())
if a<=x and a+b>=x:
print("YES")
else:
print("NO")
|
s758711844
|
p03415
|
u406114804
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 57
|
We have a 3×3 square grid, where each square contains a lowercase English letters. The letter in the square at the i-th row from the top and j-th column from the left is c_{ij}. Print the string of length 3 that can be obtained by concatenating the letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.
|
a = input()
b = input()
c = input()
print(a[0],b[1],c[2])
|
s663084571
|
Accepted
| 17
| 2,940
| 57
|
a = input()
b = input()
c = input()
print(a[0]+b[1]+c[2])
|
s751629860
|
p04043
|
u296783581
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 151
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a = list(map(int, input().split()))
n = sum(x == 7 for x in a)
k = sum(x == 5 for x in a)
if n == 2 and k == 1:
print("Yes")
else:
print("No")
|
s826883126
|
Accepted
| 18
| 2,940
| 150
|
a = list(map(int, input().split()))
n = sum(x == 7 for x in a)
k = sum(x == 5 for x in a)
if n == 1 and k == 2:
print("YES")
else:
print("NO")
|
s735573963
|
p02280
|
u317901693
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,712
| 1,664
|
A rooted binary tree is a tree with a root node in which every node has at most two children. Your task is to write a program which reads a rooted binary tree _T_ and prints the following information for each node _u_ of _T_ : * node ID of _u_ * parent of _u_ * sibling of _u_ * the number of children of _u_ * depth of _u_ * height of _u_ * node type (root, internal node or leaf) If two nodes have the same parent, they are **siblings**. Here, if _u_ and _v_ have the same parent, we say _u_ is a sibling of _v_ (vice versa). The height of a node in a tree is the number of edges on the longest simple downward path from the node to a leaf. Here, the given binary tree consists of _n_ nodes and evey node has a unique ID from 0 to _n_ -1.
|
# -*- coding: utf-8 -*-
class Node(object):
def __init__(self, N):
self.parent = [-1 for i in range(N)]
self.left = [-1 for i in range(N)]
self.right = [-1 for i in range(N)]
self.depth = [-1 for i in range(N)]
self.sibling = [-1 for i in range(N)]
self.hight = [0 for i in range(N)]
self.degree = [0 for i in range(N)]
self.max_depth = None
def set_depth(self, u):
d, U = 0, u
while T.parent[u] is not -1:
u = T.parent[u]
d += 1
self.depth[U] = d
def set_hight(self, u):
self.hight[u] = self.max_depth - self.depth[u]
N = int(input())
# create node
T = Node(N)
for j in range(N):
A = [int(i) for i in input().split()]
T.left[A[0]] = A[1]
for u in A[1:]:
if u is not -1:
T.parent[u] = A[0]
T.degree[A[0]] += 1
if (A[1] is not -1) and (A[2] is not -1):
T.right[A[1]] = A[2]
T.sibling[A[1]], T.sibling[A[2]] = A[2], A[1]
for i in range(N):
T.set_depth(i)
# set max_depth
T.max_depth = max(T.depth)
for i in range(N):
T.set_hight(i)
# ??????
for i in range(N):
if T.parent[i] is -1:
node = "root"
elif T.left[i] is -1:
node = "leaf"
else:
node = "internal node"
print("node {0}: parent = {1}, sibling = {2}, degree = {3}, depth = {4}, hight = {5}, {6}".format(i, T.parent[i], T.sibling[i], T.degree[i], T.depth[i], T.hight[i], node))
|
s085988206
|
Accepted
| 20
| 7,876
| 1,827
|
# -*- coding: utf-8 -*-
class Node(object):
def __init__(self, N):
self.parent = [-1 for i in range(N)]
self.left = [-1 for i in range(N)]
self.right = [-1 for i in range(N)]
self.depth = [-1 for i in range(N)]
self.sibling = [-1 for i in range(N)]
self.height = [0 for i in range(N)]
self.degree = [0 for i in range(N)]
def set_depth(self, u):
d, U = 0, u
while T.parent[u] is not -1:
u = T.parent[u]
d += 1
self.depth[U] = d
def set_height(self, u):
h1 = h2 = 0
if self.left[u] is not -1:
h1 = self.set_height(self.left[u]) + 1
if self.right[u] is not -1:
h2 = self.set_height(self.right[u]) + 1
self.height[u] = max(h1, h2)
return max(h1, h2)
N = int(input())
# create node
T = Node(N)
for j in range(N):
A = [int(i) for i in input().split()]
T.left[A[0]] = A[1]
T.right[A[0]] = A[2]
for u in A[1:]:
if u is not -1:
T.parent[u] = A[0]
T.degree[A[0]] += 1
if (A[1] is not -1) and (A[2] is not -1):
T.sibling[A[1]], T.sibling[A[2]] = A[2], A[1]
for i in range(N):
T.set_depth(i)
T.set_height(T.parent.index(-1))
# ??????
for i in range(N):
if T.parent[i] is -1:
node = "root"
elif T.left[i] is -1 and T.right[i] is -1:
node = "leaf"
else:
node = "internal node"
print("node {0}: parent = {1}, sibling = {2}, degree = {3}, depth = {4}, height = {5}, {6}".format(i, T.parent[i], T.sibling[i], T.degree[i], T.depth[i], T.height[i], node))
|
s122368078
|
p03693
|
u509094491
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 106
|
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
i = list(map(int, input().split()))
s=i[0]*100+i[1]*10+i[2]
if s%4==0:
print("Yes")
else :
print("No")
|
s300105855
|
Accepted
| 17
| 2,940
| 95
|
r,g,b=input().split()
sum=100*r+10*g+b
if int(sum)%4==0:
print("YES")
else:
print("NO")
|
s467345620
|
p03447
|
u352499693
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 67
|
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 = map(int, (input() for i in range(3)))
print((x-a+b-1)//b)
|
s803369691
|
Accepted
| 17
| 2,940
| 62
|
x, a, b = map(int, (input() for i in range(3)))
print((x-a)%b)
|
s708557910
|
p03577
|
u798086274
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 25
|
Rng is going to a festival. The name of the festival is given to you as a string S, which ends with `FESTIVAL`, from input. Answer the question: "Rng is going to a festival of what?" Output the answer. Here, assume that the name of "a festival of s" is a string obtained by appending `FESTIVAL` to the end of s. For example, `CODEFESTIVAL` is a festival of `CODE`.
|
s = input()
print(s[:-7])
|
s311787738
|
Accepted
| 17
| 2,940
| 25
|
s = input()
print(s[:-8])
|
s432297093
|
p03470
|
u541091793
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 189
|
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
n = int(input())
mochi = [int(input()) for i in range(n)]
mochi.reverse()
size = 0
max = 9999999999
for m in mochi:
if m < max:
size += 1
max = m
print(size)
|
s520610333
|
Accepted
| 19
| 2,940
| 213
|
n = int(input())
mochi = [int(input()) for i in range(n)]
mochi.sort(reverse=True)
size = 0
max = 9999999999
for m in mochi:
# print(m)
if m < max:
size += 1
max = m
print(size)
|
s515460077
|
p04011
|
u634248565
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 116
|
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
a = int(input())
b = int(input())
h = int(input())
one = int(a + b)
two = int(one/2)
ans = int(two*h)
print (ans)
|
s338194370
|
Accepted
| 18
| 2,940
| 118
|
N = int(input())
K = int(input())
X = int(input())
Y = int(input())
if N > K:
print(K*X+(N-K)*Y)
else:
print(N*X)
|
s335835309
|
p03160
|
u353080785
| 2,000
| 1,048,576
|
Wrong Answer
| 42
| 13,924
| 718
|
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
#!/usr/bin/env python
# coding: utf-8
dp = {}
NOT_FOUND = 99999
def sget(i):
if i in dp:
return dp[i]
return NOT_FOUND
def chmin(a, b):
if (a < b):
return a
return b
def main(N, hs):
for i in range(1, N):
dp[i] = chmin(sget(i), sget(i - 1) + abs(hs[i] - hs[i - 1]))
if i > 1:
dp[i] = chmin(sget(i), sget(i - 2) + abs(hs[i] - hs[i - 2]))
return dp[N-1]
if __name__ == '__main__':
#N = 4
#hs = [10, 30, 40, 20]
#N = 2
#hs = [10, 10]
#N = 6
#hs = [30, 10, 60, 10, 60, 50]
#N = 2
#hs = [1, 3, 2, 1]
N = int(input())
hs = list(map(int, input().split(" ")))
dp[0] = 0
#print(main(N, hs))
print(dp)
|
s528262434
|
Accepted
| 199
| 23,216
| 1,210
|
#!/usr/bin/env python
# coding: utf-8
N = int(input())
h = list(map(int, input().split()))
#dp = {}
NOT_FOUND = pow(10, 4) + 1
def sget(i):
if i in dp:
return dp[i]
return NOT_FOUND
#dp = {0: 0, 1: abs(h[0] - h[1])}
#print(dp[N-1])
dp = {0: 0}
for i in range(1, N):
dp[i] = sget(i-1) + abs(h[i] - h[i-1])
if i > 1:
dp[i] = min(sget(i), sget(i-2) + abs(h[i] - h[i-2]))
print(dp[N-1])
#def main(N, hs):
# dp[0] = 0
# #def rec(i):
# # return 0
# # res = NOT_FOUND
# # res = chmin(res, rec(i-1) + abs(hs[i] - hs[i-1]))
# # if (i > 1):
# # res = chmin(res, rec(i-2) + abs(hs[i] - hs[i-2]))
# # dp[i] = res
# #return rec(N-1)
|
s703023425
|
p04029
|
u502304480
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 103
|
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())
q, mod = divmod(n, 2)
if mod == 0:
print((1 + n) * q)
else:
print((1 + n) * q + q)
|
s285559093
|
Accepted
| 17
| 2,940
| 54
|
n = int(input())
l = list(range(1, n+1))
print(sum(l))
|
s894746894
|
p00004
|
u412890344
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,740
| 629
|
Write a program which solve a simultaneous equation: ax + by = c dx + ey = f The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.
