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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s802863883
|
p02854
|
u476435125
| 2,000 | 1,048,576 |
Wrong Answer
| 147 | 34,488 | 371 |
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters. Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times **before** he does the cut: * Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan). * Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen. Find the minimum amount of money needed before cutting the bar into two parts with the same length.
|
import copy
from bisect import bisect_left
N=int(input())
la=list(map(int,input().split()))
ll=copy.copy(la)
#lr=copy.copy(la)
for i in range(1,N):
ll[i]+=ll[i-1]
# lr[i]+=lr[i+1]
print(ll)
inl=bisect_left(ll,ll[-1]/2)
ds=ll[-1]/2-ll[inl-1]
db=ll[inl]-ll[-1]/2
if ds<=db:
print(ll[-1]-ll[inl-1]*2)
else:
print(ll[inl]*2-ll[-1])
|
s598066447
|
Accepted
| 125 | 26,172 | 361 |
import copy
from bisect import bisect_left
N=int(input())
la=list(map(int,input().split()))
ll=copy.copy(la)
#lr=copy.copy(la)
for i in range(1,N):
ll[i]+=ll[i-1]
# lr[i]+=lr[i+1]
inl=bisect_left(ll,ll[-1]/2)
ds=ll[-1]/2-ll[inl-1]
db=ll[inl]-ll[-1]/2
if ds<=db:
print(ll[-1]-ll[inl-1]*2)
else:
print(ll[inl]*2-ll[-1])
|
s356038636
|
p02397
|
u550922601
| 1,000 | 131,072 |
Wrong Answer
| 60 | 5,612 | 157 |
Write a program which reads two integers x and y, and prints them in ascending order.
|
while True :
x, y = list(map(int,input().split()))
if x == y == 0:
break
if x > y :
print(x, y)
else:
print(y, x)
|
s507282077
|
Accepted
| 60 | 5,612 | 170 |
while True:
x, y = list(map(int, input().split()))
if x == 0 and y == 0:
break
if x > y:
print(y, x)
else:
print(x, y)
|
s996137221
|
p02678
|
u244416763
| 2,000 | 1,048,576 |
Wrong Answer
| 1,387 | 47,912 | 570 |
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
n,m = map(int,input().split())
a = [0 for _ in range(m)]
b = [0 for _ in range(m)]
for i in range(m):
a[i],b[i] = map(int,input().split())
a[i] -= 1
b[i] -= 1
ans = [0 for _ in range(n-1)]
looked = [1 for _ in range(n)]
queue = [0]
looked[0] = 0
can = [[] for _ in range(n)]
for i in range(m):
can[a[i]].append(b[i])
can[b[i]].append(a[i])
while(len(queue)):
now = queue.pop(0)
for i in can[now]:
if looked[i]:
queue.append(i)
looked[i] = 0
ans[i-1] = now+1
print("\n".join(list(map(str,ans))))
|
s460786121
|
Accepted
| 1,543 | 40,544 | 639 |
n,m = map(int,input().split())
a = [0 for _ in range(m)]
b = [0 for _ in range(m)]
for i in range(m):
a[i],b[i] = map(int,input().split())
a[i] -= 1
b[i] -= 1
ans = [-1 for _ in range(n-1)]
looked = [1 for _ in range(n)]
queue = [0]
looked[0] = 0
can = [[] for _ in range(n)]
for i in range(m):
can[a[i]].append(b[i])
can[b[i]].append(a[i])
while(len(queue)):
now = queue.pop(0)
for i in can[now]:
if looked[i]:
queue.append(i)
looked[i] = 0
ans[i-1] = now+1
for i in ans:
if ans == -1:
print("No")
exit
print("Yes")
for i in ans:
print(i)
|
s385217492
|
p03575
|
u966000628
| 2,000 | 262,144 |
Wrong Answer
| 22 | 3,064 | 788 |
You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges. The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i. An edge whose removal disconnects the graph is called a _bridge_. Find the number of the edges that are bridges among the M edges.
|
N,M = list(map(int,input().split()))
set =[]
s=int(N*(N+1)/2)
count = 0
for i in range(M):
set.append(list(map(int,input().split())))
def BFS(L):
start = L.pop(0)
visited.append(start)
for item in lis[start-1]:
if (item not in visited) and (item not in L):
L.append(item)
if L:
BFS(L)
else:
global check
if sum(visited) == s:
check = 1
else:
check = 0
for num in range(N):
visited = []
lis = [[] for i in range(N)]
que = [1]
for i,item in enumerate(set):
if i == num:
continue
x,y = item
lis[x-1].append(y)
lis[y-1].append(x)
check = -1
BFS(que)
print (check)
if check == 0:
count += 1
print (count)
|
s256634282
|
Accepted
| 27 | 3,192 | 852 |
N,M = list(map(int,input().split()))
lis_s =[]
count = 0
for _ in range(M):
lis_s.append(tuple(map(int,input().split())))
graph = [[False for i in range(N+1)] for k in range(N+1)]
for tp in lis_s:
graph[tp[0]][tp[1]] = True
graph[tp[1]][tp[0]] = True
visited = [False]*(N+1)
bridge = False
def dfs(v):
visited[v] = True
for vs in range(1,N+1):
if (graph[v][vs] == True) and (visited[vs] == False):
dfs(vs)
ans = 0
for num in range(M):
graph[lis_s[num][0]][lis_s[num][1]] = False
graph[lis_s[num][1]][lis_s[num][0]] = False
for i in range(N+1):
visited[i] = False
dfs(1)
bridge = False
for check in range(1,N+1):
if visited[check] == False:
bridge = True
if bridge:
ans += 1
graph[lis_s[num][0]][lis_s[num][1]] = True
graph[lis_s[num][1]][lis_s[num][0]] = True
print(ans)
|
s764872076
|
p03132
|
u839188633
| 2,000 | 1,048,576 |
Wrong Answer
| 1,056 | 61,504 | 453 |
Snuke stands on a number line. He has L ears, and he will walk along the line continuously under the following conditions: * He never visits a point with coordinate less than 0, or a point with coordinate greater than L. * He starts walking at a point with integer coordinate, and also finishes walking at a point with integer coordinate. * He only changes direction at a point with integer coordinate. Each time when Snuke passes a point with coordinate i-0.5, where i is an integer, he put a stone in his i-th ear. After Snuke finishes walking, Ringo will repeat the following operations in some order so that, for each i, Snuke's i-th ear contains A_i stones: * Put a stone in one of Snuke's ears. * Remove a stone from one of Snuke's ears. Find the minimum number of operations required when Ringo can freely decide how Snuke walks.
|
a = [int(input()) for _ in range(int(input()))]
dp = [[0] * 5 for _ in range(len(a) + 1)]
for i, x in enumerate(a):
y = x % 2 + (x==0) * 2
z = (x + 1) % 2
dp[i+1][0] = dp[i][0] + x
dp[i+1][1] = min(dp[i][0] + y, dp[i][1] + y)
dp[i+1][2] = min(dp[i][1] + z, dp[i][0] + z, dp[i][2] + z)
dp[i+1][3] = min(dp[i][2] + y, dp[i][1] + y, dp[i][0] + y, dp[i][3] + y)
dp[i+1][4] = min(dp[i][3] + x, dp[i+1][3])
print(dp[-1][4])
|
s982430199
|
Accepted
| 993 | 67,736 | 364 |
a = [int(input()) for _ in range(int(input()))]
dp = [[0] * 5 for _ in range(len(a) + 1)]
for i, x in enumerate(a):
m = x % 2
y = m + (not x) * 2
z = 1 - m
dp[i+1][0] = dp[i][0] + x
dp[i+1][1] = min(dp[i][:2]) + y
dp[i+1][2] = min(dp[i][:3]) + z
dp[i+1][3] = min(dp[i][:4]) + y
dp[i+1][4] = min(dp[i]) + x
print(min(dp[-1]))
|
s117595247
|
p02694
|
u663710122
| 2,000 | 1,048,576 |
Wrong Answer
| 22 | 9,164 | 103 |
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
import math
X = int(input())
t = 0
m = 100
while m < X:
m = math.ceil(m * 1.01)
t += 1
print(t)
|
s421272143
|
Accepted
| 21 | 9,184 | 104 |
import math
X = int(input())
t = 0
m = 100
while m < X:
m = math.floor(m * 1.01)
t += 1
print(t)
|
s529588612
|
p03418
|
u057079894
| 2,000 | 262,144 |
Wrong Answer
| 177 | 4,464 | 159 |
Takahashi had a pair of two positive integers not exceeding N, (a,b), which he has forgotten. He remembers that the remainder of a divided by b was greater than or equal to K. Find the number of possible pairs that he may have had.
