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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s097240592
|
p02694
|
u464912173
| 2,000
| 1,048,576
|
Wrong Answer
| 20
| 9,160
| 67
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
x = int(input())
a=100
b=0
while a>x:
a +=a//100
b+=1
print(b)
|
s586509448
|
Accepted
| 22
| 9,156
| 72
|
x = int(input())
a=100
b=0
while a<x:
a =int(a*1.01)
b+=1
print(b)
|
s104152947
|
p03369
|
u013605408
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 29
|
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.
|
print(700+input().count("o"))
|
s576964705
|
Accepted
| 17
| 2,940
| 35
|
print(700+(input().count("o")*100))
|
s452905825
|
p03450
|
u785578220
| 2,000
| 262,144
|
Wrong Answer
| 2,108
| 9,012
| 1,785
|
There are N people standing on the x-axis. Let the coordinate of Person i be x_i. For every i, x_i is an integer between 0 and 10^9 (inclusive). It is possible that more than one person is standing at the same coordinate. You will given M pieces of information regarding the positions of these people. The i-th piece of information has the form (L_i, R_i, D_i). This means that Person R_i is to the right of Person L_i by D_i units of distance, that is, x_{R_i} - x_{L_i} = D_i holds. It turns out that some of these M pieces of information may be incorrect. Determine if there exists a set of values (x_1, x_2, ..., x_N) that is consistent with the given pieces of information.
|
class WeightedUnionFind:
def __init__(self, n):
self.par = [i for i in range(n+1)]
self.rank = [0] * (n+1)
self.weight = [0] * (n+1)
def find(self, x):
if self.par[x] == x:
return x
else:
y = self.find(self.par[x])
self.weight[x] += self.weight[self.par[x]]
self.par[x] = y
return y
def union(self, x, y, w):
rx = self.find(x)
ry = self.find(y)
if self.rank[rx] < self.rank[ry]:
self.par[rx] = ry
self.weight[rx] = w - self.weight[x] + self.weight[y]
else:
self.par[ry] = rx
self.weight[ry] = -w - self.weight[y] + self.weight[x]
if self.rank[rx] == self.rank[ry]:
self.rank[rx] += 1
def same(self, x, y):
return self.find(x) == self.find(y)
def diff(self, x, y):
return self.weight[x] - self.weight[y]
def main():
max_v, max_e = map(int, input().split())
uf = WeightedUnionFind(max_v)
k = []
s = 0
for i in range(max_e):
a,b,w= map(int, input().split())
if a not in k:
k.append(a)
if b not in k:
k.append(b)
if a in k and b in k:
uf.diff(a,b) == w
s =1
else:uf.union(a,b,w)
if s == 0:
print("Yes")
else:
print("No")
if __name__ == '__main__':
main()
|
s156701291
|
Accepted
| 737
| 8,948
| 1,164
|
def main():
import sys
input = sys.stdin.readline
def find(x):
if par[x] < 0:
return x
else:
px = find(par[x])
wei[x] += wei[par[x]]
par[x] = px
return px
def weight(x):
find(x)
return wei[x]
def unite(x,y,w):
w += wei[x]-wei[y]
x = find(x)
y = find(y)
if x == y:
return False
else:
if par[x] > par[y]:
x,y = y,x
w = -w
par[x] += par[y]
par[y] = x
wei[y] = w
return True
def same(x,y):
return find(x) == find(y)
def size(x):
return -par[find(x)]
def diff(x,y):
return weight(y)-weight(x)
n,m = map(int,input().split())
par = [-1]*n
wei = [0]*n
for i in range(m):
l,r,d = map(int,input().split())
l,r = l-1,r-1
if same(l,r):
if d != diff(l,r):
print('No')
exit()
else:
unite(l,r,d)
print('Yes')
if __name__ == '__main__':
main()
|
s793780446
|
p02936
|
u918601425
| 2,000
| 1,048,576
|
Wrong Answer
| 2,110
| 110,256
| 2,012
|
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
N,Q=[int(s) for s in input().split()]
e_list=[[int(s) for s in input().split()] for i in range(N-1)]
q_list=[[int(s) for s in input().split()] for i in range(Q)]
def euler_tour(edge_list,start):
reach=[[start,1,0]]
ans=[start]
path=[start]
e_list=[[e[0],e[1],0] for e in edge_list]
vertex=start
endflag=0
depth=1
while True :
for e in e_list:
if (e[0]==vertex and e[2]==0):
reach.append([e[1],0,depth])
e[2]=1
elif (e[1]==vertex and e[2]==0):
reach.append([e[0],0,depth])
e[2]=1
for i in range(len(reach)+1):
if reach[-i][1]==0 :
vertex=reach[-i][0]
if reach[-i][2]<depth:
for j in range(1,depth-reach[-i][2]+1):
path.remove(path[-1])
ans.append(path[-1])
ans.append(vertex)
path.append(vertex)
depth=reach[-i][2]+1
reach[-i][1]=1
break
if i==len(reach):
for j in range(len(path)-1):
path.remove(path[-1])
ans.append(path[-1])
return ans
break
euler=euler_tour(e_list,1)
def eu(num):
if num==1:
return [i+1 for i in range(N)]
ans=[]
i=-1
while True :
i+=1
if euler[i]==num:
ii=euler[i-1]
break
while True:
if euler[i]==ii:
break
ans.append(euler[i])
i+=1
return list(set(ans))
answer=[0 for i in range(N)]
for q in q_list:
for v in eu(q[0]):
answer[v-1]+=q[1]
print(answer)
|
s212881083
|
Accepted
| 1,860
| 68,952
| 553
|
n,q=[int(s) for s in input().split()]
graph=[[] for _ in range(n)]
for i in range(n-1):
a,b=[int(s) for s in input().split()]
a-=1
b-=1
graph[a].append(b)
graph[b].append(a)
ct=[0 for _ in range(n)]
for i in range(q):
p,x=[int(s) for s in input().split()]
p-=1
ct[p]+=x
stack=[(0, ct[0], -1)]
ans=[0 for _ in range(n)]
while stack:
vrt, cost, _from = stack.pop(-1)
ans[vrt] = cost
for _to in graph[vrt]:
if _to ==_from:
continue
stack.append((_to, cost+ct[_to],vrt))
print(*ans)
|
s083625426
|
p03611
|
u652150585
| 2,000
| 262,144
|
Wrong Answer
| 131
| 16,096
| 331
|
You are given an integer sequence of length N, a_1,a_2,...,a_N. For each 1≤i≤N, you have three choices: add 1 to a_i, subtract 1 from a_i or do nothing. After these operations, you select an integer X and count the number of i such that a_i=X. Maximize this count by making optimal choices.
|
import collections
n=int(input())
l=list(map(int,input().split()))
s=collections.Counter(l)
s=s.most_common()
#print(s)
f=[0]*len(s)
for i in range(len(s)):
f[i]=s[i][1]
if s[i-1][0]==s[i][0]-1 and i>=1:
f[i]+=s[i-1][1]
if i<=len(s)-2:
if s[i+1][0]==s[i][0]+1:
f[i]+=s[i+1][1]
print(max(f))
|
s817942745
|
Accepted
| 150
| 16,096
| 373
|
import collections
n=int(input())
l=list(map(int,input().split()))
s=collections.Counter(l)
s=list(s.items())
s=sorted(s,key=lambda x:x[0])
#print(s)
f=[0]*len(s)
for i in range(len(s)):
f[i]=s[i][1]
if i>=1:
if s[i-1][0]==s[i][0]-1:
f[i]+=s[i-1][1]
if i<=len(s)-2:
if s[i+1][0]==s[i][0]+1:
f[i]+=s[i+1][1]
print(max(f))
|
s466386587
|
p03024
|
u675757825
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 113
|
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 s:
if (i == 'o'):
cnt += 1
print('Yes' if len(s) - cnt <= 7 else 'No')
|
s241931892
|
Accepted
| 17
| 2,940
| 113
|
s = input()
cnt = 0
for i in s:
if (i == 'o'):
cnt += 1
print('YES' if len(s) - cnt <= 7 else 'NO')
|
s286123341
|
p03854
|
u457554982
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,956
| 385
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
s=list(input())
s.reverse()
judge=True
i=0
while i<=len(s)-5:
box=[]
for j in range(5):
box+=s[i+j]
if box==["m","a","e","r","d"] or box==["e","s","a","r","e"]:
print(box)
i+=5
else:
box+=s[i+5]+s[i+6]
print(box)
if box==["r","e","m","a","e","r","d"] or box==["r","e","s","a","r","e"]:
i+=7
else:
judge=False
break
if judge:
print("YES")
else:
print("NO")
|
s076494926
|
Accepted
| 68
| 3,956
| 419
|
s=list(input())
s.reverse()
judge=True
i=0
while i<=len(s)-5:
box=[]
for j in range(5):
box+=s[i+j]
if box==["m","a","e","r","d"] or box==["e","s","a","r","e"]:
i+=5
#print(box)
else:
box+=s[i+5]
#print(box)
if box==["r","e","s","a","r","e"]:
i+=6
else:
box+=s[i+6]
if box==["r","e","m","a","e","r","d"]:
i+=7
else:
judge=False
break
if judge:
print("YES")
else:
print("NO")
|
s566803198
|
p03778
|
u399721252
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 74
|
AtCoDeer the deer found two rectangles lying on the table, each with height 1 and width W. If we consider the surface of the desk as a two-dimensional plane, the first rectangle covers the vertical range of AtCoDeer will move the second rectangle horizontally so that it connects with the first rectangle. Find the minimum distance it needs to be moved.