|
def get_input():
while True:
try:
yield "".join(input())
except EOFError:
break
if __name__ == "__main__":
array = list(get_input())
#print(array)
for i in range(len(array)):
a,b,c,d,e,f = array[i].split()
if int(b)!=0:
x = (int(b)*int(f) - int(c)*int(e))/(int(b)*int(d) - int(a)*int(e))
y = -(int(a)/int(b))*x + int(c)/int(b)
elif int(b)==0 and int(a)!=0:
x = int(c)/int(a)
y = -(int(d)/int(e))*x + int(f)
print("{0} {1}".format(x,y))
|
s969931884
|
Accepted
| 30
| 7,720
| 876
|
def get_input():
while True:
try:
yield "".join(input())
except EOFError:
break
if __name__ == "__main__":
array = list(get_input())
for i in range(len(array)):
temp_a,temp_b,temp_c,temp_d,temp_e,temp_f = array[i].split()
a,b,c,d,e,f = int(temp_a),int(temp_b),int(temp_c),int(temp_d),int(temp_e),int(temp_f)
if a!=0:
if e-b*d/a !=0:
y = (f-c*d/a)/(e-b*d/a)
x = (c - b*y)/a
else:
print("cannot solve equation")
else:
if b!=0:
y = c/b
if d!=0:
x = (f -e*y)/d
else:
print("cannot solve equation")
print("{:.3f} {:.3f}".format(x,y))
|
s908246708
|
p03377
|
u948524308
| 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 = map(int,input().split())
if A <= X and X-A <= B:
print("Yes")
else:
print("No")
|
s962177377
|
Accepted
| 26
| 3,064
| 96
|
A,B,X = map(int,input().split())
if A <= X and X-A <= B:
print("YES")
else:
print("NO")
|
s556886764
|
p03457
|
u095426154
| 2,000
| 262,144
|
Wrong Answer
| 457
| 11,824
| 366
|
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=[0 for i in range(n)]
x=[0 for i in range(n)]
y=[0 for i in range(n)]
for i in range(n):
t[i],x[i],y[i]=list(map(int,input().split(" ")))
x.insert(0,0)
y.insert(0,0)
for i in range(n-1):
abx=abs(x[i]-x[i+1])
aby=abs(y[i]-y[i+1])
if abx+aby>t[i] or (abs(x[i+1])+abs(y[i+1]))%2!=t[i]%2:
print("NO")
exit(0)
print("YES")
|
s621709910
|
Accepted
| 467
| 11,824
| 387
|
n=int(input())
t=[0 for i in range(n)]
x=[0 for i in range(n)]
y=[0 for i in range(n)]
for i in range(n):
t[i],x[i],y[i]=list(map(int,input().split(" ")))
t.insert(0,0)
x.insert(0,0)
y.insert(0,0)
for i in range(n):
abx=abs(x[i]-x[i+1])
aby=abs(y[i]-y[i+1])
if abx+aby>t[i+1]-t[i] or (abs(x[i+1])+abs(y[i+1]))%2!=t[i+1]%2:
print("No")
exit(0)
print("Yes")
|
s551135046
|
p03478
|
u733132703
| 2,000
| 262,144
|
Wrong Answer
| 38
| 9,248
| 246
|
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())
def check_num(n):
s = str(n)
array = list(map(int,s))
return sum(array)
list_1 = [i for i in range(1,N+1)]
list_2 = []
for _ in list_1:
if A<=check_num(_)<=B:
list_2.append(_)
else:
pass
|
s516554186
|
Accepted
| 40
| 9,432
| 268
|
N,A,B = map(int,input().split())
def check_num(n):
s = str(n)
array = list(map(int,s))
return sum(array)
list_1 = [i for i in range(1,N+1)]
list_2 = []
for _ in list_1:
if A<=check_num(_)<=B:
list_2.append(_)
else:
pass
print(sum(list_2))
|
s872254805
|
p03478
|
u000842852
| 2,000
| 262,144
|
Wrong Answer
| 286
| 12,504
| 251
|
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).
|
import numpy as np
N, A, B = map(int, input().split())
total = 0
for i in range(N+1):
a = i
lis =[]
while i >0:
lis.append(i%10)
i //=10
lis.reverse()
lis = np.array(lis)
SUM = np.sum(lis)
if SUM >=A and SUM <=B:
total += a
|
s630836618
|
Accepted
| 294
| 12,472
| 272
|
import numpy as np
N, A, B = map(int, input().split())
total = 0
for i in range(1, N+1):
a = i
lis =[]
while i >0:
lis.append(i%10)
i //=10
lis.reverse()
lis = np.array(lis)
SUM = np.sum(lis)
if SUM >=A and SUM <=B:
total += a
print(total)
|
s449791854
|
p03433
|
u075409829
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 93
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
N = int(input())
A = int(input())
if (N-A)%500 == 0:
print("Yes")
else:
print("No")
|
s169250491
|
Accepted
| 17
| 2,940
| 88
|
N = int(input())
A = int(input())
if N%500 <= A:
print("Yes")
else:
print("No")
|
s180023767
|
p03548
|
u620846115
| 2,000
| 262,144
|
Wrong Answer
| 33
| 9,144
| 46
|
We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person. At most how many people can sit on the seat?
|
x,y,z=map(int,input().split())
print((x-z)//y)
|
s434450617
|
Accepted
| 26
| 9,160
| 50
|
x,y,z=map(int,input().split())
print((x-z)//(y+z))
|
s962387074
|
p03943
|
u143492911
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 138
|
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())
if a+b==c:
print("Yes")
exit()
if b+c==a:
print("Yes")
if a+c==b:
print("Yes")
print("No")
|
s282687654
|
Accepted
| 18
| 2,940
| 160
|
a,b,c=map(int,input().split())
if a+b==c:
print("Yes")
exit()
if b+c==a:
print("Yes")
exit()
if a+c==b:
print("Yes")
exit()
print("No")
|
s640568838
|
p04030
|
u370852395
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 254
|
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
str_list=[]
tmp=[]
s=''
str_list=list(input())
for i in str_list:
if i =='0':
tmp.append('0')
elif i == '1':
tmp.append('1')
elif i == 'B' and tmp:
tmp.pop()
for i in range(len(tmp)):
s+=tmp[i]
print(tmp)
print(s)
|
s571315879
|
Accepted
| 17
| 3,060
| 243
|
str_list=[]
tmp=[]
s=''
str_list=list(input())
for i in str_list:
if i =='0':
tmp.append('0')
elif i == '1':
tmp.append('1')
elif i == 'B' and tmp:
tmp.pop()
for i in range(len(tmp)):
s+=tmp[i]
print(s)
|
s981462393
|
p03720
|
u633450100
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,176
| 193
|
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
|
N,M = [int(i) for i in input().split()]
list = []
for i in range(M):
a,b = [int(i) for i in input().split()]
list.append(a)
list.append(b)
for i in range(N):
print(list.count(i))
|
s169857579
|
Accepted
| 27
| 9,120
| 193
|
N,M = [int(i) for i in input().split()]
list = []
for i in range(M):
a,b = [int(i) for i in input().split()]
list.append(a)
list.append(b)
for i in range(1,N+1):
print(list.count(i))
|
s806453136
|
p02612
|
u387870994
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,136
| 19
|
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.
|
print(input()[-3:])
|
s416442519
|
Accepted
| 28
| 9,160
| 79
|
n = int(input())
if n % 1000 == 0:
print(0)
else:
print(1000 - n%1000)
|
s820986982
|
p02613
|
u122743999
| 2,000
| 1,048,576
|
Wrong Answer
| 158
| 9,196
| 364
|
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.
|
score = {'AC': 0, 'WA': 0, 'TLE': 0, 'RE': 0}
N = int(input())
for _ in range(N):
result = input()
if result == 'AC':
score[result] += 1
elif result == 'WA':
score[result] += 1
elif result == 'TLE':
score[result] += 1
elif result == 'RE':
score[result] += 1
for i in score.keys():
print(i, '×', score[i])
|
s829225196
|
Accepted
| 158
| 9,196
| 362
|
score = {'AC': 0, 'WA': 0, 'TLE': 0, 'RE': 0}
N = int(input())
for _ in range(N):
result = input()
if result == 'AC':
score[result] += 1
elif result == 'WA':
score[result] += 1
elif result == 'TLE':
score[result] += 1
elif result == 'RE':
score[result] += 1
for i in score.keys():
print(i, 'x', score[i])
|
s485784400
|
p03760
|
u626881915
| 2,000
| 262,144
|
Wrong Answer
| 25
| 8,972
| 113
|
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.
|
s=input()
t=input()
l=""
i=0
j=0
while i < len(s):
l+=s[i]
i+=1
while j < len(t):
l+=t[j]
j+=1
print(l)
|
s396045687
|
Accepted
| 25
| 9,052
| 171
|
s=input()
t=input()
l=""
i=0
j=0
while True:
if i < len(s):
l+=s[i]
i+=1
if j < len(t):
l+=t[j]
j+=1
if len(l) == len(s)+len(t):
break
print(l)
|
s896151112
|
p02392
|
u933096856
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,684
| 89
|
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
a,b,c=map(int,input().split())
if a < b and b < c:
print('YES')
else:
print('NO')
|
s291658141
|
Accepted
| 50
| 7,696
| 89
|
a,b,c=map(int,input().split())
if a < b and b < c:
print('Yes')
else:
print('No')
|
s351695897
|
p03156
|
u814986259
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,064
| 264
|
You have written N problems to hold programming contests. The i-th problem will have a score of P_i points if used in a contest. With these problems, you would like to hold as many contests as possible under the following condition: * A contest has three problems. The first problem has a score not greater than A points, the second has a score between A + 1 and B points (inclusive), and the third has a score not less than B + 1 points. The same problem should not be used in multiple contests. At most how many contests can be held?
|
import bisect
N = int(input())
A,B = map(int, input().split())
P = list(map(int, input().split()))
P.sort()
ans = N
id = bisect.bisect_right(P,A)
ans = min(ans,id)
prev = id
id = bisect.bisect_right(P,B)
ans = min(ans,id - prev)
ans = min(ans,N-1 - id)
print(ans)
|
s356419642
|
Accepted
| 18
| 3,060
| 262
|
import bisect
N = int(input())
A,B = map(int, input().split())
P = list(map(int, input().split()))
P.sort()
ans = N
id = bisect.bisect_right(P,A)
ans = min(ans,id)
prev = id
id = bisect.bisect_right(P,B)
ans = min(ans,id - prev)
ans = min(ans,N - id)
print(ans)
|
s889633468
|
p03181
|
u729133443
| 2,000
| 1,048,576
|
Wrong Answer
| 994
| 31,796
| 474
|
There is a tree with N vertices, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N - 1), the i-th edge connects Vertex x_i and y_i. Taro has decided to paint each vertex in white or black, so that any black vertex can be reached from any other black vertex by passing through only black vertices. You are given a positive integer M. For each v (1 \leq v \leq N), answer the following question: * Assuming that Vertex v has to be black, find the number of ways in which the vertices can be painted, modulo M.