|
n,m = map(int,input().split())
ans = 0
for i in range(1,n+1):
print(ans)
ans += n // i * max(i-m,0) + max(n%i-m+1,0)
if m == 0:
ans -= n
print(ans)
|
s959490039
|
Accepted
| 99 | 2,940 | 144 |
n,m = map(int,input().split())
ans = 0
for i in range(1,n+1):
ans += n // i * max(i-m,0) + max(n%i-m+1,0)
if m == 0:
ans -= n
print(ans)
|
s797726964
|
p03760
|
u425762225
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 279 |
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
|
#!/usr/bin/env python3
def main():
O = input()
E = input()
s = ['x']*(len(O) + len(E))
for i in range(len(O)//2):
s[i*2] = O[i]
s[i*2+1] = E[i]
if len(O) % 2:
s[-1] = O[-1]
print(''.join(s))
if __name__ == '__main__':
main()
|
s095457633
|
Accepted
| 17 | 3,060 | 262 |
#!/usr/bin/env python3
def main():
O = input()
E = input()
s = []
for i in range(len(E)):
s.append(O[i])
s.append(E[i])
if len(O) - len(E):
s.append(O[-1])
print(''.join(s))
if __name__ == '__main__':
main()
|
s168418209
|
p03139
|
u402966520
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 487 |
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.
|
# 1 <= N < 100
# 0 <= A <= N
# 0 <= B <= N
inputstr = input()
inputary = inputstr.split(" ")
print(inputary)
N = int(inputary[0])
A = int(inputary[1])
B = int(inputary[2])
suball = N - A - B
if suball > 0:
suball = 0
else:
suball = abs(suball)
if A >= B:
print(str(B) + " " + str(suball))
elif A < B:
print(str(A) + " " + str(suball))
|
s231001666
|
Accepted
| 17 | 3,060 | 472 |
# 1 <= N < 100
# 0 <= A <= N
# 0 <= B <= N
inputstr = input()
inputary = inputstr.split(" ")
N = int(inputary[0])
A = int(inputary[1])
B = int(inputary[2])
suball = N - A - B
if suball > 0:
suball = 0
else:
suball = abs(suball)
if A >= B:
print(str(B) + " " + str(suball))
elif A < B:
print(str(A) + " " + str(suball))
|
s954621337
|
p02936
|
u556225812
| 2,000 | 1,048,576 |
Wrong Answer
| 1,882 | 109,740 | 355 |
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
N, Q = map(int, input().split())
lst = []
cnt = []
ans = [0]*N
N = N - 1
for i in range(N):
lst.append(list(map(int, input().split())))
for j in range(Q):
cnt.append(list(map(int, input().split())))
for k in range(Q):
ans[cnt[k][0]-1] += cnt[k][1]
for i in range(N):
ans[lst[i][1]-1] += ans[lst[i][0]-1]
for a in ans:
print(a, end=" ")
|
s179266461
|
Accepted
| 1,862 | 56,080 | 512 |
import collections
N, Q = map(int, input().split())
lst = [[] for _ in range(N)]
ans = [0]*N
for i in range(N-1):
a, b = map(int, input().split())
lst[b-1].append(a-1)
lst[a-1].append(b-1)
for j in range(Q):
p, x = map(int, input().split())
p -= 1
ans[p] += x
q=collections.deque()
q.append(0)
flag = [0]*N
while q:
y = q.pop()
flag[y] = 1
for k in lst[y]:
if flag[k] == 1:
continue
ans[k] += ans[y]
q.append(k)
print(*ans)
|
s290547855
|
p04029
|
u102126195
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 72 |
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())
cnt = 0
for i in range(1, N + 1):
cnt += 1
print(cnt)
|
s488850627
|
Accepted
| 17 | 2,940 | 72 |
N = int(input())
cnt = 0
for i in range(1, N + 1):
cnt += i
print(cnt)
|
s849403466
|
p03644
|
u608088992
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 179 |
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
N = int(input())
maxdiv = 0
for i in range(1, N + 1):
d = i
count = 0
while d & 1 == 0:
d >>= 1
count += 1
maxdiv = max(maxdiv, count)
print(count)
|
s166685822
|
Accepted
| 17 | 2,940 | 216 |
N = int(input())
maxdiv = 0
ans = 1
for i in range(1, N + 1):
d = i
count = 0
while d % 2 == 0:
d //= 2
count += 1
if count > maxdiv:
ans = i
maxdiv = count
print(ans)
|
s749247005
|
p02972
|
u227085629
| 2,000 | 1,048,576 |
Wrong Answer
| 719 | 14,904 | 306 |
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
n = int(input())
a = list(map(int,input().split()))
box = [0]*n
for i in range(n,0,-1):
k = a[i-1]
c = 0
for j in range(1,n//i+1):
c += box[i*j-1]
if c%2 != k:
box[i-1] = 1
ans = []
w = 0
for m in range(n):
if box[m] == 1:
w += 1
ans.append(str(i+1))
print(w)
print(' '.join(ans))
|
s992757039
|
Accepted
| 766 | 17,212 | 306 |
n = int(input())
a = list(map(int,input().split()))
box = [0]*n
for i in range(n,0,-1):
k = a[i-1]
c = 0
for j in range(1,n//i+1):
c += box[i*j-1]
if c%2 != k:
box[i-1] = 1
ans = []
w = 0
for m in range(n):
if box[m] == 1:
w += 1
ans.append(str(m+1))
print(w)
print(' '.join(ans))
|
s505309028
|
p03469
|
u408375121
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 30 |
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
S = input()
t = '2018' + S[4:]
|
s711663037
|
Accepted
| 17 | 2,940 | 39 |
S = input()
t = '2018' + S[4:]
print(t)
|
s916167210
|
p03625
|
u142415823
| 2,000 | 262,144 |
Wrong Answer
| 111 | 18,600 | 382 |
We have N sticks with negligible thickness. The length of the i-th stick is A_i. Snuke wants to select four different sticks from these sticks and form a rectangle (including a square), using the sticks as its sides. Find the maximum possible area of the rectangle.
|
import collections
N = int(input())
A = [int(i) for i in input().split()]
stick_num = collections.Counter(A)
flag = True
ans = 0
for i in sorted(stick_num, reverse=True):
if stick_num[i] % 4 == 0:
ans = i * i
elif stick_num[i] % 2 >= 0:
if flag:
a = i
flag = False
else:
ans = a * i
break
print(ans)
|
s536599783
|
Accepted
| 89 | 14,252 | 382 |
N = int(input())
A = [int(i) for i in input().split()]
A.sort(reverse=True)
flag_h = True
flag_v = True
h, v = 0, 0
for i in A:
if flag_h:
if h == i:
flag_h = False
else:
h = i
elif flag_v:
if v == i:
flag_v = False
else:
v = i
if not (flag_v and flag_h):
print(h * v)
else:
print(0)
|
s194125535
|
p03377
|
u456878170
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 106 |
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
a, b, x = list(map(int, input().split()))
if a + b >= x and a <= x:
print('Yes')
else:
print('No')
|
s163756861
|
Accepted
| 18 | 2,940 | 106 |
a, b, x = list(map(int, input().split()))
if a + b >= x and a <= x:
print('YES')
else:
print('NO')
|
s165807505
|
p03547
|
u209918867
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 102 |
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=input().split()
if sorted(x) == x:
s='='
elif sorted(x)[0] == x[0]:
s='<'
else:
s='>'
print(s)
|
s272083248
|
Accepted
| 17 | 2,940 | 102 |
s=list(input().split())
ss=sorted(s)
if s[0]==s[1]:
r='='
elif s==ss:
r='<'
else:
r='>'
print(r)
|
s550507848
|
p03729
|
u302292660
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 90 |
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
a,b,c = input().split()
if a[-1]==b[0] and b[-1]==c[0]:
print("Yes")
else:
print("No")
|
s048391377
|
Accepted
| 17 | 2,940 | 90 |
a,b,c = input().split()
if a[-1]==b[0] and b[-1]==c[0]:
print("YES")
else:
print("NO")
|
s584685316
|
p03377
|
u872793589
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 173 |
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
def main():
A,B,X = map(int, input().split(" "))
if A > X or A + B < X:
print("No")
else:
print("Yes")
if __name__ == '__main__':
main()
|
s860465460
|
Accepted
| 17 | 2,940 | 173 |
def main():
A,B,X = map(int, input().split(" "))
if A > X or A + B < X:
print("NO")
else:
print("YES")
if __name__ == '__main__':
main()
|
s219453045
|
p02742
|
u419246138
| 2,000 | 1,048,576 |
Wrong Answer
| 2,104 | 3,060 | 249 |
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
num = input().split(" ")
ki = 0
gu = 0
sum_kaku = 0
for a in range(int(num[1])):
if a % 2 == 0:
gu += 1
else:
ki += 1
a = 0
while a < int(num[1]):
if a % 2 == 0:
sum_kaku += ki
else:
sum_kaku += gu
a += 1
print(sum_kaku)
|
s783712086
|
Accepted
| 17 | 3,060 | 189 |
num = input().split(" ")
if "1" in num:
print(1)
else:
gu = int(num[1]) // 2
ki = int(num[1]) - int(gu)
gu2 = int(num[0]) // 2
ki2 = int(num[0]) - int(gu2)
print(gu * gu2 + ki * ki2)
|
s002526356
|
p02741
|
u572373398
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 23 |
Print the K-th element of the following sequence of length 32: 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51
|
print(int(input()) - 1)
|
s342697032
|
Accepted
| 17 | 3,060 | 130 |
a = [1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51]
print(a[int(input()) - 1])
|
s242500068
|
p03729
|
u168416324
| 2,000 | 262,144 |
Wrong Answer
| 26 | 9,108 | 90 |
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
a,b,c=input().split()
if a[:-1]==b[0] and b[:-1]==c[0]:
print("YES")
else:
print("NO")
|
s372072794
|
Accepted
| 29 | 8,916 | 88 |
a,b,c=input().split()
if a[-1]==b[0] and b[-1]==c[0]:
print("YES")
else:
print("NO")
|
s970667192
|
p03385
|
u538632589
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 102 |
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
s = input()
if set([s[0], s[1], s[2]]) == set(['a', 'b', 'c']):
print('YES')
else:
print('NO')
|
s925327249
|
Accepted
| 17 | 3,064 | 102 |
s = input()
if set([s[0], s[1], s[2]]) == set(['a', 'b', 'c']):
print('Yes')
else:
print('No')
|
s738544814
|
p02613
|
u749491107
| 2,000 | 1,048,576 |
Wrong Answer
| 146 | 16,264 | 377 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
row = int(input())
s = [input() for i in range(row)]
ac = 0
wa = 0
tle = 0
re = 0
for j in s:
if j == "AC":
ac += 1
elif j == "WA":
wa += 1
elif j == "TLE":
tle += 1
elif j == "re":
re += 1
ac = str(ac)
wa = str(wa)
tle = str(tle)
re = str(re)
print("AC × " + ac)
print("WA × " + wa)
print("TLE × " + tle)
print("RE × " + re)
|
s802249266
|
Accepted
| 159 | 16,276 | 373 |
row = int(input())
s = [input() for i in range(row)]
ac = 0
wa = 0
tle = 0
re = 0
for j in s:
if j == "AC":
ac += 1
elif j == "WA":
wa += 1
elif j == "TLE":
tle += 1
elif j == "RE":
re += 1
ac = str(ac)
wa = str(wa)
tle = str(tle)
re = str(re)
print("AC x " + ac)
print("WA x " + wa)
print("TLE x " + tle)
print("RE x " + re)
|
s258988009
|
p03477
|
u464912173
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 108 |
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.