|
w, a, b = [ int(v) for v in input().split() ]
print(max(abs(a-b) - 5, 0))
|
s893015695
|
Accepted
| 17
| 2,940
| 74
|
w, a, b = [ int(v) for v in input().split() ]
print(max(abs(a-b) - w, 0))
|
s840307198
|
p03386
|
u239342230
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 3,060
| 95
|
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
A,B,K=map(int,input().split())
for i in range(A,B+1):
if i<=A+K or i>=B-K:
print(i)
|
s850849283
|
Accepted
| 17
| 3,060
| 106
|
A,B,K=map(int,input().split());r=range(A,B+1)
for i in sorted(set(list(r[:K])+list(r[-K:]))):
print(i)
|
s854080803
|
p03861
|
u541610817
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 72
|
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
a, b, x = [int(x) for x in input().split()]
a //= x
b //= x
print(b - a)
|
s296980996
|
Accepted
| 17
| 2,940
| 88
|
a, b, x = [int(x) for x in input().split()]
a = (a + x - 1)//x
b //= x
print(b - a + 1)
|
s541068493
|
p02258
|
u424457654
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,568
| 173
|
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())
max = -1000000000
min = int(input())
for i in range(1, n):
r = int(input())
if max < (r - min):
max = r - min
if min > r:
max = min
print(max)
|
s263886615
|
Accepted
| 430
| 7,724
| 171
|
n = int(input())
max = -1000000000
min = int(input())
for i in range(1, n):
r = int(input())
if max < (r - min):
max = r - min
if min > r:
min = r
print(max)
|
s240485471
|
p03545
|
u632369368
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 422
|
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.
|
import itertools
A, B, C, D = [int(s) for s in list(input())]
print('A:{}, B:{}, C:{}, D:{}'.format(A, B, C, D))
symbols = ['+', '-']
for s1, s2, s3 in itertools.product(symbols, symbols, symbols):
N = 0
N = A + B if s1 == '+' else A - B
N = N + C if s2 == '+' else N - C
N = N + D if s3 == '+' else N - D
if N == 7:
print('{}{}{}{}{}{}{}={}'.format(A, s1, B, s2, C, s3, D, N))
break
|
s992710064
|
Accepted
| 17
| 3,064
| 371
|
import itertools
A, B, C, D = [int(s) for s in list(input())]
symbols = ['+', '-']
for s1, s2, s3 in itertools.product(symbols, symbols, symbols):
N = 0
N = A + B if s1 == '+' else A - B
N = N + C if s2 == '+' else N - C
N = N + D if s3 == '+' else N - D
if N == 7:
print('{}{}{}{}{}{}{}={}'.format(A, s1, B, s2, C, s3, D, N))
break
|
s013726093
|
p03730
|
u809670194
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 183
|
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
i = input().split()
a = int(i[0])
b = int(i[1])
c = int(i[2])
d =a*b
answer = "NO"
while a <= d:
if a%b==c:
answer = "YES"
break
a += a
print(answer)
|
s410813344
|
Accepted
| 17
| 2,940
| 174
|
i = input().split()
a = int(i[0])
b = int(i[1])
c = int(i[2])
answer = "NO"
count = a
for i in range(1,b+1):
if (i*a)%b==c:
answer = "YES"
print(answer)
|
s459087237
|
p04029
|
u369338402
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 67
|
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())
result=0
for i in range(1,n+1):
result+=i
print(i)
|
s529836879
|
Accepted
| 17
| 2,940
| 38
|
n=int(input())
print(int((n*(1+n))/2))
|
s551276470
|
p03386
|
u552822163
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 873
|
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
# coding=utf-8
def from_down(base, number, upper):
return_list = []
range_upper = min(upper+1, base+number)
for i in range(base, range_upper):
return_list.append(i)
return return_list
def from_up(base, number, down):
return_list = []
range_down = max(base-number+1, down)
for i in range(range_down, base+1):
return_list.append(i)
return return_list
def remove_double(object_list):
list_unique = []
for x in object_list:
if x not in list_unique:
list_unique.append(x)
return list_unique
if __name__ == '__main__':
A, B, K = map(int, input().split())
down_list = from_down(A, K, B)
upper_list = from_up(B, K, A)
print(down_list)
print(upper_list)
down_list.extend(upper_list)
print(down_list)
unique_list = remove_double(down_list)
print(unique_list)
|
s039348185
|
Accepted
| 18
| 3,064
| 843
|
# coding=utf-8
def from_down(base, number, upper):
return_list = []
range_upper = min(upper+1, base+number)
for i in range(base, range_upper):
return_list.append(i)
return return_list
def from_up(base, number, down):
return_list = []
range_down = max(base-number+1, down)
for i in range(range_down, base+1):
return_list.append(i)
return return_list
def remove_double(object_list):
list_unique = []
for x in object_list:
if x not in list_unique:
list_unique.append(x)
return list_unique
if __name__ == '__main__':
A, B, K = map(int, input().split())
down_list = from_down(A, K, B)
upper_list = from_up(B, K, A)
down_list.extend(upper_list)
unique_list = remove_double(down_list)
output = [print(element) for element in unique_list]
|
s309716532
|
p03814
|
u095969144
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,500
| 56
|
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
s = input()
a = s.find("a")
z = s.find("z")
print(z - a)
|
s184331512
|
Accepted
| 17
| 3,500
| 62
|
s = input()
a = s.find("A")
b = s.rfind("Z")
print(b - a + 1)
|
s700865459
|
p04044
|
u399721252
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 118
|
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
n, l = [ v for v in input().split() ]
string_list = sorted([ v for v in input().split() ])
print("".join(string_list))
|
s954455106
|
Accepted
| 17
| 3,060
| 122
|
n, l = [ int(v) for v in input().split() ]
string_list = sorted([ input() for v in range(n) ])
print("".join(string_list))
|
s840198232
|
p03737
|
u612975321
| 2,000
| 262,144
|
Wrong Answer
| 23
| 9,028
| 66
|
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
|
a, b, c = input().split()
print(a.upper() + b.upper() + c.upper())
|
s610154586
|
Accepted
| 24
| 9,088
| 75
|
a, b, c = input().split()
print(a[0].upper() + b[0].upper() + c[0].upper())
|
s187995638
|
p03971
|
u921773161
| 2,000
| 262,144
|
Wrong Answer
| 112
| 4,708
| 474
|
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
|
#%%
n, a, b = map(int, input().split())
s = list(input())
c = 0
d = 0
for i in range(n):
if s[i] == 'a':
if c < a+b:
c += 1
print('Yes')
else:
print('no')
elif s[i] == 'b':
if c < a+b:
if d < b:
print('Yes')
c += 1
d += 1
else:
print('No')
else:
print('No')
else:
print('No')
#%%
|
s942652340
|
Accepted
| 116
| 4,712
| 475
|
#%%
n, a, b = map(int, input().split())
s = list(input())
c = 0
d = 0
for i in range(n):
if s[i] == 'a':
if c < a+b:
c += 1
print('Yes')
else:
print('No')
elif s[i] == 'b':
if c < a+b:
if d < b:
print('Yes')
c += 1
d += 1
else:
print('No')
else:
print('No')
else:
print('No')
#%%
|
s100909731
|
p03971
|
u844895214
| 2,000
| 262,144
|
Wrong Answer
| 62
| 9,208
| 629
|
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
|
import sys
def S(): return sys.stdin.readline().rstrip()
def IL(): return map(int,sys.stdin.readline().rstrip().split())
def solve():
n,a,b = IL()
s = S()
count = 0
fs = 1
for rep in s:
if rep=='a':
if count < a+b:
count += 1
print('Yes')
else:
print('No')
elif rep=='b':
if count<a+b and fs<=b:
count += 1
fs += 1
print('yes')
else:
print('No')
else:
print('No')
return
if __name__=='__main__':
solve()
|
s188706854
|
Accepted
| 59
| 9,244
| 629
|
import sys
def S(): return sys.stdin.readline().rstrip()
def IL(): return map(int,sys.stdin.readline().rstrip().split())
def solve():
n,a,b = IL()
s = S()
count = 0
fs = 1
for rep in s:
if rep=='a':
if count < a+b:
count += 1
print('Yes')
else:
print('No')
elif rep=='b':
if count<a+b and fs<=b:
count += 1
fs += 1
print('Yes')
else:
print('No')
else:
print('No')
return
if __name__=='__main__':
solve()
|
s203297551
|
p04011
|
u900109664
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 145
|
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
N = int(input('N>'))
K = int(input('K>'))
X = int(input('X>'))
Y = int(input('Y>'))
if N >= K:
print( K*X + (N-K)*Y )
else:
print( N*X )
|
s962872565
|
Accepted
| 17
| 2,940
| 131
|
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 )
|
s006073879
|
p00015
|
u648595404
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,596
| 92
|
A country has a budget of more than 81 trillion yen. We want to process such data, but conventional integer type which uses signed 32 bit can represent up to 2,147,483,647. Your task is to write a program which reads two integers (more than or equal to zero), and prints a sum of these integers. If given integers or the sum have more than 80 digits, print "overflow".
|
n = int(input())
for i in range(n):
a = int(input())
b = int(input())
print(a+b)
|
s285971172
|
Accepted
| 20
| 7,540
| 168
|
n = int(input())
for i in range(n):
a = int(input())
b = int(input())
c = str(a+b)
if len(c) > 80:
print("overflow")
else:
print(c)
|
s623711069
|
p02257
|
u665238221
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,664
| 265
|
A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7. Write a program which reads a list of _N_ integers and prints the number of prime numbers in the list.