|
n,m,*t=map(int,open(0).read().split())
e=[[]for _ in range(n)]
for a,b in zip(*[iter(t)]*2):
e[a-1]+=b-1,
e[b-1]+=a-1,
o=[]
s=[0]
f=[1]*n
for v in s:
o+=v,
f[v]=0
t=[]
for w in e[v]:
if f[w]:
s+=w,
t+=w,
e[v]=t
dp1=[0]*n
for v in o[::-1]:
t=1
for w in e[v]:
t=t*(dp1[w]+1)%m
dp1[v]=t
dp2=[0]*n
dp2[0]=1
for v in o:
t=dp2[v]*dp1[v]%m
for w in e[v]:
dp2[w]=t*pow(dp1[w]+1,m-2,m)+1
for dp1,dp2 in zip(dp1,dp2):
print(dp1*dp2%m)
|
s225957576
|
Accepted
| 1,272
| 57,740
| 592
|
n,m,*t=map(int,open(0).read().split())
e=[[]for _ in range(n)]
for a,b in zip(*[iter(t)]*2):
e[a-1]+=b-1,
e[b-1]+=a-1,
o=[]
s=[0]
f=[1]*n
for v in s:
o+=v,
f[v]=0
t=[]
for w in e[v]:
if f[w]:
s+=w,
t+=w,
e[v]=t
dp1=[0]*n
c=[[]for _ in range(n)]
for v in o[::-1]:
c1,c2=[1],[1]
for w in e[v]:
c1+=c1[-1]*(dp1[w]+1)%m,
for w in e[v][::-1]:
c2+=c2[-1]*(dp1[w]+1)%m,
dp1[v]=c1[-1]
c[v]=c1,c2[-2::-1]
dp2=[0]*n
dp2[0]=1
for v in o:
t=dp2[v]
for w,c1,c2 in zip(e[v],*c[v]):
dp2[w]=(t*c1*c2+1)%m
for dp1,dp2 in zip(dp1,dp2):
print(dp1*dp2%m)
|
s671297858
|
p03469
|
u620846115
| 2,000
| 262,144
|
Wrong Answer
| 31
| 8,900
| 45
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
n = input()
n.replace("2017","2018")
print(n)
|
s388281216
|
Accepted
| 26
| 8,980
| 49
|
s = input()
s = s.replace("2017","2018")
print(s)
|
s354972974
|
p03605
|
u388297793
| 2,000
| 262,144
|
Wrong Answer
| 25
| 9,088
| 61
|
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
n=input()
if n in "9":
print("Yes")
else:
print("No")
|
s648270599
|
Accepted
| 29
| 9,044
| 61
|
n=input()
if "9" in n:
print("Yes")
else:
print("No")
|
s398233114
|
p03455
|
u773686010
| 2,000
| 262,144
|
Wrong Answer
| 25
| 9,156
| 106
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b = map(int, input().split())
if (a % 2 == 0) and (b % 2 ==0):
print ('Even')
else:
print('Odd')
|
s507758877
|
Accepted
| 29
| 8,852
| 145
|
##AtCoder Beginners Selection Product
a,b = map(int, input().split())
if (a % 2 == 0) or (b % 2 ==0):
print ('Even')
else:
print('Odd')
|
s578695914
|
p03155
|
u802963389
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 74
|
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board. The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns. How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?
|
N = int(input())
H = int(input())
W = int(input())
print((W-N+1)*(H-N+1))
|
s045283311
|
Accepted
| 18
| 2,940
| 75
|
N = int(input())
H = int(input())
W = int(input())
print((N-W+1)*(N-H+1))
|
s713827601
|
p03050
|
u688126754
| 2,000
| 1,048,576
|
Wrong Answer
| 277
| 12,652
| 317
|
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 numpy as np
N = int(input())
divisor_lst = []
for i in range(1, int(np.sqrt(N))+1):
if N % i == 0:
divisor_lst.append(i)
divisor_lst.append(N//i)
print(i)
arr = np.array(divisor_lst)
print(arr)
arr -= 1
print(int(np.sum(arr)) - 1)
|
s329250976
|
Accepted
| 501
| 12,424
| 209
|
import numpy as np
N = int(input())
divisor_lst = []
i = 1
while i+2 <= N//i:
if N % i == 0:
divisor_lst.append(N//i)
i += 1
arr = np.array(divisor_lst)
arr -= 1
print(int(np.sum(arr)))
|
s317613064
|
p04029
|
u095089755
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 39
|
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) / 2)
|
s861075422
|
Accepted
| 17
| 2,940
| 40
|
N = int(input())
print(N * (N + 1) // 2)
|
s653563815
|
p03860
|
u719840207
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 25
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
print("A"+input()[0]+"C")
|
s696341782
|
Accepted
| 17
| 2,940
| 25
|
print("A"+input()[8]+"C")
|
s633523526
|
p03599
|
u687053495
| 3,000
| 262,144
|
Time Limit Exceeded
| 3,156
| 3,064
| 530
|
Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the beaker. * Operation 3: Put C grams of sugar into the beaker. * Operation 4: Put D grams of sugar into the beaker. In our experimental environment, E grams of sugar can dissolve into 100 grams of water. Snuke will make sugar water with the highest possible density. The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker. Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it. If there is more than one candidate, any of them will be accepted. We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent. Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water.
|
A, B, C, D, E, F = map(int, input().split())
density = 0
ans = (0, 0)
for a in range(31):
for b in range(31):
for c in range(100):
for d in range(100):
water = 100*a*A + 100*b*B
suger = c*C + d*D
if water + suger > F or water == 0:
continue
if suger > 0 and (suger / water * 100) > E:
continue
tmp = (100 * suger) / (water + suger)
if tmp >= density:
density = tmp
ans = (water+suger, suger)
print(*ans)
|
s095009304
|
Accepted
| 316
| 3,064
| 502
|
A, B, C, D, E, F = map(int, input().split())
density = -1
ans = (0, 0)
for a in range(31):
for b in range(31):
for c in range(101):
for d in range(101):
water = 100*a*A + 100*b*B
suger = c*C + d*D
if water + suger > F or water == 0:
break
if (water / 100) * E >= suger:
tmp = (100 * suger) / (water + suger)
if tmp > density:
density = tmp
ans = (water+suger, suger)
print(*ans)
|
s560451247
|
p00728
|
u780025254
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,684
| 266
|
The International Clown and Pierrot Competition (ICPC), is one of the most distinguished and also the most popular events on earth in the show business. One of the unique features of this contest is the great number of judges that sometimes counts up to one hundred. The number of judges may differ from one contestant to another, because judges with any relationship whatsoever with a specific contestant are temporarily excluded for scoring his/her performance. Basically, scores given to a contestant's performance by the judges are averaged to decide his/her score. To avoid letting judges with eccentric viewpoints too much influence the score, the highest and the lowest scores are set aside in this calculation. If the same highest score is marked by two or more judges, only one of them is ignored. The same is with the lowest score. The average, which may contain fractions, are truncated down to obtain final score as an integer. You are asked to write a program that computes the scores of performances, given the scores of all the judges, to speed up the event to be suited for a TV program.
|
import sys
while True:
n = int(sys.stdin.readline().rstrip())
if n == 0:
break
scores = [int(sys.stdin.readline().rstrip()) for i in range(n)]
result = (sum(scores) - (min(scores) + max(scores)) // n - 2)
print(result)
|
s022082523
|
Accepted
| 30
| 7,720
| 252
|
import sys
while True:
n = int(sys.stdin.readline().rstrip())
if n == 0:
break
scores = [int(sys.stdin.readline().rstrip()) for i in range(n)]
result = (sum(scores) - (max(scores) + min(scores))) // (n - 2)
print(result)
|
s395615633
|
p03478
|
u382431597
| 2,000
| 262,144
|
Wrong Answer
| 38
| 3,060
| 176
|
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 = list(map(int, input().split()))
isum = 0
for i in range(1,n+1):
tmp = [int(j) for j in list(str(i))]
if a <= sum(tmp) <= b:
isum += sum(tmp)
print(isum)
|
s943466963
|
Accepted
| 36
| 3,060
| 194
|
n,a,b = list(map(int, input().split()))
isum = 0
for i in range(1,n+1):
tmp = [int(j) for j in list(str(i))]
#print(tmp,sum(tmp))
if a <= sum(tmp) <= b:
isum += i
print(isum)
|
s728401768
|
p02412
|
u744506422
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 360
|
Write a program which identifies the number of combinations of three integers which satisfy the following conditions: * You should select three distinct integers from 1 to n. * A total sum of the three integers is x. For example, there are two combinations for n = 5 and x = 9. * 1 + 3 + 5 = 9 * 2 + 3 + 4 = 9
|
a=[]
w=""
while(w!="0 0"):
w=input()
a.append(w)
a.pop()
for st in a:
s=st.split()
n=int(s[0])
x=int(s[1])
count=0
for a in range(1,x//3+1):
for b in range(a+1,(x-a)//2+1):
if (x-a-b)<=n and b<(x-a-b):
count+=1
print("{0} {1} {2}".format(a,b,x-a-b))
print("{0}".format(count))
|
s771592452
|
Accepted
| 30
| 5,596
| 305
|
a=[]
w=""
while(w!="0 0"):
w=input()
a.append(w)
a.pop()
for st in a:
s=st.split()
n=int(s[0])
x=int(s[1])
count=0
for a in range(1,x//3+1):
for b in range(a+1,(x-a)//2+1):
if (x-a-b)<=n and b<(x-a-b):
count+=1
print("{0}".format(count))
|
s241409241
|
p03455
|
u263830634
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 94
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
if (a*b) %2 == 0:
print ('Evev')
else:
print ('Odd')
|
s399601991
|
Accepted
| 17
| 2,940
| 94
|
a, b = map(int, input().split())
if (a*b) %2 == 0:
print ('Even')
else:
print ('Odd')
|
s227270360
|
p03477
|
u597455618
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 128
|
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.