|
a, b, c, d = map(int, input().split())
print('Right' if a+b > c+d else 'left' if a+b < c+d else 'Balanced')
|
s332195871
|
Accepted
| 18 | 2,940 | 108 |
a, b, c, d = map(int, input().split())
print('Right' if a+b < c+d else 'Left' if a+b > c+d else 'Balanced')
|
s737040926
|
p02534
|
u934119021
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,016 | 41 |
You are given an integer K. Print the string obtained by repeating the string `ACL` K times and concatenating them. For example, if K = 3, print `ACLACLACL`.
|
k = int(input())
ans = 'ACL' * k
print(k)
|
s182580777
|
Accepted
| 23 | 9,096 | 43 |
k = int(input())
ans = 'ACL' * k
print(ans)
|
s114529774
|
p00017
|
u187606290
| 1,000 | 131,072 |
Wrong Answer
| 40 | 7,948 | 555 |
In cryptography, Caesar cipher is one of the simplest and most widely known encryption method. Caesar cipher is a type of substitution cipher in which each letter in the text is replaced by a letter some fixed number of positions down the alphabet. For example, with a shift of 1, 'a' would be replaced by 'b', 'b' would become 'c', 'y' would become 'z', 'z' would become 'a', and so on. In that case, a text: this is a pen is would become: uijt jt b qfo Write a program which reads a text encrypted by Caesar Chipher and prints the corresponding decoded text. The number of shift is secret and it depends on datasets, but you can assume that the decoded text includes any of the following words: "the", "this", or "that".
|
import fileinput
import string
alphabetArray = list(string.ascii_lowercase)
for data in fileinput.input():
for n in range(-25, 26):
replacedStr = data
for i in range(0, 26):
if n > 0:
replacedStr = replacedStr.replace(alphabetArray[i], alphabetArray[(i + n) % 26])
else:
replacedStr = replacedStr.replace(alphabetArray[i], alphabetArray[i + n])
if "this" in replacedStr or "the" in replacedStr or "that" in replacedStr:
print(replacedStr)
break
|
s755092374
|
Accepted
| 40 | 7,892 | 900 |
import fileinput
import string
alphabetArray = list(string.ascii_lowercase)
for data in fileinput.input():
for n in range(-26, 27):
replacedStrAry = list(data.replace('\n', '').replace('\r', ''))
for index in range(0, len(replacedStrAry)):
if not replacedStrAry[index] in alphabetArray:
continue
if n >= 0:
replacedStrAry[index] = alphabetArray[(alphabetArray.index(replacedStrAry[index]) + n) % 26]
else:
replacedStrAry[index] = replacedStrAry[index] = alphabetArray[alphabetArray.index(replacedStrAry[index]) + n]
replacedStr = "".join(replacedStrAry)
replacedStrSplitedBySpace = replacedStr.split(' ')
if "the" in replacedStrSplitedBySpace or "this" in replacedStrSplitedBySpace or "that" in replacedStrSplitedBySpace:
print(replacedStr)
break
|
s236039123
|
p03214
|
u534308356
| 2,525 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 217 |
Niwango-kun is an employee of Dwango Co., Ltd. One day, he is asked to generate a thumbnail from a video a user submitted. To generate a thumbnail, he needs to select a frame of the video according to the following procedure: * Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video. * Select t-th frame whose representation a_t is nearest to the average of all frame representations. * If there are multiple such frames, select the frame with the smallest index. Find the index t of the frame he should select to generate a thumbnail.
|
N = int(input())
data = list(map(int, input().split()))
ave = sum(data) / len(data)
thumbnail_num = 1000
for i in range(len(data)):
if abs(data[i] - ave) < thumbnail_num:
thumbnail_num = i
print(thumbnail_num)
|
s217798955
|
Accepted
| 18 | 3,060 | 190 |
N = int(input())
data = list(map(int, input().split()))
ave = sum(data) / len(data)
ans = 0
for i in range(len(data)):
if abs(data[i] - ave) < abs(data[ans] - ave):
ans = i
print(ans)
|
s025424943
|
p02613
|
u970198631
| 2,000 | 1,048,576 |
Wrong Answer
| 148 | 16,108 | 273 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
N = int(input())
list1=[]
for i in range(N):
a = input()
list1.append(a)
a = list1.count('AC')
b = list1.count('WA')
c = list1.count('TLE')
d = list1.count('RE')
list2 = [a,b,c,d]
print('AC x'+str(a))
print('WA x'+str(b))
print('TLE x'+str(c))
print('RE x'+str(d))
|
s770562688
|
Accepted
| 149 | 16,248 | 254 |
A = int(input())
list1 = []
for i in range(A):
a = input()
list1.append(a)
b =list1.count('AC')
c =list1.count('TLE')
d =list1.count('RE')
e =list1.count('WA')
print('AC x '+ str(b))
print('WA x '+ str(e))
print('TLE x '+str(c))
print('RE x '+str(d))
|
s779949225
|
p02612
|
u850512471
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 9,080 | 55 |
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.
|
price = int(input())
bochi = price % 1000
print(bochi)
|
s618083554
|
Accepted
| 25 | 9,116 | 173 |
price = int(input())
sen = 1000
senCnt = int(price / 1000)
num = 0
if price % sen != 0:
senCnt = senCnt + 1
sen = sen * senCnt
num = sen - price
print(num)
|
s810680163
|
p03168
|
u945181840
| 2,000 | 1,048,576 |
Wrong Answer
| 233 | 47,664 | 368 |
Let N be a positive odd number. There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i. Taro has tossed all the N coins. Find the probability of having more heads than tails.