|
def is_prime(x):
if x == 2:
return True
if x < 2:
return False
for i in range(3, int(x ** .5) + 1, 2):
if x % i == 0:
return False
return True
n = int(input())
print(sum(is_prime(int(input())) for _ in range(n)))
|
s981939432
|
Accepted
| 220
| 5,688
| 279
|
def is_prime(x):
if x == 2:
return True
if x < 2 or x % 2 == 0:
return False
for i in range(3, int(x ** .5) + 1, 2):
if x % i == 0:
return False
return True
n = int(input())
print(sum(is_prime(int(input())) for _ in range(n)))
|
s298746053
|
p02646
|
u306142032
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 9,040
| 137
|
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
a, v = map(int, input().split())
b, w = map(int, input().split())
t = int(input())
if (b-a) < t*(w-v):
print('YES')
else:
print('NO')
|
s516833478
|
Accepted
| 22
| 9,176
| 141
|
a, v = map(int, input().split())
b, w = map(int, input().split())
t = int(input())
if abs(a-b) <= t*(v-w):
print('YES')
else:
print('NO')
|
s391146026
|
p03448
|
u884323674
| 2,000
| 262,144
|
Wrong Answer
| 33
| 5,656
| 356
|
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 = 30
B = 40
C = 50
X = 6000
A_list = [500*i for i in range(A+1)]
B_list = [100*i for i in range(B+1)]
C_list = [50*i for i in range(C+1)]
sum_list = []
for i in A_list:
for j in B_list:
i_j = i + j
for k in C_list:
sum_list.append(i_j + k)
result = 0
for i in sum_list:
if i == X:
result += 1
print(result)
|
s622595408
|
Accepted
| 35
| 3,060
| 258
|
A = int(input())
B = int(input())
C = int(input())
X = int(input())
result = 0
for i in range(A+1):
for j in range(B+1):
i_j = 500*i + 100*j
for k in range(C+1):
if i_j + 50*k == X:
result += 1
print(result)
|
s045593622
|
p03485
|
u485716382
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 99
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
def solve():
a, b = map(int, input().split())
x = ((a + b) // 2) + 1
print(x)
solve()
|
s090404358
|
Accepted
| 17
| 2,940
| 109
|
from math import ceil
def solve():
a, b = map(int, input().split())
print(ceil((a + b) / 2))
solve()
|
s995959785
|
p03997
|
u566321790
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 146
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
if a>b:
print((((a-b)*h)/2)+(b*h))
elif b>a:
print((((b-a)*h)/2)+(a*h))
else:
print(a*h)
|
s940821282
|
Accepted
| 17
| 3,060
| 161
|
a = int(input())
b = int(input())
h = int(input())
if a>b:
print(int((((a-b)*h)/2)+(b*h)))
elif b>a:
print(int((((b-a)*h)/2)+(a*h)))
else:
print(int(a*h))
|
s797559101
|
p03555
|
u484856305
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 81
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
a=input()
b=input()
c=b[::-1]
if a == c[::-1]:
print("YES")
else:
print("NO")
|
s272899166
|
Accepted
| 18
| 2,940
| 76
|
a=input()
b=input()
c=b[::-1]
if a == c:
print("YES")
else:
print("NO")
|
s845540191
|
p03854
|
u545368057
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,316
| 1,371
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
S = input()
strs = ["dream","dreamer","erase","eraser"]
ind = 0
while True:
# print(ind,S[ind])
if S[ind] == "e":
if S[ind:ind+5] == "erase":
if ind+5 == len(S):
print("YES")
exit()
if S[ind+5] == "r":
if ind + 6 ==len(S):
print("YES")
exit()
ind+=6
elif S[ind+5] == "d" or S[ind+5]=="e":
ind+=5
else:
print("NO")
exit()
else:
print("NO")
exit()
elif S[ind] == "d":
if S[ind:ind+5] == "dream":
if ind + 5 == len(S):
print("YES")
exit()
if S[ind+5] == "d" :ind += 5
if S[ind+5] == "e":
if S[ind+6] == "r":
if ind + 7 == len(S):
print("YES")
exit()
if S[ind+7] == "e" or S[ind+7] =="d": ind += 6
elif S[ind+7] == "a": ind += 5
else:
print("NO")
exit()
else:
print("NO")
exit()
else:
print("NO")
exit()
print(ind)
|
s148727040
|
Accepted
| 29
| 3,188
| 389
|
S = input()
strs = ["erase", "eraser", "dreamer","dream"]
rtss = [s[::-1] for s in strs]
S = S[::-1]
ind = 0
while True:
# print(ind,S[ind])
if S[ind:ind+5] in rtss:
ind += 5
elif S[ind:ind+6] in rtss:
ind += 6
elif S[ind:ind+7] in rtss:
ind +=7
else:
print("NO")
exit()
if ind == len(S):
print("YES")
exit()
|
s190885071
|
p03338
|
u945418216
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,060
| 232
|
You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.
|
#ABC098B
n = int(input())
s = input()
ans = 0
for i in range(1,n):
s1 = list(set(s[:i]))
s2 = list(set(s[i:]))
cnt = [s2.count(x) for x in s1]
count=0
for c in cnt:
count+=c
ans=max(ans,count)
print()
|
s196636677
|
Accepted
| 20
| 3,060
| 200
|
n=int(input())
s=input()
ans=0
for i in range(n):
l = list(set(s[:i]))
r = list(set(s[i:]))
count=0
for x in l:
count+=r.count(x)
if ans<count:
ans=count
print(ans)
|
s309330055
|
p03455
|
u027208253
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 107
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = input().split()
result = int(a) * int(b)
if result % 2 == 0:
print("even")
else :
print("odd")
|
s007882599
|
Accepted
| 17
| 2,940
| 108
|
a,b = input().split()
result = int(a) * int(b)
if result % 2 == 0:
print("Even")
else :
print("Odd")
|
s867551933
|
p03598
|
u586639900
| 2,000
| 262,144
|
Wrong Answer
| 32
| 9,180
| 189
|
There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i). Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N. In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B. Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i). Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N. When activated, each type of robot will operate as follows. * When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. * When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.
|
N = int(input())
K = int(input())
X = list(map(int, input().split()))
res = 0
for i in range(N):
if X[i] >= (K - X[i]):
res += 2 * X[i]
else:
res += 2 * (K - X[i])
print(res)
|
s683128610
|
Accepted
| 28
| 9,180
| 189
|
N = int(input())
K = int(input())
X = list(map(int, input().split()))
res = 0
for i in range(N):
if X[i] <= (K - X[i]):
res += 2 * X[i]
else:
res += 2 * (K - X[i])
print(res)
|
s626417296
|
p03433
|
u215643129
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,112
| 109
|
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 = list(map(int, input().split()))
a.sort(reverse=True)
print(sum(a[::2])-sum(a[1::2]))
|
s959225501
|
Accepted
| 29
| 9,120
| 83
|
n = int(input())
a = int(input())
if n%500 <= a:
print("Yes")
else:
print("No")
|
s018254056
|
p03485
|
u144242534
| 2,000
| 262,144
|
Wrong Answer
| 32
| 9,040
| 95
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a, b = map(int, input().split())
x = a + b
if x % 2 == 0:
print(x/2)
else:
print((x + 1)/2)
|
s359076722
|
Accepted
| 23
| 8,940
| 105
|
a, b = map(int, input().split())
x = a + b
if x % 2 == 0:
print(int(x/2))
else:
print(int((x + 1)/2))
|
s575509349
|
p03854
|
u390958150
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 3,188
| 271
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
s = input()
ans = 'YES'
while len(s) > 0:
if s.endswith('dreamer'):
s = s[:-7]
elif s.endswith('dream'):
s = s[:-5]
elif s.endswith('eraser'):
s = s[:-6]
elif s.endswith('erase'):
s = s[:-5]
else:
ans = 'NO'
|
s496846193
|
Accepted
| 68
| 3,188
| 298
|
s = input()
ans = 'YES'
while len(s) > 0:
if s.endswith('dreamer'):
s = s[:-7]
elif s.endswith('dream'):
s = s[:-5]
elif s.endswith('eraser'):
s = s[:-6]
elif s.endswith('erase'):
s = s[:-5]
else:
ans = 'NO'
break
print(ans)
|
s784034555
|
p03659
|
u507456172
| 2,000
| 262,144
|
Wrong Answer
| 165
| 30,656
| 180
|
Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it. They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card. Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|.
|
N = int(input())
A = list(map(int,input().split()))
S1 = A[0]
S2 = sum(A)-A[0]
B = [S1-S2]
for i in range(N-2):
S1 += A[i+1]
S2 -= A[i+1]
B.append(abs(S1-S2))
print(min(B))
|
s685515129
|
Accepted
| 165
| 30,468
| 185
|
N = int(input())
A = list(map(int,input().split()))
S1 = A[0]
S2 = sum(A)-A[0]
B = [abs(S1-S2)]
for i in range(N-2):
S1 += A[i+1]
S2 -= A[i+1]
B.append(abs(S1-S2))
print(min(B))
|
s551946561
|
p03545
|
u953868469
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 771
|
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.