|
a, b, c, d = map(int, input().split())
if a+b < c+d:
print("Left")
elif a+b == c+d:
print("Balanced")
else:
print("Right")
|
s842857642
|
Accepted
| 17
| 3,060
| 128
|
a, b, c, d = map(int, input().split())
if a+b > c+d:
print("Left")
elif a+b == c+d:
print("Balanced")
else:
print("Right")
|
s213684572
|
p03339
|
u880466014
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 89
|
There are N people standing in a row from west to east. Each person is facing east or west. The directions of the people is given as a string S of length N. The i-th person from the west is facing east if S_i = `E`, and west if S_i = `W`. You will appoint one of the N people as the leader, then command the rest of them to face in the direction of the leader. Here, we do not care which direction the leader is facing. The people in the row hate to change their directions, so you would like to select the leader so that the number of people who have to change their directions is minimized. Find the minimum number of people who have to change their directions.
|
S = list(input())
e = S.count('E')
w = S.count('W')
if e > w:
print(w)
else:
print(e)
|
s427259990
|
Accepted
| 110
| 3,700
| 466
|
N = int(input())
S = input()
E_count = S.count('E')
W_count = 0
mini_count = E_count
for i in S:
if i == 'E':
E_count -= 1
else:
W_count += 1
if mini_count > E_count + W_count:
mini_count = E_count + W_count
print(mini_count)
|
s188666720
|
p02612
|
u925242392
| 2,000
| 1,048,576
|
Wrong Answer
| 33
| 9,056
| 63
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n=int(input())
if n%1000==0:
print(0)
else:
print(n-n%1000)
|
s420455270
|
Accepted
| 28
| 9,040
| 75
|
n=int(input())
if n%1000==0:
print(0)
else:
print(((n//1000)+1)*1000-n)
|
s715780801
|
p02258
|
u716198574
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,744
| 276
|
You can obtain profits from foreign exchange margin transactions. For example, if you buy 1000 dollar at a rate of 100 yen per dollar, and sell them at a rate of 108 yen per dollar, you can obtain (108 - 100) × 1000 = 8000 yen. Write a program which reads values of a currency $R_t$ at a certain time $t$ ($t = 0, 1, 2, ... n-1$), and reports the maximum value of $R_j - R_i$ where $j > i$ .
|
N = int(input('R ?????????: '))
R = [int(input('R[%s]: ' % (str(i)))) for i in range(N)]
maxv = -20000000000
minv = R[0]
for i in range(1, N):
if R[i] - minv > maxv:
maxv = R[i] - minv
if minv > R[i]:
minv = R[i]
print('?????§??????: ' + str(maxv))
|
s035365251
|
Accepted
| 520
| 15,688
| 183
|
N = int(input())
R = [int(input()) for i in range(N)]
maxv = -20000000000
minv = R[0]
for i in range(1, N):
maxv = max(R[i] - minv, maxv)
minv = min(R[i], minv)
print(maxv)
|
s700692696
|
p02397
|
u104171359
| 1,000
| 131,072
|
Wrong Answer
| 60
| 7,624
| 430
|
Write a program which reads two integers x and y, and prints them in ascending order.
|
#!usr/bin/env python3
def string_two_numbers_spliter():
x, y = [int(i) for i in input().split()]
return x, y
def swap_two_numbers_asc(x, y):
if x > y:
x, y = y, x
return x, y
def main():
while True:
x, y = string_two_numbers_spliter()
swap_two_numbers_asc(x, y)
if x == 0 and y == 0:
break
print('%d %d' % (x, y))
if __name__ == '__main__':
main()
|
s355365052
|
Accepted
| 70
| 7,760
| 441
|
#!usr/bin/env python3
def string_two_numbers_spliter():
x, y = [int(i) for i in input().split()]
return x, y
def swap_two_numbers_asc(x, y):
x, y = y, x
return x, y
def main():
while True:
x, y = string_two_numbers_spliter()
if x == 0 and y == 0:
break
if x > y:
x, y = swap_two_numbers_asc(x, y)
print('%d %d' % (x, y))
if __name__ == '__main__':
main()
|
s085167343
|
p03380
|
u556225812
| 2,000
| 262,144
|
Wrong Answer
| 2,145
| 925,008
| 408
|
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.
|
import math
import itertools
def comb(n, r):
return math.factorial(n) // (math.factorial(n - r) * math.factorial(r))
N = int(input())
lst = list(map(int, input().split()))
ans = set(itertools.combinations(lst, 2))
max = 0
for i in ans:
if i[0] > i[1]:
n = i[0]
r = i[1]
else:
n = i[1]
r = i[0]
x = comb(n, r)
if x >= max:
pair = [n, r]
print(*pair)
|
s706236565
|
Accepted
| 81
| 14,180
| 252
|
import bisect
import math
N = int(input())
lst = list(map(int, input().split()))
lst.sort()
n = max(lst)
a = bisect.bisect_right(lst, n//2)
if abs(math.ceil(n/2)-lst[a]) >= abs(math.ceil(n/2)-lst[a-1]):
print(n, lst[a-1])
else:
print(n, lst[a])
|
s014624010
|
p03251
|
u788137651
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 377
|
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 = [int(i) for i in input().split()]
y = [int(i) for i in input().split()]
print(x, y)
if X < Y:
max_x = max(x)
min_y = min(y)
if max_x+1 < min_y:
print("No War")
else:
print("War")
else:
min_x = min(x)
max_y = max(y)
if min_x > max_y+1:
print("No War")
else:
print("War")
|
s614907253
|
Accepted
| 19
| 2,940
| 235
|
N, M, X, Y = map(int, input().split())
x = [int(i) for i in input().split()]
y = [int(i) for i in input().split()]
max_x = max(x)
min_y = min(y)
if max_x < min_y and max_x < Y and min_y > X:
print("No War")
else:
print("War")
|
s607514632
|
p02277
|
u599130514
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,608
| 1,365
|
Let's arrange a deck of cards. Your task is to sort totally n cards. A card consists of a part of a suit (S, H, C or D) and an number. Write a program which sorts such cards based on the following pseudocode: Partition(A, p, r) 1 x = A[r] 2 i = p-1 3 for j = p to r-1 4 do if A[j] <= x 5 then i = i+1 6 exchange A[i] and A[j] 7 exchange A[i+1] and A[r] 8 return i+1 Quicksort(A, p, r) 1 if p < r 2 then q = Partition(A, p, r) 3 run Quicksort(A, p, q-1) 4 run Quicksort(A, q+1, r) Here, A is an array which represents a deck of cards and comparison operations are performed based on the numbers. Your program should also 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).
|
def merge(A, left, mid, right):
n1 = mid - left
n2 = right - mid
L, R = A[left: left + n1], A[mid: mid + n2]
L.append([0,2000000000])
R.append([0,2000000000])
i, j = 0, 0
for k in range(left, right):
if L[i][1] <= R[j][1]:
A[k] = L[i]
i += 1
else:
A[k] = R[j]
j += 1
def mergeSort(A, left, right):
if left + 1 < right:
mid = (left + right) // 2
mergeSort(A, left, mid)
mergeSort(A, mid, right)
merge(A, left, mid, right)
def partition(A, p, r):
x = A[r][1]
i = p - 1
for j in range(p, r):
if A[j][1] <= x:
i += 1
A[i], A[j] = A[j], A[i]
A[i + 1], A[r] = A[r], A[i + 1]
return i + 1
def quick_sort(A, p, r):
if p < r:
q = partition(A, p, r)
quick_sort(A, p, q - 1)
quick_sort(A, q + 1, r)
n = int(input())
cards = []
merge_cards = []
for _ in range(n):
tmp = input().split(' ')
cards.append([tmp[0], int(tmp[1])])
merge_cards.append([tmp[0], int(tmp[1])])
mergeSort(merge_cards, 0, n)
quick_sort(cards, 0, n - 1)
if merge_cards == cards:
print("Stable")
else:
print("Not Stable")
for card in cards:
print(*card)
|
s869088596
|
Accepted
| 1,770
| 30,216
| 1,365
|
def merge(A, left, mid, right):
n1 = mid - left
n2 = right - mid
L, R = A[left: left + n1], A[mid: mid + n2]
L.append([0,2000000000])
R.append([0,2000000000])
i, j = 0, 0
for k in range(left, right):
if L[i][1] <= R[j][1]:
A[k] = L[i]
i += 1
else:
A[k] = R[j]
j += 1
def mergeSort(A, left, right):
if left + 1 < right:
mid = (left + right) // 2
mergeSort(A, left, mid)
mergeSort(A, mid, right)
merge(A, left, mid, right)
def partition(A, p, r):
x = A[r][1]
i = p - 1
for j in range(p, r):
if A[j][1] <= x:
i += 1
A[i], A[j] = A[j], A[i]
A[i + 1], A[r] = A[r], A[i + 1]
return i + 1
def quick_sort(A, p, r):
if p < r:
q = partition(A, p, r)
quick_sort(A, p, q - 1)
quick_sort(A, q + 1, r)
n = int(input())
cards = []
merge_cards = []
for _ in range(n):
tmp = input().split(' ')
cards.append([tmp[0], int(tmp[1])])
merge_cards.append([tmp[0], int(tmp[1])])
mergeSort(merge_cards, 0, n)
quick_sort(cards, 0, n - 1)
if merge_cards == cards:
print("Stable")
else:
print("Not stable")
for card in cards:
print(*card)
|
s422499200
|
p00003
|
u477023447
| 1,000
| 131,072
|
Wrong Answer
| 30
| 5,920
| 305
|
Write a program which judges wheather given length of three side form a right triangle. Print "YES" if the given sides (integers) form a right triangle, "NO" if not so.
|
tri_list = []
N = int(input())
for i in range(N):
tri = input()
tri_list.append(tri.split())
for i in range(N):
tri_list[i].sort()
print(tri_list[i])
if int(tri_list[i][2])^2 == int(tri_list[i][0])^2 + int(tri_list[i][1])^2 :
print("YES")
else :
print("NO")
|
s126417528
|
Accepted
| 40
| 5,600
| 256
|
n = int(input())
for i in range(n):
tri_list = list(map(int,input().split(" ")))
tri_list.sort()
a = int(tri_list[2]) ** 2
b = int(tri_list[0]) ** 2 + int(tri_list[1]) ** 2
if a == b:
print("YES")
else:
print("NO")
|
s281048358
|
p03943
|
u904945034
| 2,000
| 262,144
|
Wrong Answer
| 31
| 9,160
| 112
|
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 = list(map(int,input().split()))
b = a.pop(a.index(max(a)))
if b == sum(a):
print("YES")
else:
print("NO")
|
s485277227
|
Accepted
| 28
| 9,064
| 112
|
a = list(map(int,input().split()))
b = a.pop(a.index(max(a)))
if b == sum(a):
print("Yes")
else:
print("No")
|
s517752580
|
p03545
|
u212328220
| 2,000
| 262,144
|
Wrong Answer
| 31
| 9,200
| 322
|
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.