|
import numpy as np
N = int(input())
p = np.array([0.0] + list(map(float, input().split())))
dp = np.zeros((N + 1, N + 1), np.float32)
dp[0, 0] = 1
dp[:, 0] = np.cumprod(1 - p).T
for i in range(1, N + 1):
dp[i][1:] = dp[i - 1][:-1] * p[i] + dp[i - 1][1:] * (1 - p[i])
print(dp[-1][N // 2 + 1:].sum())
|
s304771764
|
Accepted
| 248 | 82,736 | 370 |
import numpy as np
N = int(input())
p = np.array([0.0] + list(map(float, input().split())))
dp = np.zeros((N + 1, N + 1), np.float64)
dp[0, 0] = 1
dp[:, 0] = np.cumprod(1 - p).T
for i in range(1, N + 1):
dp[i][1:] = dp[i - 1][:-1] * p[i] + dp[i - 1][1:] * (1 - p[i])
print(dp[-1][(N + 1) // 2:].sum())
|
s899102627
|
p02678
|
u932868243
| 2,000 | 1,048,576 |
Wrong Answer
| 2,207 | 49,956 | 249 |
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
n,m=map(int,input().split())
ab=[list(map(int,input().split())) for i in range(m)]
l=[[]]*n
for a,b in ab:
l[a-1].append(b)
l[b-1].append(a)
ans=[[0]]*(n)
for j in range(n):
for lll in l[j]:
ans[lll-1]=j+1
for t in range(1,n):
print(ans)
|
s242081611
|
Accepted
| 1,388 | 38,120 | 429 |
n,m=map(int,input().split())
l=[[] for _ in range(n)]
for i in range(m):
a,b=map(int,input().split())
l[a-1].append(b-1)
l[b-1].append(a-1)
queue=[0]
visit=[0]*n
visit[0]=1
ans=[0]*n
while len(queue)>0:
now=queue.pop(0)
for p in l[now]:
if visit[p]==0:
visit[p]+=1
queue.append(p)
ans[p]=now+1
k=ans[1:]
if not all(kk!=0 for kk in k):
print('No')
exit()
print('Yes')
for kk in k:
print(kk)
|
s948936022
|
p03338
|
u076764813
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 147 |
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.
|
n=int(input())
s=list(input())
num=0
for i in range(n):
x=set(s[:n])
y=set(s[n:])
if len(x&y)>num:
num=len(x&y)
print(num)
|
s428817412
|
Accepted
| 18 | 2,940 | 147 |
n=int(input())
s=list(input())
num=0
for i in range(n):
x=set(s[:i])
y=set(s[i:])
if len(x&y)>num:
num=len(x&y)
print(num)
|
s142071421
|
p03959
|
u579299997
| 2,000 | 262,144 |
Wrong Answer
| 23 | 3,064 | 74 |
Mountaineers Mr. Takahashi and Mr. Aoki recently trekked across a certain famous mountain range. The mountain range consists of N mountains, extending from west to east in a straight line as Mt. 1, Mt. 2, ..., Mt. N. Mr. Takahashi traversed the range from the west and Mr. Aoki from the east. The height of Mt. i is h_i, but they have forgotten the value of each h_i. Instead, for each i (1 ≤ i ≤ N), they recorded the maximum height of the mountains climbed up to the time they reached the peak of Mt. i (including Mt. i). Mr. Takahashi's record is T_i and Mr. Aoki's record is A_i. We know that the height of each mountain h_i is a positive integer. Compute the number of the possible sequences of the mountains' heights, modulo 10^9 + 7. Note that the records may be incorrect and thus there may be no possible sequence of the mountains' heights. In such a case, output 0.
|
from itertools import groupby
print(sum(1 for _ in groupby(input())) - 1)
|
s718065882
|
Accepted
| 159 | 19,012 | 1,065 |
N = int(input())
T = list(map(int, input().split()))
A = list(map(int, input().split()))
R = []
M = 1000000007
def main():
global R
global A
p = -1
for t in T:
if t > p: # new
R.append(t)
p = t
else:
R.append(-t)
R = list(reversed(R))
A = list(reversed(A))
p = -1
for i, t in enumerate(A):
r = R[i]
if r < 0:
max_h = -r
if t > p: # new
p = t
if t <= max_h:
R[i] = t
else:
return 0
else:
R[i] = -min(max_h, t)
else:
if t > p: # new
p = t
if t != r:
return 0
else:
if r > t:
return 0
answer = 1
for r in R:
if r > 0:
answer *= 1
else:
max_h = -r
answer *= max_h
answer %= M
return answer
print(main())
|
s851981588
|
p04029
|
u364741711
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 31 |
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
n=int(input())
print(n*(n+1)/2)
|
s385461886
|
Accepted
| 17 | 2,940 | 36 |
n=int(input())
print(int(n*(n+1)/2))
|
s671145511
|
p04011
|
u430223993
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 125 |
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())
k = int(input())
x = int(input())
y = int(input())
if n <= k:
print(n*x)
else:
print(n*x + (k-n)*y)
|
s445193069
|
Accepted
| 17 | 3,064 | 125 |
n = int(input())
k = int(input())
x = int(input())
y = int(input())
if n <= k:
print(n*x)
else:
print(k*x + (n-k)*y)
|
s628572240
|
p02612
|
u528484796
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 9,140 | 32 |
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())
k=n%1000
print(k)
|
s423274857
|
Accepted
| 31 | 9,152 | 68 |
from math import *
n=int(input())
k=n/1000
p=ceil(k)*1000
print(p-n)
|
s935387458
|
p02409
|
u879471116
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,600 | 476 |
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
A = []
for _ in range(4):
floor = []
for _ in range(3):
room = []
for _ in range(10):
room.append(0)
floor.append(room)
A.append(floor)
n = int(input())
for _ in range(n):
b, f, r, v = map(int, input().split())
A[b-1][f-1][r-1] += v
for b in A:
for f in b:
f = list(map(str, f))
print(' '.join(f))
print('#'*20)
|
s226781565
|
Accepted
| 20 | 5,600 | 266 |
A = []
for _ in range(12):
A.append([0]*10)
n = int(input())
for _ in range(n):
b, f, r, v = map(int, input().split())
A[3*b-(3-f)-1][r-1] += v
for i in range(12):
print(' ', end='')
print(' '.join(map(str, A[i])))
if i in [2, 5, 8]:
print('#'*20)
|
s711526601
|
p03474
|
u266874640
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 464 |
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.
|
A,B = map(int,input().split())
S = list(input())
c = 0
if len(S) != A+B+1:
print("No")
else:
for i in range(A):
if S[i] == "-":
print("No")
break
else:
c += 1
if c == 1 :
if S[A+1] != "-":
print("No")
c += 1
else:
for j in range(A+1,A+B+2):
if S[j] == "-":
print("No")
c += 1
break
if c == 1:
print("Yes")
|
s536904360
|
Accepted
| 20 | 3,188 | 462 |
A,B = map(int,input().split())
S = list(input())
c = 0
if len(S) != A+B+1:
print("No")
else:
for i in range(A):
if S[i] == "-":
print("No")
break
else:
c += 1
if c == A:
if S[A] != "-":
print("No")
c += 1
else:
for j in range(A+1,A+B+1):
if S[j] == "-":
print("No")
c += 1
break
if c == A:
print("Yes")
|
s927129706
|
p00003
|
u036236649
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,572 | 186 |
Write a program which judges wheather given length of three side form a right triangle. Print "YES" if the given sides (integers) form a right triangle, "NO" if not so.
|
import sys
input()
for line in sys.stdin:
a, b, c = map(int, line.split())
d = (a + b + c) / 2
if(a > d or b > d or c > d):
print("NO")
else:
print("YES")
|
s117988567
|
Accepted
| 40 | 7,616 | 225 |
N = int(input())
for i in range(N):
a, b, c = map(int, input().split())
if(a > c):
a, c = c, a
if(b > c):
b, c = c, b
if(c**2 == a**2 + b**2):
print("YES")
else:
print("NO")
|
s738902447
|
p03495
|
u142415823
| 2,000 | 262,144 |
Wrong Answer
| 189 | 32,540 | 654 |
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.
|
import collections
def calc(A, K):
dist = {}
tmp = collections.Counter(A)
type_ = len(tmp)
if type_ <= K:
return 0
for a, count in tmp.items():
if count not in dist.keys():
dist[count] = [a]
else:
dist[count].append(a)
rewrite = 0
print(dist)
for i, c in enumerate(sorted(dist.keys()), 1):
for _ in dist[c]:
rewrite += c
if i == type_ - K:
return rewrite
def main():
N, K = map(int, input().split())
A = list(map(int, input().split()))
ans = calc(A, K)
print(ans)
if __name__ == "__main__":
main()
|
s927464753
|
Accepted
| 138 | 20,184 | 179 |
N, K = map(int, input().split())
cnt = [0 for _ in range(N)]
for i in map(int, input().split()):
cnt[i-1] += 1
ans = 0
for i in sorted(cnt)[:N - K]:
ans += i
print(ans)
|
s861945184
|
p02601
|
u119015607
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,192 | 234 |
M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful.
|
A,B, C= list(map(int,input().split()))
K = int(input())
for i in range((K)):
if(B>C):
C=C*2
print(A,B,C)
elif(A>B):
B=B*2
print(A,B,C)
if (C>B >A):
print('Yes')
else:
print('No')
|
s859477171
|
Accepted
| 25 | 9,044 | 227 |
A,B, C= list(map(int,input().split()))
K = int(input())
for i in range((K)):
if(A>=B):
B=B*2
elif(B>=C):
C=C*2
else:
break
if (C>B >A):
print('Yes')
else:
print('No')
|
s954208814
|
p03228
|
u223904637
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 117 |
In the beginning, Takahashi has A cookies, and Aoki has B cookies. They will perform the following operation alternately, starting from Takahashi: * If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other person. Find the numbers of cookies Takahashi and Aoki respectively have after performing K operations in total.