|
n = list(map(int,input()))
if n[0] + n[1] + n[2] + n[3] == 7:
print("{}+{}+{}+{}".format(n[0],n[1],n[2],n[3]))
elif n[0] + n[1] + n[2] - n[3] == 7:
print("{}+{}+{}+{}".format(n[0],n[1],n[2],n[3]))
elif n[0] + n[1] - n[2] + n[3] == 7:
print("{}+{}+{}+{}".format(n[0],n[1],n[2],n[3]))
elif n[0] + n[1] - n[2] - n[3] == 7:
print("{}+{}+{}+{}".format(n[0],n[1],n[2],n[3]))
elif n[0] - n[1] + n[2] + n[3] == 7:
print("{}+{}+{}+{}".format(n[0],n[1],n[2],n[3]))
elif n[0] - n[1] + n[2] - n[3] == 7:
print("{}+{}+{}+{}".format(n[0],n[1],n[2],n[3]))
elif n[0] - n[1] - n[2] + n[3] == 7:
print("{}+{}+{}+{}".format(n[0],n[1],n[2],n[3]))
elif n[0] - n[1] - n[2] - n[3] == 7:
print("{}+{}+{}+{}".format(n[0],n[1],n[2],n[3]))
else:
print("error")
|
s964925332
|
Accepted
| 18
| 3,188
| 786
|
n = list(map(int,input()))
if n[0] + n[1] + n[2] + n[3] == 7:
print("{}+{}+{}+{}=7".format(n[0],n[1],n[2],n[3]))
elif n[0] + n[1] + n[2] - n[3] == 7:
print("{}+{}+{}-{}=7".format(n[0],n[1],n[2],n[3]))
elif n[0] + n[1] - n[2] + n[3] == 7:
print("{}+{}-{}+{}=7".format(n[0],n[1],n[2],n[3]))
elif n[0] + n[1] - n[2] - n[3] == 7:
print("{}+{}-{}-{}=7".format(n[0],n[1],n[2],n[3]))
elif n[0] - n[1] + n[2] + n[3] == 7:
print("{}-{}+{}+{}=7".format(n[0],n[1],n[2],n[3]))
elif n[0] - n[1] + n[2] - n[3] == 7:
print("{}-{}+{}-{}=7".format(n[0],n[1],n[2],n[3]))
elif n[0] - n[1] - n[2] + n[3] == 7:
print("{}-{}-{}+{}=7".format(n[0],n[1],n[2],n[3]))
elif n[0] - n[1] - n[2] - n[3] == 7:
print("{}-{}-{}-{}=7".format(n[0],n[1],n[2],n[3]))
else:
print("error")
|
s925372873
|
p02612
|
u617037231
| 2,000
| 1,048,576
|
Wrong Answer
| 30
| 9,132
| 33
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
print(N%1000)
|
s794703458
|
Accepted
| 27
| 9,152
| 99
|
N = int(input())
if N%1000 == 0:
print(0)
else:
while N>=1000:
N-=1000
print(abs(N-1000))
|
s814839697
|
p02406
|
u589886885
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,480
| 126
|
In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }
|
n = int(input())
for i in range(3, n + 1):
if i % 3 == 0:
print(i)
elif '3' in list(str(i)):
print(i)
|
s962766238
|
Accepted
| 20
| 8,112
| 172
|
n = int(input())
result = ['']
for i in range(3, n + 1):
if i % 3 == 0:
result.append(i)
elif '3' in list(str(i)):
result.append(i)
print(*result)
|
s748050805
|
p03494
|
u871867619
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 300
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
arr = [int(i) for i in input().split()]
count = 0
while(True):
for i, j in enumerate(arr):
if j % 2 == 0:
arr[i] = j / 2
if i == 2:
count += 1
else:
break
else:
continue
break
print(count)
|
s020446284
|
Accepted
| 160
| 12,420
| 347
|
import numpy as np
N = int(input())
arr = np.array([int(i) for i in input().split()])
count = 0
while(True):
exits_odd = False
for i in range(N):
if arr[i] % 2 != 0:
exits_odd = True
else:
continue
if exits_odd:
break
else:
count += 1
arr = arr / 2
print(count)
|
s248992789
|
p03814
|
u663101675
| 2,000
| 262,144
|
Wrong Answer
| 46
| 3,560
| 187
|
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
A = input()
N = len(A)
print(A)
for i in range(N):
if A[i] == 'A':
S = i
break
for i in range(N):
if A[N-i-1] == 'Z':
E = i
break
print(N - S - E)
|
s677463445
|
Accepted
| 45
| 3,516
| 188
|
A = input()
N = len(A)
#print(A)
for i in range(N):
if A[i] == 'A':
S = i
break
for i in range(N):
if A[N-i-1] == 'Z':
E = i
break
print(N - S - E)
|
s451730786
|
p03494
|
u009219947
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 3,064
| 236
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
_ = input()
num_list = input().split(" ")
num_list = list(map(lambda x:int(x), num_list))
min_num = min(num_list)
shift_counter = 0
while(True):
if min_num == 1:
break
min_num /= 2
shift_counter += 1
print(shift_counter)
|
s600349580
|
Accepted
| 18
| 3,064
| 352
|
_ = input()
num_list = input().split(" ")
num_list = list(map(lambda x:int(x), num_list))
shift_counter = 0
def checker(num_list):
for n in num_list:
if n % 2 != 0:
return True
return False
while(True):
if checker(num_list):
print(shift_counter)
break
else:
num_list = [int(n/2) for n in num_list]
shift_counter += 1
|
s169595696
|
p03567
|
u476044037
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 22
|
Snuke built an online judge to hold a programming contest. When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.) Determine whether the judge can return the string `AC` as the verdict to a program.
|
print("AC" in input())
|
s395716624
|
Accepted
| 19
| 2,940
| 83
|
answer = "AC" in input()
if answer == False:
print("No")
else:
print("Yes")
|
s171927806
|
p02806
|
u000840710
| 2,525
| 1,048,576
|
Wrong Answer
| 18
| 3,064
| 224
|
Niwango created a playlist of N songs. The title and the duration of the i-th song are s_i and t_i seconds, respectively. It is guaranteed that s_1,\ldots,s_N are all distinct. Niwango was doing some work while playing this playlist. (That is, all the songs were played once, in the order they appear in the playlist, without any pause in between.) However, he fell asleep during his work, and he woke up after all the songs were played. According to his record, it turned out that he fell asleep at the very end of the song titled X. Find the duration of time when some song was played while Niwango was asleep.
|
N = int(input().rstrip())
src = [list(input().split()) for _ in range(N)]
for row in src:
row[1] = int(row[1])
X = input()
ans = 0
for row in src:
if row[0] == X:
break
else:
ans += row[1]
print(ans)
|
s707147116
|
Accepted
| 18
| 3,064
| 315
|
N = int(input().rstrip())
src = [list(input().split()) for _ in range(N)]
for row in src:
row[1] = int(row[1])
X = input()
total = 0
for row in src:
total += row[1]
awake = 0
for row in src:
if row[0] != X:
awake += row[1]
else:
awake += row[1]
break
ans = total - awake
print(ans)
|
s028630182
|
p02694
|
u321881571
| 2,000
| 1,048,576
|
Wrong Answer
| 2,205
| 9,076
| 129
|
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())
i=1
Y=100
while True:
Z=Y+0.01
if Z>=X:
print(i)
break
else:
i=i+1
Y=Z
|
s553853438
|
Accepted
| 21
| 9,172
| 130
|
X=int(input())
i=0
Y=100
while True:
Z=int(Y*1.01)
i=i+1
if Z>=X:
print(i)
break
else:
Y=Z
|
s875074237
|
p02612
|
u364363327
| 2,000
| 1,048,576
|
Wrong Answer
| 124
| 27,136
| 52
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
import numpy as np
N = int(input())
print(N%1000)
|
s379521382
|
Accepted
| 115
| 27,124
| 124
|
import numpy as np
N = int(input())
for i in range(10000):
if ( N <= 1000*i ):
print(1000*i - N)
break
|
s864604382
|
p03385
|
u662449766
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 203
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
import sys
input = sys.stdin.readline
def main():
s = input()
if "a" in s and "b" in s and "c" in s:
print("YES")
else:
print("NO")
if __name__ == "__main__":
main()
|
s040299259
|
Accepted
| 17
| 2,940
| 203
|
import sys
input = sys.stdin.readline
def main():
s = input()
if "a" in s and "b" in s and "c" in s:
print("Yes")
else:
print("No")
if __name__ == "__main__":
main()
|
s015579499
|
p03836
|
u462703607
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 164
|
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
sx,sy,tx,ty=map(int,input().split())
x=tx-sx
y=ty-sy
path=""
path+="U"*x+"R"*y+"D"*x+"L"*y
path+="L"+"R"*(x+1)+"U"*(y+1)+"L"+"U"+"L"*(x+1)+"D"*(y+1)+"R"
print(path)
|
s620613556
|
Accepted
| 18
| 3,060
| 164
|
sx,sy,tx,ty=map(int,input().split())
y=tx-sx
x=ty-sy
path=""
path+="U"*x+"R"*y+"D"*x+"L"*y
path+="L"+"U"*(x+1)+"R"*(y+1)+"D"+"R"+"D"*(x+1)+"L"*(y+1)+"U"
print(path)
|
s116068109
|
p03547
|
u379692329
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 38
|
In programming, hexadecimal notation is often used. In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively. In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`. When X and Y are seen as hexadecimal numbers, which is larger?
|
X, Y = input().split()
print(max(X,Y))
|
s146053614
|
Accepted
| 19
| 3,060
| 89
|
X, Y = input().split()
if X > Y:
print(">")
elif X < Y:
print("<")
else:
print("=")
|
s719119572
|
p03699
|
u185034753
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 311
|
You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade?