|
s = input()
n = 3
for i in range(2**n):
ans = s[0]
sum = int(s[0])
for j in range(3):
if 1 & (i >> j):
ans += f'+{s[j+1]}'
sum += int(s[j+1])
else:
ans += f'-{s[j+1]}'
sum -= int(s[j+1])
if sum == 7:
print(f'{ans}+7')
exit()
|
s164350221
|
Accepted
| 29
| 8,996
| 322
|
s = input()
n = 3
for i in range(2**n):
ans = s[0]
sum = int(s[0])
for j in range(3):
if 1 & (i >> j):
ans += f'+{s[j+1]}'
sum += int(s[j+1])
else:
ans += f'-{s[j+1]}'
sum -= int(s[j+1])
if sum == 7:
print(f'{ans}=7')
exit()
|
s764921437
|
p03543
|
u395894569
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 56
|
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=list(input())
print('YES' if len(set(n))<=2 else 'NO')
|
s181070020
|
Accepted
| 17
| 2,940
| 119
|
n=list(input())
for i in range(2):
if n[i] == n[i + 1] == n[i + 2]:
print('Yes')
exit()
print('No')
|
s964176122
|
p03697
|
u077019541
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 82
|
You are given two integers A and B as the input. Output the value of A + B. However, if A + B is 10 or greater, output `error` instead.
|
A,B = map(int,input().split())
if A+B<=10:
print("error")
else:
print(A+B)
|
s496451427
|
Accepted
| 17
| 2,940
| 82
|
A,B = map(int,input().split())
if A+B>=10:
print("error")
else:
print(A+B)
|
s704455252
|
p03720
|
u204523044
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 221
|
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
|
N, M = map(int, input().split())
a = []
b = []
for i in range(M):
ai, bi = map(int, input().split())
a.append(ai)
b.append(bi)
for i in range(N):
r = 0
r += a.count(i+1)
r += b.count(i+1)
print(r)
|
s775103691
|
Accepted
| 18
| 3,060
| 225
|
N, M = map(int, input().split())
a = []
b = []
for i in range(M):
ai, bi = map(int, input().split())
a.append(ai)
b.append(bi)
for i in range(N):
r = 0
r += a.count(i+1)
r += b.count(i+1)
print(r)
|
s481408102
|
p03369
|
u608726540
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 33
|
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
x=input()
print(700+x.count('o'))
|
s397669508
|
Accepted
| 17
| 2,940
| 48
|
s=list(input())
n=s.count('o')
print(700+100*n)
|
s424496641
|
p02665
|
u935241425
| 2,000
| 1,048,576
|
Wrong Answer
| 58
| 20,064
| 555
|
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.
|
n = int( input() )
p = [ int( pi ) for pi in input().split() ]
sump = int( 0 )
for pi in p:
sump += pi
ch = int( -1 )
ans = int( 0 )
for i in range( n+1 ):
if i == 0:
ans += 1
px = 1
elif i == 1:
ans = -1
break
else:
if px*2 <= sump:
ans += px*2
px = px*2 - p[ i ]
sump = sump - p[ i ]
else:
ans += sump
px = sump - p[ i ]
sump = sump - p[ i ]
if px < 0:
ans = -1
break
print( ans )
|
s070400953
|
Accepted
| 103
| 20,008
| 828
|
n = int( input() )
p = [ int( pi ) for pi in input().split() ]
sump = int( 0 )
for pi in p:
sump += pi
ans = int( 0 )
for i in range( n+1 ):
if i == 0:
ans += 1
px = 1
if p[ i ] == 1 and n == 0:
ans = 1
break
if p[ i ] != 0 and p[ i ] != 1:
ans = -1
break
else:
if px*2 <= sump:
ans += px*2
px = px*2 - p[ i ]
sump = sump - p[ i ]
else:
ans += sump
px = sump - p[ i ]
sump = sump - p[ i ]
# print( px, sump )
if i != n and px <= 0:
ans = -1
break
elif i == n and sump != 0:
ans = -1
break
elif i == n and px != 0:
ans = -1
break
print( ans )
|
s712416594
|
p03471
|
u728498511
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 485
|
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
n, y = map(int, input().split())
y /= 1000
a = [0]*3
a[1], a[2] = divmod(y, 5)
ans = [-1]*3
while sum(a)!=n:
if n-sum(a) >= 4:
tmp = (n-sum(a))//4
a[1] -= tmp
a[2] += tmp*5
elif n-sum(a) <= -1:
tmp = sum(a)-n
a[1] -= tmp*2
a[0] += tmp
else:
tmp = n-sum(a)
a[1] -= tmp*3
a[2] += tmp*5
a[0] += tmp
if a[1]<0:break
else: ans = a
ans = list(map(int, ans))
print(*ans)
|
s526870228
|
Accepted
| 432
| 2,940
| 148
|
n,y=map(int,input().split())
ans=[-1]*3
y//=1000
for i in range(n+1):
for j in range(n-i+1):
if i*9+j*4+n==y:ans=[i,j,n-i-j]
print(*ans)
|
s976189493
|
p03455
|
u226176079
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 101
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
x = a * b
if x % 2 == 0:
print('EVEN')
else:
print('Odd')
|
s651220666
|
Accepted
| 17
| 2,940
| 102
|
a, b = map(int, input().split())
x = a * b
if x % 2 == 0:
print('Even')
else:
print('Odd')
|
s465286314
|
p03067
|
u476124554
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 110
|
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
a,b,c = map(int,input().split())
if a < c and b < c or c < a and b < c:
print('Yes')
else:
print('No')
|
s775563421
|
Accepted
| 17
| 2,940
| 110
|
a,b,c = map(int,input().split())
if a < c and c < b or c < a and b < c:
print('Yes')
else:
print('No')
|
s050143963
|
p02742
|
u664907598
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 94
|
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
h,w = map(int,input().split())
if h*w % 2 == 0:
print(h*w/2)
else:
print((h*w+1)/2)
|
s242569024
|
Accepted
| 17
| 2,940
| 138
|
h,w = map(int,input().split())
if h ==1 or w ==1:
print(1)
elif h*w % 2 == 0:
print(int(h*w/2))
else:
print(int((h*w+1)/2))
|
s667578334
|
p00050
|
u546285759
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,340
| 155
|
福島県は果物の産地としても有名で、その中でも特に桃とりんごは全国でも指折りの生産量を誇っています。ところで、ある販売用の英文パンフレットの印刷原稿を作ったところ、手違いでりんごに関する記述と桃に関する記述を逆に書いてしまいました。 あなたは、apple と peach を修正する仕事を任されましたが、なにぶん面倒です。１行の英文を入力して、そのなかの apple という文字列を全て peach に、peach という文字列を全てapple に交換した英文を出力するプログラムを作成してください。
|
d = {"apple": "peach", "peach": "apple"}
ans = []
for c in input().split():
try:
ans.append(d[c])
except:
ans.append(c)
print(*ans)
|
s370787563
|
Accepted
| 50
| 7,368
| 236
|
ans = []
for word in input().split():
if "apple" in word:
ans.append(word.replace("apple", "peach"))
elif "peach" in word:
ans.append(word.replace("peach", "apple"))
else:
ans.append(word)
print(*ans)
|
s142515986
|
p03486
|
u193927973
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 189
|
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=list(input())
t=list(input())
s.sort()
t.sort(reverse=True)
l=min(len(s), len(t))
for i in range(l):
if s[i]>t[i]:
print("No")
exit()
if s==t:
print("No")
else:
print("Yes")
|
s662418305
|
Accepted
| 17
| 3,064
| 271
|
s=list(input())
t=list(input())
s.sort()
t.sort(reverse=True)
l=min(len(s), len(t))
for i in range(l):
if s[i]==t[i]:
continue
if s[i]<t[i]:
print("Yes")
exit()
else:
print("No")
exit()
if len(s)>=len(t):
print("No")
else:
print("Yes")
|
s679611284
|
p03448
|
u352623442
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,180
| 315
|
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
a = int(input())
b = int(input())
c = int(input())
x = int(input())
ans = 0
for i in range(a+1):
if i*500 <= x:
for k in range(b+1):
if k*100 <= (x-i*500):
if (x-i*500-k*100)/50 <= c:
print(i,k,(x-i*500-k*100)/50)
ans += 1
print(ans)
|
s289497159
|
Accepted
| 18
| 3,060
| 265
|
a = int(input())
b = int(input())
c = int(input())
x = int(input())
ans = 0
for i in range(a+1):
if i*500 <= x:
for k in range(b+1):
if k*100 <= (x-i*500):
if (x-i*500-k*100)/50 <= c:
ans += 1
print(ans)
|
s333922783
|
p03502
|
u079022116
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 76
|
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number.
|
n=input()
a = []
for i in range(len(n)):
a.append(int(n[i]))
print(sum(a))
|
s500131934
|
Accepted
| 18
| 2,940
| 119
|
n=input()
a = []
for i in range(len(n)):
a.append(int(n[i]))
if int(n) % sum(a) == 0:
print('Yes')
else:print('No')
|
s758576386
|
p03860
|
u785205215
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 204
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
from sys import stdin, stdout
def readLine_str_list():return list(map(str, stdin.readline().split()))
def main():
s = readLine_str_list()
print("A"+s[0]+"C")
if __name__ == "__main__":
main()
|
s936515210
|
Accepted
| 17
| 2,940
| 207
|
from sys import stdin, stdout
def readLine_str_list():return list(map(str, stdin.readline().split()))
def main():
s = readLine_str_list()
print("A"+s[1][0]+"C")
if __name__ == "__main__":
main()
|
s554633431
|
p02612
|
u005569385
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,136
| 38
|
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())
a = N % 1000
print(a)
|
s176589415
|
Accepted
| 30
| 9,092
| 84
|
N = int(input())
a = 1000-(N % 1000)
if N%1000 == 0:
print(0)
else:
print(a)
|
s829411113
|
p02865
|
u177388368
| 2,000
| 1,048,576
|
Wrong Answer
| 20
| 3,316
| 30
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
n=int(input())
n=n//2
print(n)
|
s776669817
|
Accepted
| 17
| 2,940
| 58
|
n=int(input())
if n%2==0:n=n/2-1
else:n=n//2
print(int(n))
|
s816474498
|
p02285
|
u370086573
| 2,000
| 131,072
|
Wrong Answer
| 30
| 7,784
| 2,703
|
Write a program which performs the following operations to a binary search tree $T$ by adding delete operation to B: Binary Search Tree II. * insert $k$: Insert a node containing $k$ as key into $T$. * find $k$: Report whether $T$ has a node containing $k$. * delete $k$: Delete a node containing $k$. * print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively. The operation delete $k$ for deleting a given node $z$ containing key $k$ from $T$ can be implemented by an algorithm which considers the following cases: 1. If $z$ has no children, we modify its parent $z.p$ to replace $z$ with NIL as its child (delete $z$). 2. If $z$ has only a single child, we "splice out" $z$ by making a new link between its child and its parent. 3. If $z$ has two children, we splice out $z$'s successor $y$ and replace $z$'s key with $y$'s key.