|
a,b,k=map(int,input().split())
for i in range(k):
a-=a%2
b-=b%2
tmp=a//2
a=b//2
b=tmp
print(a,b)
|
s505499809
|
Accepted
| 17 | 2,940 | 178 |
a,b,k=map(int,input().split())
for i in range(k):
if i%2==0:
a-=a%2
b+=a//2
a=a//2
else:
b-=b%2
a+=b//2
b=b//2
print(a,b)
|
s195873281
|
p03388
|
u439542873
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 482 |
10^{10^{10}} participants, including Takahashi, competed in two programming contests. In each contest, all participants had distinct ranks from first through 10^{10^{10}}-th. The _score_ of a participant is the product of his/her ranks in the two contests. Process the following Q queries: * In the i-th query, you are given two positive integers A_i and B_i. Assuming that Takahashi was ranked A_i-th in the first contest and B_i-th in the second contest, find the maximum possible number of participants whose scores are smaller than Takahashi's.
|
from math import sqrt
q = int(input())
query = [tuple(map(int, input().split())) for _ in range(q)]
def sqrtfloor(x):
s = int(sqrt(x))
if s ** 2 == x:
s -= 1
return s
def solve(a, b):
a, b = min(a, b), max(a, b)
if a == b or a + 1 == b:
return 2 * a - 1
else:
s = sqrtfloor(a * b)
if s * (s + 1) < a * b:
return 2 * s - 1
else:
return 2 * s - 2
for a, b in query:
print(solve(a, b))
|
s049201298
|
Accepted
| 18 | 3,064 | 482 |
from math import sqrt
q = int(input())
query = [tuple(map(int, input().split())) for _ in range(q)]
def sqrtfloor(x):
s = int(sqrt(x))
if s ** 2 == x:
s -= 1
return s
def solve(a, b):
a, b = min(a, b), max(a, b)
if a == b or a + 1 == b:
return 2 * a - 2
else:
s = sqrtfloor(a * b)
if s * (s + 1) < a * b:
return 2 * s - 1
else:
return 2 * s - 2
for a, b in query:
print(solve(a, b))
|
s720207522
|
p00003
|
u841567836
| 1,000 | 131,072 |
Wrong Answer
| 40 | 5,612 | 351 |
Write a program which judges wheather given length of three side form a right triangle. Print "YES" if the given sides (integers) form a right triangle, "NO" if not so.
|
def judge(lists):
lists.sort()
if lists[0] ** 2 + lists[1] ** 2 - lists[2] ** 2:
return False
else:
return True
def run():
N = int(input())
for _ in range(N):
if judge([int(x) for x in input().split()]):
print("Yes")
else:
print("NO")
if __name__ == '__main__':
run()
|
s088669244
|
Accepted
| 40 | 5,612 | 351 |
def judge(lists):
lists.sort()
if lists[0] ** 2 + lists[1] ** 2 - lists[2] ** 2:
return False
else:
return True
def run():
N = int(input())
for _ in range(N):
if judge([int(x) for x in input().split()]):
print("YES")
else:
print("NO")
if __name__ == '__main__':
run()
|
s873907373
|
p02406
|
u539789745
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 147 |
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; }
|
def main():
n = int(input())
for i in range(3, n + 1, 3):
print("", i, end="")
print("")
if __name__ == "__main__":
main()
|
s510899902
|
Accepted
| 20 | 6,176 | 401 |
def main():
"""
n=30
3 6 9 12 13 15 18 21 23 24 27 30
"""
n = int(input())
for i in range(1, n + 1):
x = i
if x % 3 == 0:
print("", i, end="")
continue
while x > 0:
if x % 10 == 3:
print("", i, end="")
break
x //= 10
print("")
if __name__ == "__main__":
main()
|
s632719924
|
p03455
|
u291628833
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 20 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
s = input()
print(s)
|
s076235911
|
Accepted
| 17 | 2,940 | 82 |
a,b = map(int,input().split())
if a*b%2 == 0:
print("Even")
else:
print("Odd")
|
s948319094
|
p02694
|
u143051858
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 9,248 | 111 |
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())
num=100
cnt=0
while num < x:
num*=1.01
num=int(num)
print(num)
cnt+=1
print(cnt)
|
s318232138
|
Accepted
| 21 | 9,124 | 96 |
x=int(input())
num=100
cnt=0
while num < x:
num*=1.01
num=int(num)
cnt+=1
print(cnt)
|
s971698498
|
p02742
|
u962309487
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 76 |
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
import math
h, w = map(int, input().split())
ans = math.ceil(h*w)
print(ans)
|
s270070414
|
Accepted
| 18 | 2,940 | 122 |
import math
h, w = map(int, input().split())
if h == 1 or w == 1:
ans = 1
else:
ans = math.ceil(h*w/2)
print(ans)
|
s087135251
|
p03370
|
u076543276
| 2,000 | 262,144 |
Wrong Answer
| 41 | 3,064 | 199 |
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
|
N, X = [int(i) for i in input().split()]
m = [int(input()) for i in range(N)]
con = 0
for i in m:
X -= i
m.sort()
print(X)
while X > m[0]:
X -= m[0]
con+=1
print(len(m)+con)
|
s073801404
|
Accepted
| 40 | 3,060 | 193 |
N, X = [int(i) for i in input().split()]
m = [int(input()) for i in range(N)]
con = 0
for i in m:
X -= i
m.sort()
while X >= m[0]:
X -= m[0]
con+=1
print(len(m)+con)
|
s184496523
|
p04030
|
u223133214
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 211 |
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
s=list(input())
print(s)
ans=''
for a in s:
if ans=='' and a=='B':
continue
elif a=='B':
ans = ans.rstrip(ans[-1])
elif a=='1':
ans+='1'
else:
ans+='0'
print(ans)
|
s658698767
|
Accepted
| 20 | 2,940 | 215 |
s=list(input())
ans=[]
for a in s:
if len(ans)==0 and a=='B':
continue
elif a=='B':
del ans[-1]
elif a=='1':
ans.append('1')
else:
ans.append('0')
print(''.join(ans))
|
s940321603
|
p03599
|
u052332717
| 3,000 | 262,144 |
Wrong Answer
| 61 | 3,188 | 450 |
Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the beaker. * Operation 3: Put C grams of sugar into the beaker. * Operation 4: Put D grams of sugar into the beaker. In our experimental environment, E grams of sugar can dissolve into 100 grams of water. Snuke will make sugar water with the highest possible density. The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker. Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it. If there is more than one candidate, any of them will be accepted. We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent. Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water.
|
a,b,c,d,e,f = map(int,input().split())
w = set()
for x in range(0,f+1,100*a):
for y in range(0,f+1-x,100*b):
w.add(x+y)
s = set()
for x in range(0,f+1,c):
for y in range(0,f+1-x,d):
s.add(x+y)
ans = 0
sugerwater = 0
suger = 0
for x in w:
for y in s:
if 0<x+y<=f and 100*y/(x+y) <= e:
if ans < y/(x+y):
ans = y/(x+y)
sugerwater = x+y
suger = y
print(x,y)
|
s946438014
|
Accepted
| 63 | 3,188 | 461 |
a,b,c,d,e,f = map(int,input().split())
w = set()
for x in range(0,f+1,100*a):
for y in range(0,f+1-x,100*b):
w.add(x+y)
s = set()
for x in range(0,f+1,c):
for y in range(0,f+1-x,d):
s.add(x+y)
ans = -1
sugerwater = 0
suger = 0
for x in w:
for y in s:
if 0<x+y<=f and y <= x*e/100:
if ans < y/(x+y):
ans = y/(x+y)
sugerwater = x+y
suger = y
print(sugerwater,suger)
|
s728872674
|
p03836
|
u780698286
| 2,000 | 262,144 |
Wrong Answer
| 26 | 9,088 | 167 |
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
sx, sy, tx, ty = map(int, input().split())
print("U"*(ty-sy) + "R"*(tx-sx+1) + "D"*(ty-sx+1) + "U" + "L" + "U"*(ty-sy+1) + "R"*(tx-sx+1) + "D"*(ty-sy+1) + "L"*(tx-sx))
|
s503953983
|
Accepted
| 25 | 9,204 | 228 |
xy = list(map(int, input().split()))
print("U"*(xy[3]-xy[1]) + "R"*(xy[2]-xy[0]) + "D"*(xy[3]-xy[1]) + "L"*(xy[2]-xy[0]+1) + "U"*(xy[3]-xy[1]+1) + "R"*(xy[2]-xy[0]+1) + "D" + "R"+ "D"*(xy[3]-xy[1]+1) + "L"*(xy[2]-xy[0]+1) + "U")
|
s114058636
|
p03795
|
u886902015
| 2,000 | 262,144 |
Wrong Answer
| 26 | 9,096 | 32 |
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
n=int(input())
print(800*n-n%15)
|
s342632619
|
Accepted
| 24 | 9,128 | 39 |
n=int(input())
print(800*n-200*(n//15))
|
s670983134
|
p03997
|
u983918956
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,316 | 77 |
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())
ans = (a+b) * h
print(ans)
|
s177801054
|
Accepted
| 18 | 2,940 | 78 |
a = int(input())
b = int(input())
h = int(input())
s = (a+b) * h // 2
print(s)
|
s699206819
|
p03730
|
u766407523
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 119 |
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
A, B, C = map(int, input().split())
for i in range(1, B+1):
if A*i%B==C:
print("YES")
else:
print("NO")
|
s263260932
|
Accepted
| 17 | 2,940 | 134 |
A, B, C = map(int, input().split())
for i in range(1, B+1):
if A*i%B==C:
print("YES")
break
else:
print("NO")
|
s366452366
|
p03759
|
u786020649
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,084 | 79 |
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
a,b,c=map(int,input().split())
if b-c==b-a:
print('YES')
else:
print('NO')
|
s460121217
|
Accepted
| 27 | 9,008 | 80 |
a,b,c=map(int,input().split())
if b-c==a-b:
print('YES')
else:
print('NO')
|
s089423647
|
p02612
|
u171255092
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 9,144 | 32 |
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)
|
s801821270
|
Accepted
| 28 | 9,176 | 41 |
n = int(input())
print((1000 - n) % 1000)
|
s325381229
|
p03759
|
u931489673
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 82 |
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
a,b,c=map(int,input().split())
if (a-b)==(c-b):
print("YES")
else:
print("NO")
|
s938914536
|
Accepted
| 17 | 2,940 | 83 |
a,b,c=map(int,input().split())
if (b-a)==(c-b):
print("YES")
else:
print("NO")
|
s832474773
|
p02606
|
u529737989
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,180 | 67 |
How many multiples of d are there among the integers between L and R (inclusive)?