|
def solve():
N = int(input())
a = sorted([int(input()) for _ in range(N)])
s = sum(a)
if s % 10 != 0:
return s
for x in a:
if (s-x)%10 != 0:
print(x)
return s - x
return 0
def main():
print(solve())
if __name__ == '__main__':
main()
|
s458640191
|
Accepted
| 17
| 3,064
| 290
|
def solve():
N = int(input())
a = sorted([int(input()) for _ in range(N)])
s = sum(a)
if s % 10 != 0:
return s
for x in a:
if (s-x)%10 != 0:
return s - x
return 0
def main():
print(solve())
if __name__ == '__main__':
main()
|
s823970929
|
p03623
|
u561828236
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 103
|
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x,a,b = map(int,input().split())
if abs(x-a) >= abs(x-b):
print(abs(x-b))
else:
print(abs(x-a))
|
s990932462
|
Accepted
| 18
| 2,940
| 94
|
x,a,b = map(int,input().split())
if abs(x-a) >= abs(x-b):
print("B")
else:
print("A")
|
s979101755
|
p02263
|
u404682284
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,592
| 341
|
An expression is given in a line. Two consequtive symbols (operand or operator) are separated by a space character. You can assume that +, - and * are given as the operator and an operand is a positive integer less than 106
|
input_line = input().split()
stack = []
for i in input_line:
if i == '+':
stack.append(stack.pop() + stack.pop())
elif i == '-':
a = stack.pop()
b = stack.pop()
stack.append(b - a)
elif i =='*':
stack.append(stack.pop() * stack.pop())
else:
stack.append(int(i))
print(stack)
|
s388582645
|
Accepted
| 20
| 5,600
| 344
|
input_line = input().split()
stack = []
for i in input_line:
if i == '+':
stack.append(stack.pop() + stack.pop())
elif i == '-':
a = stack.pop()
b = stack.pop()
stack.append(b - a)
elif i =='*':
stack.append(stack.pop() * stack.pop())
else:
stack.append(int(i))
print(stack[0])
|
s001119532
|
p04029
|
u149752754
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 233
|
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?
|
c = list(input())
ans = []
for i in range(len(c)):
if c[i] == '0':
ans.append('0')
elif c[i] == '1':
ans.append('1')
elif len(ans) != 0:
del ans[-1]
else:
continue
print (''.join(ans))
|
s658610180
|
Accepted
| 18
| 2,940
| 37
|
N = int(input())
print((N*(N+1))//2)
|
s941881176
|
p03623
|
u287431190
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 90
|
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x,a,b = map(int, input().split())
if abs(x-a) > abs(x-b):
print('A')
else:
print('B')
|
s603692768
|
Accepted
| 17
| 2,940
| 90
|
x,a,b = map(int, input().split())
if abs(x-a) < abs(x-b):
print('A')
else:
print('B')
|
s828216874
|
p03079
|
u503228842
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 95
|
You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C.
|
a,b,c=map(int,input().split())
if a==b:
if b==c:
print("yes")
else:
print("No")
|
s400452417
|
Accepted
| 17
| 2,940
| 91
|
a,b,c = map(int,input().split())
if (a==b) and b==c:
print("Yes")
else:
print("No")
|
s059331451
|
p03815
|
u641406334
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 32
|
Snuke has decided to play with a six-sided die. Each of its six sides shows an integer 1 through 6, and two numbers on opposite sides always add up to 7. Snuke will first put the die on the table with an arbitrary side facing upward, then repeatedly perform the following operation: * Operation: Rotate the die 90° toward one of the following directions: left, right, front (the die will come closer) and back (the die will go farther). Then, obtain y points where y is the number written in the side facing upward. For example, let us consider the situation where the side showing 1 faces upward, the near side shows 5 and the right side shows 4, as illustrated in the figure. If the die is rotated toward the right as shown in the figure, the side showing 3 will face upward. Besides, the side showing 4 will face upward if the die is rotated toward the left, the side showing 2 will face upward if the die is rotated toward the front, and the side showing 5 will face upward if the die is rotated toward the back. Find the minimum number of operation Snuke needs to perform in order to score at least x points in total.
|
n = int(input())
print(n//5.5+1)
|
s078497324
|
Accepted
| 17
| 2,940
| 81
|
n = int(input())
a = n//11
print(2*a if n%11==0 else 2*a+1 if n%11<=6 else 2*a+2)
|
s338369648
|
p03673
|
u776237437
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 26,180
| 117
|
You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.
|
N = int(input())
A = list(map(int, input().split()))
L = []
for a in A:
L.append(a)
L.reverse()
print(L)
|
s977197782
|
Accepted
| 219
| 25,416
| 364
|
from collections import deque
N = int(input())
A = list(map(int, input().split()))
B = deque()
if N % 2 == 0:
for i in range(N):
if i % 2 == 0:
B.append(A[i])
else:
B.appendleft(A[i])
else:
for i in range(N):
if i % 2 == 0:
B.appendleft(A[i])
else:
B.append(A[i])
print(*B)
|
s290731156
|
p02842
|
u994502918
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 142
|
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
import math
n = int(input())
x = math.ceil(n / 1.08)
result = math.floor(x * 1.08)
if (result == n):
print(result)
else:
print(":(")
|
s469316875
|
Accepted
| 18
| 2,940
| 137
|
import math
n = int(input())
x = math.ceil(n / 1.08)
result = math.floor(x * 1.08)
if (result == n):
print(x)
else:
print(":(")
|
s372996343
|
p02612
|
u576320075
| 2,000
| 1,048,576
|
Wrong Answer
| 30
| 9,084
| 59
|
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())
while n >= 1000:
n -= 1000
print(n)
|
s295317788
|
Accepted
| 31
| 9,116
| 118
|
n = int(input())
# while n >= 1000:
# n -= 1000
# n = max(n, 0)
print((1000 - n % 1000) % 1000)
# print(n)
|
s272233525
|
p00251
|
u766477342
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,724
| 61
|
選手の皆さん、パソコン甲子園にようこそ。パソコン甲子園は今年で10周年になりますが、出題される問題数や合計得点は年によって異なります。各問題には難易度に応じて得点が決められています。問題数が10問で、それぞれの問題の得点が与えられるとき、それらの合計を出力するプログラムを作成して下さい。
|
sum = 0
for i in range(10):sum = int(input());
print(sum)
|
s980515083
|
Accepted
| 30
| 6,724
| 56
|
s = 0
for i in range(10):s += int(input());
print(s)
|
s493964843
|
p03636
|
u492447501
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 45
|
The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.
|
S = input()
print(S[0]+S[1:-1]+S[len(S)-1])
|
s902118848
|
Accepted
| 17
| 2,940
| 55
|
S = input()
print(S[0]+str(len(S[1:-1]))+S[len(S)-1])
|
s975316846
|
p03730
|
u588081069
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 127
|
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
A, B, C = list(map(int, input().split()))
if A == 1:
print("YES")
elif A % B == C:
print("YES")
else:
print("NO")
|
s761892691
|
Accepted
| 17
| 2,940
| 137
|
A, B, C = list(map(int, input().split()))
for i in range(100):
if (A * i) % B == C:
print("YES")
exit()
print("NO")
|
s706438608
|
p03945
|
u699699071
| 2,000
| 262,144
|
Wrong Answer
| 52
| 3,316
| 130
|
Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa. In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white. Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.
|
S=input()
print(S)
result=0
for i in range(len(S)-1):
state=S[i]
if S[i]!=S[i+1]:
result+=1
else:
pass
print(result)
|
s621995349
|
Accepted
| 50
| 3,188
| 121
|
S=input()
result=0
for i in range(len(S)-1):
state=S[i]
if S[i]!=S[i+1]:
result+=1
else:
pass
print(result)
|
s664046628
|
p03997
|
u028014940
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 63
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a=int(input())
b=int(input())
h=int(input())
print((a+b)*h/2)
|
s934203221
|
Accepted
| 17
| 2,940
| 68
|
a=int(input())
b=int(input())
h=int(input())
print(int((a+b)*h/2))
|
s216062776
|
p03495
|
u050708958
| 2,000
| 262,144
|
Wrong Answer
| 115
| 35,996
| 147
|
Takahashi has N balls. Initially, an integer A_i is written on the i-th ball. He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls. Find the minimum number of balls that Takahashi needs to rewrite the integers on them.
|
from collections import Counter
n, k = map(int, input().split())
c = sum(map(lambda x: x[1], Counter(input().split()).most_common(k)))
print(k - c)
|
s280777155
|
Accepted
| 131
| 35,996
| 147
|
from collections import Counter
n, k = map(int, input().split())
c = sum(map(lambda x: x[1], Counter(input().split()).most_common(k)))
print(n - c)
|
s977693308
|
p02417
|
u340503368
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,560
| 350
|
Write a program which counts and reports the number of each alphabetical letter. Ignore the case of characters.
|
sentence = input()
count = [0 for i in range(26)]
alpha = "abcdefghijklmnopqrstuvwxyz"
for letter in sentence:
letter.lower()
num = 0
for j in alpha:
if letter == j:
count[num] += 1
break
else:
num += 1
num = 0
for k in alpha:
print(k + " : " + str(count[num]))
num += 1
|
s489374011
|
Accepted
| 20
| 5,568
| 233
|
import sys
s = sys.stdin.read().lower()
count = [0 for i in range(26)]
for j in s:
num = ord(j) - ord('a')
if num >= 0 and num <= 25:
count[num] += 1
for k in range(26):
print(chr(k + 97) + " : " + str(count[k]))
|
s639176991
|
p02408
|
u737311644
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,576
| 64
|
Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers). The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond.