|
class Tree:
def __init__(self):
self.root = None
def insert(self, key):
z = Node(key)
y = None
x = self.root
while x:
y = x
if z.key < x.key:
x = x.left
else:
x = x.right
z.parent = y
if y is None:
self.root = z
elif z.key < y.key:
y.left = z
else:
y.right = z
def find(self, key):
x = self.root
while x and key != x.key:
if key < x.key:
x = x.left
else:
x = x.right
return x
def getSuccessor(self, x):
if x.right is not None:
return self.getMinimum(x.right)
y = x.parent
while y and x == y.right:
x = y
y = y.parent
return y
def getMinimum(self, x):
while x.left is not None:
x = x.left
return x
def delete(self, z):
if z.left is None or z.right is None:
y = z
else:
y = getSuccessor(z)
# y??????x????????????
if y.left is not None:
x = y.left
else:
x = y.right
if x is not None:
x.parent = y.parent
if y.parent is None:
self.root = x
elif y == y.parent.left:
y.parent.left = x
else:
y.parent.right = x
if y != z:
z.key = y.key
def show(self):
print(" ", end="")
print(*list(map(str, self.root.inwalk())))
print(" ", end="")
print(*list(map(str, self.root.prewalk())))
class Node:
def __init__(self, key):
self.key = key
self.parent = self.left = self.right = None
def prewalk(self):
nodeList = [self.key]
if self.left:
nodeList += self.left.prewalk()
if self.right:
nodeList += self.right.prewalk()
return nodeList
def inwalk(self):
nodeList = []
if self.left:
nodeList += self.left.inwalk()
nodeList += [self.key]
if self.right:
nodeList += self.right.inwalk()
return nodeList
tree = Tree()
n = int(input())
for i in range(n):
cmd = list(input().split())
if cmd[0] == 'insert':
tree.insert(int(cmd[1]))
elif cmd[0] == 'find':
if tree.find(int(cmd[1])):
print("yes")
else:
print("no")
elif cmd[0] == 'print':
tree.show()
|
s242016225
|
Accepted
| 8,040
| 156,584
| 2,799
|
class Tree:
def __init__(self):
self.root = None
def insert(self, key):
z = Node(key)
y = None
x = self.root
while x:
y = x
if z.key < x.key:
x = x.left
else:
x = x.right
z.parent = y
if y is None:
self.root = z
elif z.key < y.key:
y.left = z
else:
y.right = z
def find(self, key):
x = self.root
while x and key != x.key:
if key < x.key:
x = x.left
else:
x = x.right
return x
def getSuccessor(self, x):
if x.right is not None:
return self.getMinimum(x.right)
y = x.parent
while y and x == y.right:
x = y
y = y.parent
return y
def getMinimum(self, x):
while x.left is not None:
x = x.left
return x
def delete(self, key):
z = self.find(key)
if z.left is None or z.right is None:
y = z
else:
y = self.getSuccessor(z)
# y??????x????????????
if y.left is not None:
x = y.left
else:
x = y.right
if x is not None:
x.parent = y.parent
if y.parent is None:
self.root = x
elif y == y.parent.left:
y.parent.left = x
else:
y.parent.right = x
if y != z:
z.key = y.key
def show(self):
print(" ", end="")
print(*list(map(str, self.root.inwalk())))
print(" ", end="")
print(*list(map(str, self.root.prewalk())))
class Node:
def __init__(self, key):
self.key = key
self.parent = self.left = self.right = None
def prewalk(self):
nodeList = [self.key]
if self.left:
nodeList += self.left.prewalk()
if self.right:
nodeList += self.right.prewalk()
return nodeList
def inwalk(self):
nodeList = []
if self.left:
nodeList += self.left.inwalk()
nodeList += [self.key]
if self.right:
nodeList += self.right.inwalk()
return nodeList
tree = Tree()
n = int(input())
for i in range(n):
cmd = list(input().split())
if cmd[0] == 'insert':
tree.insert(int(cmd[1]))
elif cmd[0] == 'find':
if tree.find(int(cmd[1])):
print("yes")
else:
print("no")
elif cmd[0] == 'print':
tree.show()
elif cmd[0] == 'delete':
tree.delete(int(cmd[1]))
|
s271894920
|
p03455
|
u021548497
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 67
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
print("Odd" if a*b//2 else "Even")
|
s328530289
|
Accepted
| 17
| 2,940
| 81
|
a, b = map(int, input().split())
if (a*b)%2:
print("Odd")
else:
print("Even")
|
s734789980
|
p03493
|
u257856080
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 324
|
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.
|
n=int(input())
count = 0
if n == 0:
print(0)
elif n == 1:
print(1)
elif n > 10 and n <100:
count = count + 1
n = n%10
if n == 1:
count = count + 1
print(count)
elif n > 100:
count = count + 1
n = n%100
if n > 10:
count = count + 1
n = n%10
if n == 1:
count = count + 1
print(count)
|
s785208890
|
Accepted
| 17
| 3,064
| 489
|
def ten_decision(n,count):
tmp = count
if n >= 10 and n<100:
count = count + 1
n = n%10
if n == 1:
count = count + 1
if tmp == count:
return 0
return count
n=int(input())
count = 0
if n == 0:
print(0)
elif n == 1:
print(1)
elif ten_decision(n,count):
print(ten_decision(n,count))
elif n >= 100:
count = count + 1
n = n%100
if ten_decision(n,count):
print(ten_decision(n,count))
else :
if n ==1:
count = count + 1
print(count)
|
s886369104
|
p03671
|
u306142032
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 113
|
Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately. He has very high awareness of safety, and decides to buy two bells, one for each hand. The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the minimum total price of two different bells.
|
a, b, c = map(int, input().split())
p = []
p.append(a+b)
p.append(b+c)
p.append(c+a)
p.sort()
print(p[0] + p[1])
|
s523113116
|
Accepted
| 17
| 2,940
| 106
|
a, b, c = map(int, input().split())
p = []
p.append(a+b)
p.append(b+c)
p.append(c+a)
p.sort()
print(p[0])
|
s150274588
|
p03385
|
u740047492
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 61
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
a = input()
if "abc" in a:
print("Yes")
else:
print("No")
|
s799619880
|
Accepted
| 17
| 2,940
| 85
|
a = input()
if "a" in a and "b" in a and "c" in a:
print("Yes")
else:
print("No")
|
s196408276
|
p03047
|
u496762077
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 44
|
Snuke has N integers: 1,2,\ldots,N. He will choose K of them and give those to Takahashi. How many ways are there to choose K consecutive integers?
|
k,n = map(int, input().split())
print(n-k+1)
|
s009431322
|
Accepted
| 48
| 2,940
| 44
|
n,k = map(int, input().split())
print(n-k+1)
|
s609730305
|
p03795
|
u560988566
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 49
|
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())
x = n * 800
y = n//15
print(x-y)
|
s163347074
|
Accepted
| 19
| 2,940
| 71
|
n = int(input())
x = n * 800
y = (n // 15) * 200
ans = x - y
print(ans)
|
s800841068
|
p02407
|
u811841526
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,396
| 48
|
Write a program which reads a sequence and prints it in the reverse order.
|
input()
xs = input().split()
print(reversed(xs))
|
s236072505
|
Accepted
| 20
| 5,540
| 62
|
n = input()
l = input().split()
print(' '.join(reversed(l)))
|
s132066915
|
p03477
|
u131881594
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,136
| 111
|
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.
|
a,b,c,d=map(int,input().split())
if a+b>c+d: print("left")
elif a+b<c+d: print("right")
else: print("balanced")
|
s524530497
|
Accepted
| 29
| 9,184
| 112
|
a,b,c,d=map(int,input().split())
if a+b>c+d: print("Left")
elif a+b<c+d: print("Right")
else: print("Balanced")
|
s477777209
|
p03377
|
u133936772
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,088
| 62
|
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=map(int,input().split())
print('YNeos'[x<a or x>a+b::2])
|
s719802701
|
Accepted
| 29
| 9,060
| 61
|
a,b,x=map(int,input().split())
print(['NO','YES'][a<=x<=a+b])
|
s297824103
|
p03456
|
u202877219
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 139
|
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
import math
x,y = map(str, input().split())
a = x+y
y = math.sqrt(int(a))
print (a,y)
if a == y:
print ("Yes")
else:
print ("No")
|
s970633650
|
Accepted
| 17
| 2,940
| 122
|
import math
x,y = map(str, input().split())
a = x+y
if math.sqrt(int(a)) % 1 == 0:
print("Yes")
else:
print("No")
|
s924297868
|
p03959
|
u648212584
| 2,000
| 262,144
|
Wrong Answer
| 272
| 17,120
| 1,256
|
Mountaineers Mr. Takahashi and Mr. Aoki recently trekked across a certain famous mountain range. The mountain range consists of N mountains, extending from west to east in a straight line as Mt. 1, Mt. 2, ..., Mt. N. Mr. Takahashi traversed the range from the west and Mr. Aoki from the east. The height of Mt. i is h_i, but they have forgotten the value of each h_i. Instead, for each i (1 ≤ i ≤ N), they recorded the maximum height of the mountains climbed up to the time they reached the peak of Mt. i (including Mt. i). Mr. Takahashi's record is T_i and Mr. Aoki's record is A_i. We know that the height of each mountain h_i is a positive integer. Compute the number of the possible sequences of the mountains' heights, modulo 10^9 + 7. Note that the records may be incorrect and thus there may be no possible sequence of the mountains' heights. In such a case, output 0.