|
L,R,N = map(int,input().split())
Ln = L//N
Rn = R//N
print(Rn-Ln+1)
|
s534354038
|
Accepted
| 31 | 9,152 | 69 |
L,R,N = map(int,input().split())
Ln = (L-1)//N
Rn = R//N
print(Rn-Ln)
|
s038180152
|
p03836
|
u100641536
| 2,000 | 262,144 |
Wrong Answer
| 29 | 9,164 | 154 |
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())
dx=tx-sx
dy=ty-sy
s="U"*dy+"R"*dx+"D"*dy+"L"*(dx+1)+"U"*(dy+1)+"R"*(dx+1)+"D"+"R"+"D"*(dy+1)+"L"*(dx+1)
print(s)
|
s281698814
|
Accepted
| 28 | 9,156 | 158 |
sx,sy,tx,ty = map(int,input().split())
dx=tx-sx
dy=ty-sy
s="U"*dy+"R"*dx+"D"*dy+"L"*(dx+1)+"U"*(dy+1)+"R"*(dx+1)+"D"+"R"+"D"*(dy+1)+"L"*(dx+1)+"U"
print(s)
|
s483950038
|
p01304
|
u798803522
| 8,000 | 131,072 |
Wrong Answer
| 30 | 7,812 | 1,360 |
平安京は、道が格子状になっている町として知られている。 平安京に住んでいるねこのホクサイは、パトロールのために毎日自宅から町はずれの秘密の場所まで行かなければならない。しかし、毎日同じ道を通るのは飽きるし、後を付けられる危険もあるので、ホクサイはできるだけ毎日異なる経路を使いたい。その一方で、ホクサイは面倒臭がりなので、目的地から遠ざかるような道は通りたくない。 平安京のあちこちの道にはマタタビが落ちていて、ホクサイはマタタビが落ちている道を通ることができない。そのような道を通るとめろめろになってしまうからである。幸いなことに、交差点にはマタタビは落ちていない。 ホクサイは、自宅から秘密の場所までの可能な経路の数を知りたい。ここで、ホクサイの自宅は (0, 0) にあり、秘密の場所は(gx, gy)にある。道は x = i (i は整数), y = j (j は整数) に格子状に敷かれている。
|
trial = int(input())
for t in range(trial):
targ = [int(n) for n in input().split(' ')]
root = [[0 for n in range(targ[0] + 1)] for m in range(targ[1] + 1)]
matanum = int(input())
for m in range(matanum):
matax,matay,secx,secy = (int(n) for n in input().split(' '))
if matax == secx:
root[max(matay,secy)][matax] = 'y'
else:
root[matay][max(secx,matax)] = 'x'
root[0][0] = 1
for yaxis in range(targ[1] + 1):
for xaxis in range(targ[0] + 1):
if xaxis == 0:
if root[yaxis][xaxis] == 'y':
root[yaxis][xaxis] = 0
else:
root[yaxis][xaxis] = 1
elif yaxis == 0:
if root[yaxis][xaxis] == 'x':
root[yaxis][xaxis] = 0
else:
root[yaxis][xaxis] = root[yaxis][xaxis-1]
else:
if root[yaxis][xaxis] == 'y':
root[yaxis][xaxis] = root[yaxis][xaxis - 1]
elif root[yaxis][xaxis] == 'x':
root[yaxis][xaxis] = root[yaxis - 1][xaxis]
else:
root[yaxis][xaxis] = root[yaxis - 1][xaxis] + root[yaxis][xaxis - 1]
if root[targ[1]][targ[0]] == 0:
print("Miserable Hokusai!")
else:
print(root[targ[1]][targ[0]])
|
s415333471
|
Accepted
| 40 | 7,752 | 1,884 |
trial = int(input())
for t in range(trial):
targ = [int(n) for n in input().split(' ')]
root = [[0 for n in range(targ[0] + 1)] for m in range(targ[1] + 1)]
matanum = int(input())
for m in range(matanum):
matax,matay,secx,secy = (int(n) for n in input().split(' '))
if matax == secx:
if root[max(matay,secy)][matax] == 'x':
root[max(matay,secy)][matax] = 'xy'
else:
root[max(matay,secy)][matax] = 'y'
else:
if root[matay][max(matax,secx)] == 'y':
root[matay][max(matax,secx)] = 'xy'
else:
root[matay][max(matax,secx)] = 'x'
for yaxis in range(targ[1] + 1):
for xaxis in range(targ[0] + 1):
if xaxis == 0:
if root[yaxis][xaxis] == 'y' or root[yaxis][xaxis] == 'xy':
root[yaxis][xaxis] = 0
else:
if yaxis == 0:
root[0][0] = 1
else:
root[yaxis][xaxis] = root[yaxis- 1][xaxis]
elif yaxis == 0:
if root[yaxis][xaxis] == 'x' or root[yaxis][xaxis] == 'xy':
root[yaxis][xaxis] = 0
else:
root[yaxis][xaxis] = root[yaxis][xaxis-1]
else:
if root[yaxis][xaxis] == 'xy':
root[yaxis][xaxis] = 0
elif root[yaxis][xaxis] == 'y':
root[yaxis][xaxis] = root[yaxis][xaxis - 1]
elif root[yaxis][xaxis] == 'x':
root[yaxis][xaxis] = root[yaxis - 1][xaxis]
else:
root[yaxis][xaxis] = root[yaxis - 1][xaxis] + root[yaxis][xaxis - 1]
if root[targ[1]][targ[0]] == 0:
print("Miserable Hokusai!")
else:
print(root[targ[1]][targ[0]])
|
s150230032
|
p00019
|
u011621222
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,480 | 115 |
Write a program which reads an integer n and prints the factorial of n. You can assume that n ≤ 20\.
|
factorial= 1
number= int(input("Enter a number:"))
for r in range (1,number+1):
factorial*=r
print(factorial)
|
s758632174
|
Accepted
| 20 | 5,560 | 136 |
def fac(n):
ans=1
for i in range(n+1):
if i is not 0:
ans*=i
return ans
n = eval(input())
print(fac(n))
|
s580349496
|
p03447
|
u947327691
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 58 |
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)
|
s745207444
|
Accepted
| 17 | 2,940 | 60 |
x=int(input())
a=int(input())
b=int(input())
print((x-a)%b)
|
s068583403
|
p03005
|
u671869417
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 114 |
Takahashi is distributing N balls to K persons. If each person has to receive at least one ball, what is the maximum possible difference in the number of balls received between the person with the most balls and the person with the fewest balls?