|
n=int(input())
a=input().split()
a.reverse()
print(" ".join(a))
|
s603887861
|
Accepted
| 20
| 5,608
| 471
|
n=int(input())
spade=set(range(1,14,1))
heart=set(range(1,14,1))
club=set(range(1,14,1))
diamond=set(range(1,14,1))
for a in range(n):
e,b=input().split()
if e=="S":
spade.remove(int(b))
if e=="H":
heart.remove(int(b))
if e=="C":
club.remove(int(b))
if e=="D":
diamond.remove(int(b))
for i in spade:
print("S",i)
for i in heart:
print("H",i)
for i in club:
print("C", i)
for i in diamond:
print("D",i)
|
s567393879
|
p03455
|
u652109955
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,168
| 119
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
# coding: utf-8
# Your code here!
a,b=(map(int,input().split()))
if a*b/2==0:
print("Even")
else:
print("Odd")
|
s793936768
|
Accepted
| 22
| 9,148
| 119
|
# coding: utf-8
# Your code here!
a,b=(map(int,input().split()))
if a*b%2==0:
print("Even")
else:
print("Odd")
|
s780354652
|
p00032
|
u647694976
| 1,000
| 131,072
|
Time Limit Exceeded
| 10,000
| 5,588
| 218
|
機械に辺・対角線の長さのデータを入力し、プラスティック板の型抜きをしている工場があります。この工場では、サイズは様々ですが、平行四辺形の型のみを切り出しています。あなたは、切り出される平行四辺形のうち、長方形とひし形の製造個数を数えるように上司から命じられました。 「機械に入力するデータ」を読み込んで、長方形とひし形の製造個数を出力するプログラムを作成してください。
|
hisi=0
tyou=0
while True:
try:
a,b,c=map(int,input().split(","))
if a**2+b**2==c**2:
tyou +=1
elif a==b:
hisi +=1
except:
print(tyou)
print(hisi)
|
s623724001
|
Accepted
| 20
| 5,592
| 233
|
hisi=0
tyou=0
while True:
try:
a,b,c=map(int,input().split(","))
if a**2+b**2==c**2:
tyou +=1
elif a==b:
hisi +=1
except:
print(tyou)
print(hisi)
break
|
s616146740
|
p02414
|
u628732336
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,628
| 208
|
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula: \\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\] where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
|
n, m, l = [int(i) for i in input().split()]
A = []
B = []
C = []
for ni in range(n):
A.append([int(i) for i in input().split()])
for mi in range(m):
B.append([int(i) for i in input().split()])
|
s994596180
|
Accepted
| 490
| 8,980
| 420
|
n, m, l = [int(i) for i in input().split()]
A = []
B = []
C = []
for ni in range(n):
A.append([int(i) for i in input().split()])
for mi in range(m):
B.append([int(i) for i in input().split()])
for i in range(n):
C.append([])
for j in range(l):
C[i].append(0)
for k in range(m):
C[i][j] += A[i][k] * B[k][j]
for ni in range(n):
print(" ".join([str(s) for s in C[ni]]))
|
s388021952
|
p03470
|
u729133443
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 42
|
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?
|
print(len(set(open(0).readlines()[1:]))-1)
|
s113273036
|
Accepted
| 17
| 2,940
| 31
|
_,*s=open(0);print(len(set(s)))
|
s782858704
|
p03814
|
u782330257
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,516
| 75
|
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
s=input()
rev=s[::-1]
n=len(s)
print((n-rev.find('Z')) - (s.find('A') + 1))
|
s315236067
|
Accepted
| 18
| 3,512
| 89
|
s=input().strip()
rev=s[::-1]
n=len(s)
#print(n)
print((n-rev.find('Z')) - (s.find('A')))
|
s339767885
|
p02743
|
u395202850
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 83
|
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
|
a, b, c = map(int, input().split())
print("YNeos"[(a * b) >= (c - a - b) ** 2::2])
|
s131838252
|
Accepted
| 17
| 2,940
| 112
|
a, b, c = map(int, input().split())
print("Yes" if 4 * (a * b) < (c - a - b) ** 2 and c - a - b > 0 else "No")
|
s081143891
|
p02665
|
u761529120
| 2,000
| 1,048,576
|
Wrong Answer
| 921
| 674,256
| 883
|
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.
|
def main():
N = int(input())
A = list(map(int, input().split()))
B = [0] * (N + 1)
if A[0] >= 2:
print(-1)
exit()
if A[0] == 1:
if len(A) == 1:
print(1)
exit()
else:
print(-1)
exit()
B[0] = 1
ans = 1
for i in range(1,N+1):
tmp = 2 * B[i-1]
if tmp < A[i]:
print(-1)
exit()
elif tmp == A[i]:
if i != N:
print(-1)
exit()
B[i] = tmp - A[i]
ans += A[N]
C = [0] * (N+1)
C[N] += A[N]
for i in range(N-1,0,-1):
if C[i+1] <= B[i] + A[i]:
ans += A[i] + C[i+1]
C[i] = A[i] + C[i+1]
else:
ans += B[i] + A[i]
C[i] = B[i]
print(ans)
if __name__ == "__main__":
main()
|
s740155500
|
Accepted
| 770
| 679,908
| 892
|
def main():
N = int(input())
A = list(map(int, input().split()))
B = [0] * (N + 1)
if A[0] >= 2:
print(-1)
exit()
if A[0] == 1:
if len(A) == 1:
print(1)
exit()
else:
print(-1)
exit()
B[0] = 1
ans = 1
for i in range(1,N+1):
tmp = 2 * (B[i-1] - A[i-1])
if tmp < A[i]:
print(-1)
exit()
elif tmp == A[i]:
if i != N:
print(-1)
exit()
B[i] = tmp
ans += A[N]
C = [0] * (N+1)
C[N] += A[N]
for i in range(N-1,0,-1):
if C[i+1] + A[i] <= B[i]:
ans += A[i] + C[i+1]
C[i] = A[i] + C[i+1]
else:
ans += B[i]
C[i] = B[i]
print(ans)
if __name__ == "__main__":
main()
|
s551082571
|
p00010
|
u659302741
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,640
| 707
|
Write a program which prints the central coordinate $(p_x, p_y)$ and the radius $r$ of a circumscribed circle of a triangle which is constructed by three points $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ on the plane surface.
|
def simultaneous_equasion(a, b, c, d, e, f):
"??£???????¨????"
det = a * d - b * c
a11 = d / det
a12 = - b / det
a21 = - c / det
a22 = a / det
return a11 * e + a12 * f, a21 * e + a22 * f
n = int(input())
for i in range(n):
x1, y1, x2, y2, x3, y3 = map(float, input().split())
a = 2 * (x2 - x1)
b = 2 * (y2 - y1)
c = 2 * (x3 - x1)
d = 2 * (y3 - y1)
e = (y2 - y1) ** 2 + (x2 - x1) ** 2
f = (y3 - y1) ** 2 + (x3 - x1) ** 2
px, py = simultaneous_equasion(a, b, c, d, e, f)
print("%.3f %3f" % (round(px, 3), round(py, 3)))
|
s466402039
|
Accepted
| 30
| 7,776
| 970
|
import math
def simultaneous_equasion(a, b, c, d, e, f):
"??£???????¨????"
det = a * d - b * c
a11 = d / det
a12 = - b / det
a21 = - c / det
a22 = a / det
return a11 * e + a12 * f, a21 * e + a22 * f
n = int(input())
for i in range(n):
x1, y1, x2, y2, x3, y3 = map(float, input().split())
# (x - x1) ^ 2 + (y - y1) ^ 2 = (x - x2) ^ 2 + (y - y2) ^ 2
a = 2 * (x1 - x2)
b = 2 * (y1 - y2)
c = 2 * (x1 - x3)
d = 2 * (y1 - y3)
e = x1 ** 2 + y1 ** 2 - x2 ** 2 - y2 ** 2
f = x1 ** 2 + y1 ** 2 - x3 ** 2 - y3 ** 2
px, py = simultaneous_equasion(a, b, c, d, e, f)
r = math.sqrt((px - x1) ** 2 + (py - y1) ** 2)
print("%.3f %.3f %.3f" % (round(px, 3), round(py, 3), round(r, 3)))
|
s010372607
|
p03478
|
u578501242
| 2,000
| 262,144
|
Wrong Answer
| 26
| 3,060
| 195
|
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).
|
a =list(map(int,input().split()))
x=a[0]
y=a[1]
z=a[2]
t=0
for i in range(1,x):
if i//1000+((i//100)%10)+((i//10)%10)+(i%10)>=y and i//1000+((i//100)%10)+((i//10)%10)+(i%10)<=z:
t=t+i
print(t)
|
s933155068
|
Accepted
| 26
| 3,064
| 229
|
a =list(map(int,input().split()))
x=a[0]
y=a[1]
z=a[2]
t=0
for i in range(1,x+1):
if i//10000+((i//1000)%10)+((i//100)%10)+((i//10)%10)+(i%10)>=y and i//10000+((i//1000)%10)+((i//100)%10)+((i//10)%10)+(i%10)<=z:
t=t+i
print(t)
|
s044255484
|
p02613
|
u691576679
| 2,000
| 1,048,576
|
Wrong Answer
| 137
| 16,268
| 165
|
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.
|
a=int(input())
s=[input() for i in range(a)]
print("AC ×",s.count("AC"))
print("WA ×",s.count("WA"))
print("TLE ×",s.count("TLE"))
print("RE ×",s.count("RE"))
|
s812164493
|
Accepted
| 141
| 16,160
| 213
|
a=int(input())
s=[input() for i in range(a)]
print("AC x {:d}".format(s.count("AC")))
print("WA x {:d}".format(s.count("WA")))
print("TLE x {:d}".format(s.count("TLE")))
print("RE x {:d}".format(s.count("RE")))
|
s015063033
|
p03448
|
u587213169
| 2,000
| 262,144
|
Wrong Answer
| 50
| 3,060
| 235
|
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())
count=0
for i in range(a+1):
for i in range(b+1):
for i in range(c+1):
if x == 500*a + 100*b + 50*c:
count+=1
print(count)
|
s063934338
|
Accepted
| 50
| 3,064
| 237
|
a=int(input())
b=int(input())
c=int(input())
x=int(input())
count=0
for i in range(a+1):
for j in range(b+1):
for k in range(c+1):
if x == (i*500 + j*100 + k*50):
count+=1
print(count)
|
s000611537
|
p03997
|
u620238824
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 68
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h/2)
|
s757453831
|
Accepted
| 16
| 2,940
| 73
|
a = int(input())
b = int(input())
h = int(input())
print(int((a+b)*h/2))
|
s769321803
|
p02613
|
u466916194
| 2,000
| 1,048,576
|
Wrong Answer
| 162
| 16,152
| 340
|
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.