|
import sys
input = sys.stdin.buffer.readline
from operator import itemgetter
def main():
N = int(input())
t = list(map(int,input().split()))
a = list(map(int,input().split()))
MOD = 10**9+7
tt,tl = 0,[False for _ in range(N)]
for x,num in enumerate(t):
if num > tt:
tl[x] = True
tt = num
at,al = 0,[False for _ in range(N)]
for x,num in enumerate(a[::-1]):
if num > at:
al[-x-1] = True
at = num
print(tl,al)
ans,now = 1,0
for i in range(N):
print(i,now)
if tl[i] == True and al[i] == True:
if now < t[i] == a[i]:
now = min(t[i],a[i])
else:
print(-1)
exit()
elif tl[i] == True and al[i] == False:
if now > t[i]:
print(-1)
exit()
else:
now = t[i]
elif tl[i] == False and al[i] == True:
if now < a[i]:
print(-1)
exit()
else:
now = a[i]
else:
ans *= min(t[i],a[i])
ans %= MOD
print(ans)
if __name__ == "__main__":
main()
|
s628527614
|
Accepted
| 140
| 17,120
| 1,195
|
import sys
input = sys.stdin.buffer.readline
from operator import itemgetter
def main():
N = int(input())
t = list(map(int,input().split()))
a = list(map(int,input().split()))
MOD = 10**9+7
tt,tl = 0,[False for _ in range(N)]
for x,num in enumerate(t):
if num > tt:
tl[x] = True
tt = num
at,al = 0,[False for _ in range(N)]
for x,num in enumerate(a[::-1]):
if num > at:
al[-x-1] = True
at = num
ans,now = 1,0
for i in range(N):
if tl[i] == True and al[i] == True:
if now < t[i] == a[i]:
now = a[i]
else:
print(0)
exit()
elif tl[i] == True and al[i] == False:
if t[i] > a[i]:
print(0)
exit()
else:
now = t[i]
elif tl[i] == False and al[i] == True:
if t[i] < a[i]:
print(0)
exit()
else:
now = a[i]
else:
ans *= min(t[i],a[i])
ans %= MOD
print(ans)
if __name__ == "__main__":
main()
|
s016754098
|
p03502
|
u143492911
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 121
|
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number.
|
n=input()
x=int(n)
y=0
for i in n:
y+=int(i)
print(y)
print(x)
if x%y==0:
print("Yes")
else:
print("No")
|
s611522005
|
Accepted
| 18
| 3,060
| 152
|
n=list(map(int,input()))
moji=""
ans=0
for i in n:
moji+=str(i)
ans+=i
moji_i=int(moji)
if moji_i%ans==0:
print("Yes")
else:
print("No")
|
s225555262
|
p03140
|
u325264482
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 281
|
You are given three strings A, B and C. Each of these is a string of length N consisting of lowercase English letters. Our objective is to make all these three strings equal. For that, you can repeatedly perform the following operation: * Operation: Choose one of the strings A, B and C, and specify an integer i between 1 and N (inclusive). Change the i-th character from the beginning of the chosen string to some other lowercase English letter. What is the minimum number of operations required to achieve the objective?
|
N = int(input())
A = input()
B = input()
C = input()
cntAB = N
cntBC = N
cntCA = N
for i in range(N):
if A[i] == B[i]:
cntAB -= 1
if B[i] == C[i]:
cntBC -= 1
if C[i] == A[i]:
cntCA -= 1
cnt = [cntAB, cntBC, cntCA]
print(sum(cnt) - max(cnt))
|
s265250111
|
Accepted
| 17
| 3,060
| 363
|
N = int(input())
A = input()
B = input()
C = input()
ans = 0
for i in range(N):
if A[i] == B[i]:
if B[i] == C[i]:
ans += 0
else:
ans += 1
else:
if B[i] == C[i]:
ans += 1
else:
if C[i] != A[i]:
ans += 2
else:
ans += 1
print(ans)
|
s915328227
|
p00003
|
u842823276
| 1,000
| 131,072
|
Wrong Answer
| 40
| 5,600
| 249
|
Write a program which judges wheather given length of three side form a right triangle. Print "YES" if the given sides (integers) form a right triangle, "NO" if not so.
|
def ifTriangle(a, b, c):
if a**2 + b**2 == c: print("YES")
else: print("NO")
n=int(input())
for i in range(n):
lines=list(map(int, input().split()))
lines.sort()
ifTriangle(lines[0], lines[1], lines[2])
#ifTriangle(a, b, c)
|
s360977183
|
Accepted
| 40
| 5,600
| 230
|
def ifTriangle(a, b, c):
if (a**2) + (b**2) == c**2: print("YES")
else: print("NO")
n=int(input())
for i in range(n):
lines=list(map(int, input().split()))
lines.sort()
ifTriangle(lines[0], lines[1], lines[2])
|
s663862085
|
p03449
|
u268279636
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,060
| 195
|
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j). You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down. The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them. At most how many candies can you collect when you choose the best way to travel?
|
N=int(input())
l1=input().split(' ')
l2=input().split(' ')
max=0
for i in range(N):
sum=0
for j in range(i+1):
sum+=int(l1[j])
for k in range(N-i):
sum+=int(l2[k])
if max<sum:
max=sum
|
s964000583
|
Accepted
| 20
| 3,064
| 207
|
N=int(input())
l1=input().split(' ')
l2=input().split(' ')
max=0
for i in range(N):
sum=0
for j in range(i+1):
sum+=int(l1[j])
for k in range(i,N):
sum+=int(l2[k])
if max<sum:
max=sum
print(max)
|
s360183056
|
p03434
|
u627417051
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 172
|
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()))
alice = 0
bob = 0
A.sort()
for i in range(N):
if i % 2 ==0:
alice += A[-i]
else:
bob += A[-i]
print(alice - bob)
|
s643107955
|
Accepted
| 17
| 3,060
| 180
|
N = int(input())
A = list(map(int, input().split()))
alice = 0
bob = 0
A.sort()
for i in range(N):
if i % 2 ==0:
alice += A[-i - 1]
else:
bob += A[-i - 1]
print(alice - bob)
|
s180946466
|
p02402
|
u731896389
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,532
| 77
|
Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.
|
input()
a = list(map(int,input().split()))
print(min(a),max(a),sum(a)/len(a))
|
s455522516
|
Accepted
| 20
| 8,568
| 70
|
input()
a = list(map(int,input().split()))
print(min(a),max(a),sum(a))
|
s138115840
|
p02646
|
u932370518
| 2,000
| 1,048,576
|
Wrong Answer
| 25
| 9,192
| 184
|
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
a, v = [int(a) for a in input().split()]
b, w = [int(a) for a in input().split()]
t = int(input())
ret = w-v
ans = ""
if (ret*t) <= (b-a):
ans = "NO"
else:
ans = "YES"
print(ans)
|
s095928534
|
Accepted
| 21
| 9,180
| 214
|
a, v = [int(a) for a in input().split()]
b, w = [int(a) for a in input().split()]
t = int(input())
distance = abs(b-a)
ds = v - w
ans = ""
if (ds * t) < (distance):
ans = "NO"
else:
ans = "YES"
print(ans)
|
s556653360
|
p02407
|
u150984829
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,536
| 39
|
Write a program which reads a sequence and prints it in the reverse order.
|
input()
print(*sorted(input().split()))
|
s176023963
|
Accepted
| 20
| 5,556
| 37
|
input()
print(*input().split()[::-1])
|
s663680843
|
p02401
|
u019678978
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,660
| 292
|
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.
|
alist = [ x for x in input().split() ]
if alist[1] == "*" :
print(int(alist[0]) * int(alist[2]))
elif alist[1] == "/" :
print(int(alist[0]) / int(alist[2]))
elif alist[1] == "+" :
print(int(alist[0]) + int(alist[2]))
elif alist[1] == "-" :
print(int(alist[0]) - int(alist[2]))
|
s704876789
|
Accepted
| 30
| 7,804
| 406
|
import math
while True :
alist = [ x for x in input().split() ]
if alist[1] == "?" :
break
elif alist[1] == "*" :
print(int(alist[0]) * int(alist[2]))
elif alist[1] == "/" :
print(math.floor(int(alist[0]) / int(alist[2])))
elif alist[1] == "+" :
print(int(alist[0]) + int(alist[2]))
elif alist[1] == "-" :
print(int(alist[0]) - int(alist[2]))
|
s157421360
|
p02645
|
u695474809
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,560
| 175
|
When you asked some guy in your class his name, he called himself S, where S is a string of length between 3 and 20 (inclusive) consisting of lowercase English letters. You have decided to choose some three consecutive characters from S and make it his nickname. Print a string that is a valid nickname for him.
|
import random
name = list(input())
print(name[random.randint(0,len(name)-1)],end="")
print(name[random.randint(0,len(name)-1)],end="")
print(name[random.randint(0,len(name))])
|
s207956792
|
Accepted
| 29
| 8,848
| 33
|
S = input()
print(S[0]+S[1]+S[2])
|
s546836700
|
p03024
|
u902576227
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 103
|
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
s=input()
cnt=0
for i in range(len(s)):
if s[i] == 'o':
cnt+=1
print("YES" if cnt >= 8 else "NO")
|
s366027745
|
Accepted
| 17
| 2,940
| 116
|
s=input()
cnt=0
for i in range(len(s)):
if s[i] == 'o':
cnt+=1
print("YES" if cnt >= 8-(15-len(s)) else "NO")
|
s506450434
|
p02261
|
u612243550
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,668
| 928
|
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).