|
N, K = input().split()
N, K = int(N), int(K)
if K==1:
print(0)
if K==2:
print(N-1)
else: print((N-(K-1))-1)
|
s580574146
|
Accepted
| 17 | 2,940 | 91 |
N, K = input().split()
N, K = int(N), int(K)
if K==1:
print(0)
else: print((N-(K-1))-1)
|
s164548910
|
p04011
|
u480568292
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 172 |
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_ = int(input())
m = int(input())
m_ = int(input())
count = 0
N = n-n_
for i in range(N):
count += m
for x in range(n_):
count += m_
print(count)
|
s544951979
|
Accepted
| 17 | 2,940 | 68 |
n,k,x,y = (int(input()) for i in [0]*4)
print(n*x-(x-y)*max(n-k,0))
|
s876284303
|
p03448
|
u490623664
| 2,000 | 262,144 |
Wrong Answer
| 52 | 3,060 | 186 |
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 j in range(b+1):
for k in range(c+1):
if x==500*a+100*b+50*c:
count+=1
print(count)
|
s353147281
|
Accepted
| 49 | 3,060 | 186 |
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==500*i+100*j+50*k:
count+=1
print(count)
|
s757160623
|
p00275
|
u724548524
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 324 |
百人一首の札を使った遊戯の１つに、「坊主めくり」というものがあります。絵札だけを使う簡単な遊戯なので広く楽しまれています。きまりには様々な派生型がありますが、ここで考える坊主めくりはN人の参加者で、以下のようなルールで行います。 * ６４枚の「男」、１５枚の「坊主」、２１枚の「姫」、計１００枚の札を使う。 * 絵が見えないように札を裏がえしにしてよく混ぜ、「札山」をつくる。 * 参加者の一人目から順番に１枚ずつ札山の札を引く。N人目の次は、また一人目から繰り返す。 * 引いた札が男なら、引いた人はその札を手に入れる。 * 引いた札が坊主なら、引いた人はその札を含め、持っている札をすべて「場」に出す。 * 引いた札が姫なら、引いた人はその札を含め、場にある札をすべて手に入れる。 * 札山の札がなくなったら終了で、一番多くの札を持っている人の勝ち。 参加者数と札山に積まれた札の順番が与えられたとき、遊戯が終了した時点で各参加者が持っている札数を昇順で並べたものと、場に残っている札数を出力するプログラムを作成してください。
|
while 1:
n = int(input())
if n == 0:break
p = [0] * n
s = 0
f = input()
for i in range(100):
if f[i] == "M":p[i % n] += 1
elif f[i] == "S":
s += p[i % n] + 1
p[i % n] = 0
else:
p[i % n] += s + 1
s = 0
p += [s]
print(*p)
|
s347330727
|
Accepted
| 30 | 5,608 | 337 |
while 1:
n = int(input())
if n == 0:break
p = [0] * n
s = 0
f = input()
for i in range(100):
if f[i] == "M":p[i % n] += 1
elif f[i] == "S":
s += p[i % n] + 1
p[i % n] = 0
else:
p[i % n] += s + 1
s = 0
p.sort()
p += [s]
print(*p)
|
s385016308
|
p02399
|
u485986915
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 69 |
Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)
|
a,b = map(int,input().split())
d = a/b
r = a%b
f = a/b
print(d,r,f)
|
s778282683
|
Accepted
| 20 | 5,600 | 86 |
a,b = map(int,input().split())
d = a//b
r = a%b
f = a/b
print('%s %s %.5f'%(d,r,f))
|
s686227064
|
p03962
|
u367373844
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 252 |
AtCoDeer the deer recently bought three paint cans. The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c. Here, the color of each paint can is represented by an integer between 1 and 100, inclusive. Since he is forgetful, he might have bought more than one paint can in the same color. Count the number of different kinds of colors of these paint cans and tell him.
|
candy_bag = list(map(int, input().split()))
max_bag_num = max(candy_bag)
bag = 0
small_bag_sum = 0
for bag in candy_bag:
if bag == max_bag_num:
continue
else:
small_bag_sum += bag
if max_bag_num == small_bag_sum:
print("Yes")
else:
print("No")
|
s410255828
|
Accepted
| 17 | 3,060 | 211 |
a,b,c = map(int,input().split())
same_count = 0
if a == b:
same_count += 1
if a == c:
same_count += 1
if b == c:
same_count += 1
if same_count == 0:
print(3)
elif same_count == 1:
print(2)
else:
print(1)
|
s059202245
|
p02612
|
u189397279
| 2,000 | 1,048,576 |
Wrong Answer
| 32 | 9,136 | 56 |
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)
|
s117013839
|
Accepted
| 29 | 9,156 | 115 |
N = int(input())
for i in range(13):
if 1000 * i < N and N <= 1000 * (i + 1):
print(1000 * (i + 1) - N)
|
s809856141
|
p02850
|
u691018832
| 2,000 | 1,048,576 |
Wrong Answer
| 530 | 44,892 | 928 |
Given is a tree G with N vertices. The vertices are numbered 1 through N, and the i-th edge connects Vertex a_i and Vertex b_i. Consider painting the edges in G with some number of colors. We want to paint them so that, for each vertex, the colors of the edges incident to that vertex are all different. Among the colorings satisfying the condition above, construct one that uses the minimum number of colors.
|
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
n = int(readline())
ab = [list(map(int, readline().split())) for i in range(n - 1)]
gragh = [[] for j in range(n + 1)]
for a, b in ab:
gragh[a].append(b)
gragh[b].append(a)
root = 1
parent = [0] * (n + 1)
stack = [root]
order = []
while stack:
x = stack.pop()
order.append(x)
for y in gragh[x]:
if y == parent[x]:
continue
parent[y] = x
stack.append(y)
color = [-1] * (n + 1)
for x in order:
ng = color[x]
cnt = 1
for y in gragh[x]:
if y == parent[x]:
continue
if cnt == ng:
cnt += 1
color[y] = cnt
cnt += 1
ans = []
append = ans.append
for a, b in ab:
if parent[a] == b:
append(color[a])
else:
append(color[b])
print(max(ans))
print('/n'.join(map(str, ans)))
|
s933105721
|
Accepted
| 412 | 45,704 | 826 |
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)
from collections import deque
n = int(readline())
graph = [[] for _ in range(n + 1)]
for i in range(n - 1):
a, b = map(int, readline().split())
graph[a].append((b, i))
graph[b].append((a, i))
color = [-1] * (n + 1)
color[1] = -2
ans = [0] * (n - 1)
def bfs(s):
stack = deque([s])
while stack:
num = 1
now = stack.pop()
for next, idx in graph[now]:
if color[next] != -1:
continue
if color[now] == num:
num += 1
color[next] = num
ans[idx] = num
num += 1
stack.append(next)
bfs(1)
print(max(ans))
print('\n'.join(map(str, ans)))
|
s849387707
|
p03371
|
u974935538
| 2,000 | 262,144 |
Wrong Answer
| 122 | 8,772 | 318 |
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
|
A,B,C,X,Y = map(int,input().split())
cnt_A = X
cnt_B = Y
cnt_C = 0
ans = []
for _ in range(X+Y):
if 2*C<=A+B:
cnt_C += 2
cnt_A -= 1
cnt_B -= 1
ans.append(cnt_A*A+cnt_B*B+cnt_C*C)
if ((cnt_A<0) or (cnt_B<0)) or ((cnt_C==2*X)or(cnt_C ==Y*22)):
break
ans.sort()
print(ans[0])
|
s955515432
|
Accepted
| 113 | 3,060 | 171 |
a, b, c, x, y = map(int, input().split())
ans = 10000000007
for i in range(10**5 + 1):
ans = min(ans, i * (2 * c) + max(0, x - i) * a + max(0, y - i) * b)
print(ans)
|
s751448900
|
p00046
|
u546285759
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,400 | 111 |
今まで登ったことのある山の標高を記録したデータがあります。このデータを読み込んで、一番高い山と一番低い山の標高差を出力するプログラムを作成してください。
|
import sys
ms = []
try:
for v in sys.stdin:
ms.append(float(v))
except:
print(max(ms)-min(ms))
|
s697695862
|
Accepted
| 20 | 7,332 | 207 |
maxv, minv = 0, 10**6
while True:
try:
height = float(input())
except:
break
maxv = height if maxv < height else maxv
minv = height if minv > height else minv
print(maxv-minv)
|
s178940969
|
p03377
|
u413165887
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 93 |
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
a, b, x = map(int, input().split())
if x<=a or a+b<=x:
print("YES")
else:
print("NO")
|
s329604838
|
Accepted
| 18 | 2,940 | 91 |
a, b, x = map(int, input().split())
if a+b<x or a>x:
print("NO")
else:
print("YES")
|
s084155396
|
p03796
|
u026788530
| 2,000 | 262,144 |
Wrong Answer
| 45 | 2,940 | 70 |
Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.
|
n=int(input())
ans=1
for i in range(n):
ans*=(i+1)
ans%=1000000007
|
s584837501
|
Accepted
| 43 | 2,940 | 84 |
n=int(input())
ans=1
for i in range(n):
ans*=(i+1)
ans%=1000000007
print(ans)
|
s065974721
|
p03455
|
u384935968
| 2,000 | 262,144 |
Wrong Answer
| 24 | 9,008 | 84 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b=map(int,input().split())
if (a*b) % 2 == 0:
print('Odd')
else:
print('Even')
|
s891315227
|
Accepted
| 21 | 8,984 | 83 |
a,b=map(int,input().split())
if (a*b) % 2 == 0:
print('Even')
else:
print('Odd')
|
s108892461
|
p03739
|
u223663729
| 2,000 | 262,144 |
Wrong Answer
| 213 | 14,084 | 335 |
You are given an integer sequence of length N. The i-th term in the sequence is a_i. In one operation, you can select a term and either increment or decrement it by one. At least how many operations are necessary to satisfy the following conditions? * For every i (1≤i≤n), the sum of the terms from the 1-st through i-th term is not zero. * For every i (1≤i≤n-1), the sign of the sum of the terms from the 1-st through i-th term, is different from the sign of the sum of the terms from the 1-st through (i+1)-th term.