|
numberotestcase = int(input())
verdicat=[]
for step in range(numberotestcase):
exp = input().upper()
verdicat.append(exp)
first = verdicat.count('AC')
sec = verdicat.count('WA')
thir = verdicat.count('TLE')
fou = verdicat.count('RE')
print('AC * '+str(first))
print('WA * '+str(sec))
print('TLE * '+str(thir))
print('RE * '+str(fou))
|
s842650365
|
Accepted
| 156
| 16,276
| 341
|
numberotestcase = int(input())
verdicat=[]
for step in range(numberotestcase):
exp = input().upper()
verdicat.append(exp)
first = verdicat.count('AC')
sec = verdicat.count('WA')
thir = verdicat.count('TLE')
fou = verdicat.count('RE')
print('AC x '+str(first))
print('WA x '+str(sec))
print('TLE x '+str(thir))
print('RE x '+str(fou))
|
s332000296
|
p03673
|
u519923151
| 2,000
| 262,144
|
Wrong Answer
| 130
| 26,184
| 353
|
You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.
|
n = int(input())
al = list(map(int, input().split()))
res = [0]*n
m = n //2
if n % 2 ==0:
for i in range(m):
res[i] = al[-1-i*2]
for j in range(m):
res[m+j]= al[j*2]
print(res)
else:
for i in range(m+1):
res[i] = al[-1-i*2]
for j in range(m):
res[m+1+j] = al[1+j*2]
print(res)
|
s848459003
|
Accepted
| 231
| 26,180
| 405
|
n = int(input())
al = list(map(int, input().split()))
res = [0]*n
m = n //2
if n % 2 ==0:
for i in range(m):
res[i] = al[-1-i*2]
for j in range(m):
res[m+j]= al[j*2]
else:
for i in range(m+1):
res[i] = al[-1-i*2]
for j in range(m):
res[m+1+j] = al[1+j*2]
ress=""
for k in range(n):
ress +=str(res[k])
ress +=" "
print(ress[:-1])
|
s467707077
|
p03369
|
u855985627
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 31
|
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.
|
print(100*input().count('○'))
|
s086526467
|
Accepted
| 17
| 2,940
| 33
|
print(700+100*input().count('o'))
|
s229289554
|
p03416
|
u627147604
| 2,000
| 262,144
|
Wrong Answer
| 70
| 2,940
| 188
|
Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.
|
def is_palindromic(n):
if str(n) == str(n)[::-1]:
True
else:
False
A,B = map(int, input().split())
for i in range(A, B):
if is_palindromic(i):
print(i)
|
s857863525
|
Accepted
| 69
| 2,940
| 223
|
def is_palindromic(n):
if str(n) == str(n)[::-1]:
return True
else:
return False
res = 0
A,B = map(int, input().split())
for i in range(A, B+1):
if is_palindromic(i):
res += 1
print(res)
|
s634841195
|
p03827
|
u517309493
| 2,000
| 262,144
|
Wrong Answer
| 28
| 8,752
| 308
|
You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).
|
import os
import sys
import math
if os.environ.get("DEBUG") is not None:
sys.stdin = open("in.txt", "r")
rl = sys.stdin.readline
n, s = [rl().rstrip("\n") for _ in range(2)]
print(n, s)
x = 0
ans = 0
for it in s:
if it == "I":
x += 1
else:
x -= 1
ans = max(ans, x)
print(ans)
|
s675944553
|
Accepted
| 26
| 9,060
| 296
|
import os
import sys
import math
if os.environ.get("DEBUG") is not None:
sys.stdin = open("in.txt", "r")
rl = sys.stdin.readline
n, s = [rl().rstrip("\n") for _ in range(2)]
x = 0
ans = 0
for it in s:
if it == "I":
x += 1
else:
x -= 1
ans = max(ans, x)
print(ans)
|
s870444628
|
p03795
|
u174536291
| 2,000
| 262,144
|
Wrong Answer
| 31
| 9,132
| 84
|
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())
a = n % 15
my_result = n * 800 - (n - a) / 15
print(int(my_result))
|
s769905368
|
Accepted
| 23
| 9,088
| 90
|
n = int(input())
a = n % 15
my_result = n * 800 - (n - a) * 200 / 15
print(int(my_result))
|
s444117206
|
p03069
|
u130900604
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 3,620
| 169
|
There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.
|
n=int(input())
s=input()
cnt=0
for i in range(n):
a=s.count(".#")
if a==0:
break
cnt+=a
s=s.replace(".#","..")
print(s,cnt)
|
s897060519
|
Accepted
| 81
| 11,260
| 170
|
n=int(input())
s=input()
w_cnt=s.count(".")
ans=[w_cnt]
b=0
w=w_cnt
for i in s:
if i=="#":
b+=1
else:
w-=1
ans.append(b+w)
print(min(ans))
|
s596749782
|
p03409
|
u882359130
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,064
| 466
|
On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (a_i, b_i), and the coordinates of the i-th blue point are (c_i, d_i). A red point and a blue point can form a _friendly pair_ when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point. At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs.
|
N = int(input())
Red, Blue = [], []
for n in range(N):
Red.append([int(ab) for ab in input().split()])
Red.sort(key=lambda x: x[1], reverse=True)
for n in range(N):
Blue.append([int(cd) for cd in input().split()])
Blue.sort(key=lambda x: x[0])
pairs = 0
for B in Blue:
c, d = B
for R in Red:
a, b = R
inclination = (d-b) / abs(c-a)
if inclination > 0:
pairs += 1
print([R, B])
Red.remove(R)
break
print(pairs)
|
s336905705
|
Accepted
| 19
| 3,064
| 411
|
N = int(input())
Red, Blue = [], []
for n in range(N):
Red.append([int(ab) for ab in input().split()])
Red.sort(key=lambda x: x[1], reverse=True)
for n in range(N):
Blue.append([int(cd) for cd in input().split()])
Blue.sort(key=lambda x: x[0])
pairs = 0
for B in Blue:
c, d = B
for R in Red:
a, b = R
if a < c and b < d:
pairs += 1
Red.remove(R)
break
print(pairs)
|
s844510708
|
p03155
|
u752898745
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 61
|
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,h,w=[int(input()) for _ in range(3)];print((n-1+h)*(n-1+w))
|
s882342635
|
Accepted
| 17
| 2,940
| 61
|
n,h,w=[int(input()) for _ in range(3)];print((n+1-h)*(n+1-w))
|
s207917933
|
p03494
|
u320098990
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,176
| 330
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
N = int(input())
in_list = input().split(' ')
judge_list = max([int(s)%2 for s in in_list])
if judge_list==1:
print(0)
else:
min_num = min([int(s) for s in in_list])
amari = min_num%2
counter = 0
while amari == 0 :
min_num = min_num/2
amari = min_num%2
counter += 1
print(counter)
|
s812781556
|
Accepted
| 27
| 9,188
| 288
|
N = int(input())
in_list = input().split(' ')
target_list = [int(s) for s in in_list]
max_num = max([s%2 for s in target_list])
counter = 0
while max_num==0:
target_list = [int(s)/2 for s in target_list]
max_num = max([s%2 for s in target_list])
counter += 1
print(counter)
|
s233403887
|
p03659
|
u038408819
| 2,000
| 262,144
|
Wrong Answer
| 404
| 30,604
| 361
|
Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it. They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card. Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|.
|
N = int(input())
a = list(map(int, input().split()))
s_right = [0] * (N + 1)
s_left = [0] * (N + 1)
for i in range(N):
s_left[i + 1] = s_left[i] + a[i]
s_right[i + 1] = s_right[i] + a[N - 1 - i]
ans = float('Inf')
for i in range(1, N):
print(ans)
if ans > abs(s_left[i] - s_right[N - i]):
ans = abs(s_left[i] - s_right[N - i])
print(ans)
|
s183137696
|
Accepted
| 191
| 24,832
| 293
|
N = int(input())
a = list(map(int, input().split()))
s = [0] * (N + 1)
for i in range(N):
s[i + 1] = s[i] + a[i]
ans = float('Inf')
for i in range(1, N):
#print(ans)
kouho = abs(s[i] - ((s[-1]) - s[i]))
if ans > kouho:
ans = kouho
#ans = min(ans, kouho)
print(ans)
|
s728358186
|
p03380
|
u320567105
| 2,000
| 262,144
|
Wrong Answer
| 670
| 19,884
| 314
|
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.
|
n = int(input())
a = list(map(int,input().split()))
a.sort()
a_i = a[-1]
a_i_2 = a_i / 2.0
b = a[0]
for i in a:
if i < a_i_2:
b = i
else:
c = i
break
from scipy.misc import comb
if comb(a_i, b) > comb(a_i,c):
print("{} {}".format(a_i, b))
else:
print("{} {}".format(a_i, c))
|
s788052902
|
Accepted
| 80
| 14,648
| 465
|
ri = lambda: int(input())
rl = lambda: list(map(int,input().split()))
rr = lambda N: [ri() for _ in range(N)]
YN = lambda b: print('YES') if b else print('NO')
yn = lambda b: print('Yes') if b else print('No')
OE = lambda x: print('Odd') if x%2 else print('Even')
INF = 10**18
n=ri()
a=rl()
a.sort()
import bisect
p = a[-1]
p2 = p/2
q_i = bisect.bisect_left(a,p2)
if ((a[q_i]-p2) > (p2 - a[q_i-1]) or p==a[q_i]) and q_i != 0:
q_i -= 1
q = a[q_i]
print(p,q)
|
s790299717
|
p02927
|
u382176416
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 3,064
| 288
|
Today is August 24, one of the five Product Days in a year. A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day): * d_1 \geq 2 * d_{10} \geq 2 * d_1 \times d_{10} = m Takahashi wants more Product Days, and he made a new calendar called Takahashi Calendar where a year consists of M month from Month 1 to Month M, and each month consists of D days from Day 1 to Day D. In Takahashi Calendar, how many Product Days does a year have?