|
def select(S):
for i in range(0, n):
minj = i
for j in range(i, n):
if int(S[j][1]) < int(S[minj][1]):
minj = j
(S[i], S[minj]) = (S[minj], S[i])
return S
def bubble(B):
flag = True
while flag:
flag = False
for j in reversed(range(1, n)):
if int(B[j][1]) < int(B[j-1][1]):
(B[j-1], B[j]) = (B[j], B[j-1])
flag = True
return B
def isStable(inA, out):
for i in range(0, n):
for j in range(i+1, n):
for a in range(0, n):
for b in range(a+1, n):
if inA[i][1] == inA[j][1] and inA[i] == out[b] and inA[j] == out[a]:
return "Not Stable"
return "Stable"
n = int(input())
A = input().split(' ')
B = bubble(A[:])
print(" ".join(B))
print(isStable(A, B))
S = select(A[:])
print(" ".join(S))
print(isStable(A, S))
|
s453678351
|
Accepted
| 130
| 7,764
| 928
|
def select(S):
for i in range(0, n):
minj = i
for j in range(i, n):
if int(S[j][1]) < int(S[minj][1]):
minj = j
(S[i], S[minj]) = (S[minj], S[i])
return S
def bubble(B):
flag = True
while flag:
flag = False
for j in reversed(range(1, n)):
if int(B[j][1]) < int(B[j-1][1]):
(B[j-1], B[j]) = (B[j], B[j-1])
flag = True
return B
def isStable(inA, out):
for i in range(0, n):
for j in range(i+1, n):
for a in range(0, n):
for b in range(a+1, n):
if inA[i][1] == inA[j][1] and inA[i] == out[b] and inA[j] == out[a]:
return "Not stable"
return "Stable"
n = int(input())
A = input().split(' ')
B = bubble(A[:])
print(" ".join(B))
print(isStable(A, B))
S = select(A[:])
print(" ".join(S))
print(isStable(A, S))
|
s966935341
|
p03359
|
u119578112
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 93
|
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?
|
N = list(map(int, input().split()))
if N[0] < N[1]:
print(N[0])
else :
print(N[0]-1)
|
s794315793
|
Accepted
| 17
| 2,940
| 94
|
N = list(map(int, input().split()))
if N[0] <= N[1]:
print(N[0])
else :
print(N[0]-1)
|
s476070840
|
p03643
|
u723721005
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 43
|
This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer.
|
str=input();
print ("ABC", str)
print('\n')
|
s735832300
|
Accepted
| 17
| 2,940
| 23
|
print("ABC" + input())
|
s890802845
|
p03589
|
u646352133
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 124
|
You are given an integer N. Find a triple of positive integers h, n and w such that 4/N = 1/h + 1/n + 1/w. If there are multiple solutions, any of them will be accepted.
|
n = int(input())
if n % 4 == 0:
out = n/4*3
print(out,out,out)
elif n % 2 == 0:
print(n/2,n,n)
else:
pass
|
s284178143
|
Accepted
| 1,968
| 3,064
| 280
|
N = int(input())
for a in range(1,3501):
for b in range(1,3501):
if (4*a*b - a*N - b*N) > 0:
c = (a*b*N) // (4*a*b - a*N - b*N)
else:
continue
if c != 0 and 4/N == (1/a + 1/b + 1/c):
print(a,b,c)
exit()
|
s496915076
|
p02694
|
u012955130
| 2,000
| 1,048,576
|
Wrong Answer
| 22
| 9,164
| 115
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
X = int(input())
_sum = 100
year = 0
while _sum <= X:
_sum = int(_sum * 1.01)
year += 1
print(int(year))
|
s617432525
|
Accepted
| 21
| 9,168
| 162
|
import math
X = int(input())
_sum = 100
year = 0
while True:
if _sum >= X :
break
_sum = math.floor(_sum * 1.01)
year += 1
print(int(year))
|
s319258104
|
p03504
|
u375616706
| 2,000
| 262,144
|
Wrong Answer
| 386
| 27,400
| 256
|
Joisino is planning to record N TV programs with recorders. The TV can receive C channels numbered 1 through C. The i-th program that she wants to record will be broadcast from time s_i to time t_i (including time s_i but not t_i) on Channel c_i. Here, there will never be more than one program that are broadcast on the same channel at the same time. When the recorder is recording a channel from time S to time T (including time S but not T), it cannot record other channels from time S-0.5 to time T (including time S-0.5 but not T). Find the minimum number of recorders required to record the channels so that all the N programs are completely recorded.
|
N, C = list(map(int, input().split()))
prog = []
for _ in range(N):
prog.append((list)(map(int, input().split())))
T = 10**5
for i in range(T):
a = 0
for i in range(T):
a = 1
for i in range(T):
a = 2
for i in range(T):
a = 6
print(a)
|
s537718256
|
Accepted
| 1,600
| 77,904
| 670
|
from itertools import accumulate
# -*- coding: utf-8 -*-
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline
N, C = map(int, input().split())
TV = dict()
programs = [list(map(int, input().split())) for _ in range(N)]
last_time = max([p[1] for p in programs])
for i in range(1, C+1):
TV[i] = [0]*(last_time*2+2)
for s, t, c in programs:
s *= 2
t *= 2
TV[c][s] += 1
TV[c][t+1] -= 1
for c in range(1, C+1):
TV[c] = list(accumulate(TV[c]))
ans = 0
for i in range(last_time*2+1):
tmp = 0
for c in range(1, C+1):
if TV[c][i] > 0:
tmp += 1
ans = max(ans, tmp)
print(ans)
|
s764745563
|
p02271
|
u022407960
| 5,000
| 131,072
|
Wrong Answer
| 60
| 7,776
| 713
|
Write a program which reads a sequence _A_ of _n_ elements and an integer _M_ , and outputs "yes" if you can make _M_ by adding elements in _A_ , otherwise "no". You can use an element only once. You are given the sequence _A_ and _q_ questions where each question contains _M i_.
|
# encoding: utf-8
import sys
__input = sys.stdin.readlines()
length_1 = int(__input[0])
array_1 = list(map(int, __input[1].split()))
length_2 = int(__input[2])
array_2 = list(map(int, __input[3].split()))
def solution():
print(length_1, array_1, length_2, array_2)
assert length_1 == len(array_1) and length_2 == len(array_2)
for i in range(length_2):
ans = solve(0, array_2[i])
if ans:
print("yes")
else:
print("no")
return None
def solve(i, m):
if m == 0:
return True
if i >= length_1:
return False
res = solve(i + 1, m) or solve(i + 1, m - array_1[i])
return res
if __name__ == '__main__':
solution()
|
s480602819
|
Accepted
| 20
| 7,768
| 738
|
# encoding: utf-8
import sys
__input = sys.stdin.readlines()
length_1 = int(__input[0])
array_1 = list(map(int, __input[1].split()))
length_2 = int(__input[2])
array_2 = list(map(int, __input[3].split()))
record = [False] * 2000
def solution():
assert length_1 == len(array_1) and length_2 == len(array_2)
for i in array_1:
# record[i] = True
for j in range(2000 - i, 0, -1):
if record[j]:
record[i + j] = True
record[i] = True
for index in array_2:
if record[index]:
print("yes")
else:
print("no")
if __name__ == '__main__':
solution()
|
s555369105
|
p02747
|
u159994501
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 72
|
A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string.
|
s = input()
s.replace('hi', '')
if s:
print('No')
else:
print('Yes')
|
s875787102
|
Accepted
| 19
| 2,940
| 77
|
s = input()
s = s.replace('hi', '')
if s:
print('No')
else:
print('Yes')
|
s013197757
|
p02613
|
u765776018
| 2,000
| 1,048,576
|
Wrong Answer
| 152
| 9,212
| 300
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n = int(input())
k = {}
k['AC\r'] =0
k["TLE\r"]=0
k["WA\r"]=0
k["RE\r"]=0
for i in range(n):
a = str(input())
#print(a)
if a in k:
k[a]+=1
print("AC x {}".format(k['AC\r']))
print("WA x {}".format(k['WA\r']))
print("TLE x {}".format(k['TLE\r']))
print("RE x {}".format(k['RE\r']))
|
s526222708
|
Accepted
| 159
| 9,188
| 284
|
n = int(input())
k = {}
k['AC'] =0
k["TLE"]=0
k["WA"]=0
k["RE"]=0
for i in range(n):
a = str(input())
#print(a)
if a in k:
k[a]+=1
print("AC x {}".format(k['AC']))
print("WA x {}".format(k['WA']))
print("TLE x {}".format(k['TLE']))
print("RE x {}".format(k['RE']))
|
s259236166
|
p03557
|
u411858517
| 2,000
| 262,144
|
Wrong Answer
| 2,105
| 23,360
| 1,052
|
The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower. He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i. To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar. How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.
|
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
C = list(map(int, input().split()))
A.sort()
C.sort()
res = 0
for i in range(N):
low = 0
high = N - 1
t1 = (low + high) // 2
while (low <= high):
if ( A[t1-1] <= B[i] < A[t1]):
break
elif (B[i] > A[t1]):
low = t1 + 1
elif (B[i] < A[t1]):
high = t1- 1
elif (t1 == N-1):
break
t1 = (low + high) // 2
low = 0
high = N - 1
t2 = (low + high) // 2
while (low <= high):
if ( C[t2-1] <= B[i] < C[t2] ):
break
elif (B[i] > C[t2]):
low = t2 + 1
elif (B[i] < C[t2]):
high = t2 - 1
elif (t2 == 0):
break
t2 = (low + high) // 2
res += (t1+1) * (N - t2 - 1)
print(res, t1, t2)
|
s015877798
|
Accepted
| 327
| 23,328
| 272
|
import bisect
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
C = list(map(int, input().split()))
A.sort()
B.sort()
C.sort()
res = 0
for b in B:
res += bisect.bisect_left(A, b) * (N - bisect.bisect_right(C, b))
print(res)
|
s666787404
|
p03965
|
u415905784
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,188
| 53
|
AtCoDeer the deer and his friend TopCoDeer is playing a game. The game consists of N turns. In each turn, each player plays one of the two _gestures_ , _Rock_ and _Paper_ , as in Rock-paper-scissors, under the following condition: (※) After each turn, (the number of times the player has played Paper)≦(the number of times the player has played Rock). Each player's score is calculated by (the number of turns where the player wins) - (the number of turns where the player loses), where the outcome of each turn is determined by the rules of Rock-paper-scissors. _(For those who are not familiar with Rock-paper-scissors: If one player plays Rock and the other plays Paper, the latter player will win and the former player will lose. If both players play the same gesture, the round is a tie and neither player will win nor lose.)_ With his supernatural power, AtCoDeer was able to foresee the gesture that TopCoDeer will play in each of the N turns, before the game starts. Plan AtCoDeer's gesture in each turn to maximize AtCoDeer's score. The gesture that TopCoDeer will play in each turn is given by a string s. If the i-th (1≦i≦N) character in s is `g`, TopCoDeer will play Rock in the i-th turn. Similarly, if the i-th (1≦i≦N) character of s in `p`, TopCoDeer will play Paper in the i-th turn.
|
s = input()
print((s.count('p') - s.count('g')) // 2)
|
s919008049
|
Accepted
| 18
| 3,188
| 53
|
s = input()
print((s.count('g') - s.count('p')) // 2)
|
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