|
m,d=map(int, input().split())
count=0
if d <= 10:
print(0)
else:
for i in range(1,m+1):
for j in range(11,d+1):
dl = list(str(j))
if int(dl[0]) >=2 and int(dl[1]) >= 2 and int(dl[0])*int(dl[1]) == i:
print(str(i)+ ' ' + str(j))
count+=1
print(count)
|
s420735400
|
Accepted
| 29
| 3,064
| 252
|
m,d=map(int, input().split())
count=0
if d <= 10:
print(0)
else:
for i in range(1,m+1):
for j in range(11,d+1):
dl = list(str(j))
if int(dl[0]) >=2 and int(dl[1]) >= 2 and int(dl[0])*int(dl[1]) == i:
count+=1
print(count)
|
s629082072
|
p03845
|
u167647458
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 220
|
Joisino is about to compete in the final round of a certain programming competition. In this contest, there are N problems, numbered 1 through N. Joisino knows that it takes her T_i seconds to solve problem i(1≦i≦N). Also, there are M kinds of drinks offered to the contestants, numbered 1 through M. If Joisino takes drink i(1≦i≦M), her brain will be stimulated and the time it takes for her to solve problem P_i will become X_i seconds. It does not affect the time to solve the other problems. A contestant is allowed to take exactly one of the drinks before the start of the contest. For each drink, Joisino wants to know how many seconds it takes her to solve all the problems if she takes that drink. Here, assume that the time it takes her to solve all the problems is equal to the sum of the time it takes for her to solve individual problems. Your task is to write a program to calculate it instead of her.
|
n = int(input())
t = list(map(int, input().split()))
m = int(input())
px = [list(map(int, input().split())) for _ in range(m)]
sum_t = sum(t)
for i in range(m):
print(min(sum_t, sum_t - t[px[i][0] - 1] + px[i][1]))
|
s912717749
|
Accepted
| 18
| 3,060
| 208
|
n = int(input())
t = list(map(int, input().split()))
m = int(input())
px = [list(map(int, input().split())) for _ in range(m)]
sum_t = sum(t)
for i in range(m):
print(sum_t - t[px[i][0] - 1] + px[i][1])
|
s915653401
|
p00423
|
u900033552
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,672
| 187
|
A と B の 2 人のプレーヤーが， 0 から 9 までの数字が書かれたカードを使ってゲームを行う．最初に， 2 人は与えられた n 枚ずつのカードを，裏向きにして横一列に並べる．その後， 2 人は各自の左から 1 枚ずつカードを表向きにしていき，書かれた数字が大きい方のカードの持ち主が，その 2 枚のカードを取る．このとき，その 2 枚のカードに書かれた数字の合計が，カードを取ったプレーヤーの得点となるものとする．ただし，開いた 2 枚のカードに同じ数字が書かれているときには，引き分けとし，各プレーヤーが自分のカードを 1 枚ずつ取るものとする． 例えば， A，B の持ち札が，以下の入力例 1 から 3 のように並べられている場合を考えよう．ただし，入力ファイルは n + 1 行からなり， 1 行目には各プレーヤのカード枚数 n が書かれており， i + 1 行目（i = 1，2，... ，n）には A の左から i 枚目のカードの数字と B の左から i 枚目の カードの数字が，空白を区切り文字としてこの順で書かれている．すなわち，入力ファイルの 2 行目以降は，左側の列が A のカードの並びを，右側の列が B のカードの並びを，それぞれ表している．このとき，ゲーム終了後の A と B の得点は，それぞれ，対応する出力例に示したものとなる． 入力ファイルに対応するゲームが終了したときの A の得点と B の得点を，この順に空白を区切り文字として 1 行に出力するプログラムを作成しなさい．ただし， n ≤ 10000 とする． 入力例１ | 入力例２ | 入力例３ ---|---|--- 3| 3| 3 9 1| 9 1| 9 1 5 4| 5 4| 5 5 0 8| 1 0| 1 8 出力例１ | 出力例２ | 出力例３ 19 8| 20 0| 15 14
|
#coding: utf-8
n = int(input())
s = [list(map(int, input().split(' '))) for i in range(n)]
win = {True: 0, False: 0}
for a, b in s:
win[a>b] += (a + b)
print(win[True], win[False])
|
s862680058
|
Accepted
| 110
| 9,436
| 483
|
#coding: utf-8
def one_game():
n = int(input())
if n == 0:
return False
s = [list(map(int, input().split(' '))) for i in range(n)]
win = {True: 0, False: 0}
for a, b in s:
if a == b:
win[True] += a
win[False] += a
continue
win[a>b] += (a + b)
print(win[True], win[False])
return True
while one_game() is True:
a = 0
|
s025601691
|
p03387
|
u488127128
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 810
|
You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order: * Choose two among A, B and C, then increase both by 1. * Choose one among A, B and C, then increase it by 2. It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.
|
def solve():
a,b,c = map(int, input().split())
count = 0
l = (a%2, b%2, c%2)
print(l)
if l == (0,0,0) or l == (1,1,1):
count += abs(a-b) // 2
count += abs(max(a,b) - c) // 2
elif l == (1,0,0) or l == (0,1,1):
count += abs(b-c) // 2
if max(b,c) <= a:
count += a - max(b,c)
else:
count += max(b,c) - a
elif l == (0,1,0) or l == (1,0,1):
count += abs(a-c) // 2
if max(a,c) <= b:
count += b - max(a,c)
else:
count += max(a,c) - b
elif l == (0,0,1) or l == (1,1,0):
count += abs(a-b) // 2
if max(a,b) <= c:
count += c - max(a,b)
else:
count += max(a,b) - c
return count
if __name__ == '__main__':
print(solve())
|
s704043263
|
Accepted
| 17
| 3,064
| 768
|
def solve():
T = list(map(int, input().split()))
l = [t%2 for t in T]
count = 0
if sum(l) == 1:
odd = T.pop(l.index(1))
count += abs(T[0] - T[1]) // 2
if max(T[0],T[1]) <= odd:
count += odd - max(T[0],T[1])
else:
count += (max(T[0],T[1]) - odd)//2 + 2
elif sum(l) == 2:
even = T.pop(l.index(0))
count += abs(T[0] - T[1]) // 2
if max(T[0],T[1]) <= even:
count += even - max(T[0],T[1])
else:
count += (max(T[0],T[1]) - even)//2 + 2
else:
m = T.pop(T.index(max(T)))
count += abs(T[0] - T[1]) // 2
count += abs(m - max(T[0],T[1]))
return count
if __name__ == '__main__':
print(solve())
|
s625475188
|
p03680
|
u658993896
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 7,080
| 312
|
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym. The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it. Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up. Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
|
N=int(input())
buttun=[0]
for i in range(N):
buttun.append(int(input()))
light=1
ans=0
while True:
if buttun[light]==2:
break
elif light==buttun[light]:
ans=-1
break
else:
light=buttun[light]
ans+=1
if light==1:
ans=-1
break
print(ans)
|
s083943716
|
Accepted
| 221
| 7,080
| 344
|
N=int(input())
buttun=[0]
for i in range(N):
buttun.append(int(input()))
light=1
ans=0
while True:
if light==2:
break
elif light==buttun[light]:
ans=-1
break
else:
tmp=light
light=buttun[light]
ans+=1
buttun[tmp]=1
if light==1:
ans=-1
break
print(ans)
|
s178187564
|
p03680
|
u750838232
| 2,000
| 262,144
|
Wrong Answer
| 189
| 7,084
| 167
|
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym. The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it. Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up. Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
|
n = int(input())
a= [int(input()) for _ in range(n)]
count = 0
now = a[0]
while now != 2 and count < n:
now = a[now-1]
count += 1
print(count if count<n else -1)
|
s826182810
|
Accepted
| 193
| 7,084
| 167
|
n = int(input())
a= [int(input()) for _ in range(n)]
count = 1
now = a[0]
while now != 2 and count < n:
now = a[now-1]
count += 1
print(count if count<n else -1)
|
s599859050
|
p03447
|
u106342872
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 68
|
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 = int(input())
a = int(input())
b = int(input())
print((x-a) // b)
|
s333222619
|
Accepted
| 19
| 2,940
| 85
|
x = int(input())
a = int(input())
b = int(input())
k = (x-a) // b
print(x-a-b*k)
|
s921182777
|
p03699
|
u371763408
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 100
|
You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade?
|
n=int(input())
S=[int(input()) for i in range(n)]
if sum(S)%10==0:
print(0)
else:
print(sum(S))
|
s350439369
|
Accepted
| 18
| 3,060
| 178
|
n=int(input())
S=sorted([int(input()) for i in range(n)])
not_10=[i for i in S if i%10!=0]
sums=sum(S)
if sums%10!=0:
print(sums)
else:
print(sums-not_10[0] if not_10 else 0)
|
s529317723
|
p03597
|
u972892985
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 49
|
We have an N \times N square grid. We will paint each square in the grid either black or white. If we paint exactly A squares white, how many squares will be painted black?
|
n = int(input())
a = int(input())
print((n**n)-1)
|
s787218508
|
Accepted
| 17
| 2,940
| 49
|
n = int(input())
a = int(input())
print((n**2)-a)
|
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