|
n, *A = map(int, open(0).read().split())
def sgn(n):
return 0 if n==0 else 1 if n>0 else -1
C = [0, 0]
S = [1, -1]
for a in A:
for i, s in enumerate(S):
sgn_sum = sgn(s)
if sgn(s+a) != sgn_sum:
S[i] += a
else:
C[i] += abs(s+a+sgn_sum)
S[i] = -sgn_sum
print(min(C))
|
s481400204
|
Accepted
| 213 | 14,084 | 398 |
n, *A = map(int, open(0).read().split())
def sgn(n):
return 0 if n==0 else 1 if n>0 else -1
a = A[0]
C = [0 if a>0 else -a+1, 0 if a<0 else a+1]
S = [max(1, a), min(-1, a)]
for a in A[1:]:
for i, s in enumerate(S):
sgn_sum = sgn(s)
if sgn(s+a) == -sgn_sum:
S[i] += a
else:
C[i] += abs(s+a+sgn_sum)
S[i] = -sgn_sum
print(min(C))
|
s715259121
|
p04029
|
u479788474
| 2,000 | 262,144 |
Wrong Answer
| 26 | 8,944 | 116 |
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())
# 1 <= N <= 100
total = N * ( N + 1)/ 2
print(total)
|
s000983576
|
Accepted
| 26 | 9,072 | 118 |
N = int(input())
# 1 <= N <= 100
total = N * ( N + 1) // 2
print(total)
|
s642011604
|
p03943
|
u478888559
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 227 |
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a, b, c = map(int, input().split())
candies = []
candies.append(a)
candies.append(b)
candies.append(c)
max_candy = max(candies)
candies.remove(max_candy)
if sum(candies) == max_candy:
print("YES")
else:
print("NO")
|
s817300677
|
Accepted
| 17 | 2,940 | 227 |
a, b, c = map(int, input().split())
candies = []
candies.append(a)
candies.append(b)
candies.append(c)
max_candy = max(candies)
candies.remove(max_candy)
if sum(candies) == max_candy:
print("Yes")
else:
print("No")
|
s991494708
|
p03730
|
u212328220
| 2,000 | 262,144 |
Wrong Answer
| 2,297 | 64,220 | 274 |
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`.
|
import itertools
A, B, C = map(int,input().split())
lst = []
for i in range(1,100):
lst.append(A * i)
for i in range(1,5):
for v in itertools.combinations(lst,i):
print(v)
if sum(v) % B == C:
print('YES')
exit()
print('NO')
|
s857588609
|
Accepted
| 31 | 9,024 | 257 |
import itertools
A, B, C = map(int,input().split())
lst = []
for i in range(1,100):
lst.append(A * i)
for i in range(1,3):
for v in itertools.combinations(lst,i):
if sum(v) % B == C:
print('YES')
exit()
print('NO')
|
s594578966
|
p03418
|
u466331465
| 2,000 | 262,144 |
Wrong Answer
| 250 | 4,456 | 216 |
Takahashi had a pair of two positive integers not exceeding N, (a,b), which he has forgotten. He remembers that the remainder of a divided by b was greater than or equal to K. Find the number of possible pairs that he may have had.
|
N,K = [int(x) for x in input().split()]
ans =0
for i in range(K+1,N+1):
print((N//i)*(i-K),N%i+1-K)
ans += (N//i)*(i-K)
if N%i!=0 and K!=0:
ans +=(N%i+1-K)
elif N%i!=0 and K==0:
ans +=(N%i)
print(ans)
|
s992130624
|
Accepted
| 96 | 3,188 | 224 |
N,K = [int(x) for x in input().split()]
ans =0
for i in range(K+1,N+1):
ans += (N//i)*(i-K)
if N%i!=0 and K!=0:
ans +=max((N%i+1-K),0)
elif N%i!=0 and K==0:
ans +=(N%i)
print(ans)
|
s234409705
|
p03575
|
u133377913
| 2,000 | 262,144 |
Wrong Answer
| 170 | 12,596 | 1,149 |
You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges. The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i. An edge whose removal disconnects the graph is called a _bridge_. Find the number of the edges that are bridges among the M edges.
|
# -*- coding: utf-8 -*-
import numpy as np
N, M = map(int, input().split())
L = [list(map(int, input().split())) for _ in range(M)]
adjacent = np.zeros((N, N))
for l in L:
adjacent[l[0] - 1, l[1] - 1] = 1
adjacent[l[1] - 1, l[0] - 1] = 1
print(adjacent, end='\n\n')
pre = [-1 for _ in range(N)]
low = [-1 for _ in range(N)]
v = k = 0
ans = 0
def dfs(v, pre, low, k, parent):
global ans
pre[v] = k
k += 1
low[v] = pre[v]
for to, n in enumerate(adjacent[v, :]):
if n == 1:
if pre[to] == -1:
# destination has not been visited
# visit destination and update low[v]
# print('-----', v, '-------')
# print(pre, low)
low[v] = min(low[v], dfs(to, pre, low, k, v))
if low[to] == pre[to]:
ans += 1
# print(v + 1, to + 1)
else:
# visited
if to == parent:
pass
else:
low[v] = min(low[v], low[to])
return low[v]
dfs(v, pre, low, k, -1)
# print(pre, low)
print(ans)
|
s332385394
|
Accepted
| 152 | 12,476 | 1,151 |
# -*- coding: utf-8 -*-
import numpy as np
N, M = map(int, input().split())
L = [list(map(int, input().split())) for _ in range(M)]
adjacent = np.zeros((N, N))
for l in L:
adjacent[l[0] - 1, l[1] - 1] = 1
adjacent[l[1] - 1, l[0] - 1] = 1
# print(adjacent, end='\n\n')
pre = [-1 for _ in range(N)]
low = [-1 for _ in range(N)]
v = k = 0
ans = 0
def dfs(v, pre, low, k, parent):
global ans
pre[v] = k
k += 1
low[v] = pre[v]
for to, n in enumerate(adjacent[v, :]):
if n == 1:
if pre[to] == -1:
# destination has not been visited
# visit destination and update low[v]
# print('-----', v, '-------')
# print(pre, low)
low[v] = min(low[v], dfs(to, pre, low, k, v))
if low[to] == pre[to]:
ans += 1
# print(v + 1, to + 1)
else:
# visited
if to == parent:
pass
else:
low[v] = min(low[v], low[to])
return low[v]
dfs(v, pre, low, k, -1)
# print(pre, low)
print(ans)
|
s094651095
|
p03370
|
u476048753
| 2,000 | 262,144 |
Wrong Answer
| 2,103 | 3,064 | 277 |
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
|
N, X = map(int, input().split())
powder_list = [int(input()) for i in range(N)]
min_powder = min(powder_list)
ans = 0
for i in range(N):
X -= powder_list[i]
if X >= 0:
ans += 1
else:
break
while X < min_powder:
X -= min_powder
ans += 1
print(ans)
|
s676882834
|
Accepted
| 18 | 2,940 | 101 |
N, X = map(int, input().split())
m = [int(input()) for i in range(N)]
print(N + (X-sum(m))//min(m))
|
s037226643
|
p04029
|
u681502232
| 2,000 | 262,144 |
Wrong Answer
| 26 | 8,964 | 150 |
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?
|
#043A
x=input()
i=int(x)
sum=0
while i>0:
sum=sum+i
i=i-1
print(sum)
|
s132361421
|
Accepted
| 24 | 9,044 | 160 |
#043A
x=input()
i=int(x)
sum=0
while i>0:
sum=sum+i
i=i-1
if i==0 :
print(sum)
|
s890645865
|
p03999
|
u403331159
| 2,000 | 262,144 |
Wrong Answer
| 22 | 3,444 | 221 |
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion. All strings that can be obtained in this way can be evaluated as formulas. Evaluate all possible formulas, and print the sum of the results.
|
s=input()
n=len(s)-1
ans=0
for bit in range(1<<n):
f=s[0]
for i in range(n):
if ((bit>>i)&1):
f+="+"
f+=s[i+1]
for i in f.split("+"):
ans+=int(i)
print(i)
print(ans)
|
s441514804
|
Accepted
| 21 | 3,060 | 204 |
s=input()
n=len(s)-1
ans=0
for bit in range(1<<n):
f=s[0]
for i in range(n):
if ((bit>>i)&1):
f+="+"
f+=s[i+1]
for i in f.split("+"):
ans+=int(i)
print(ans)
|
s942582541
|
p03486
|
u382303205
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 1,239 |
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
n = input()
m = input()
flag = True
if len(n) < len(m):
for i in range(len(n)):
if n[i] == m[i]:
continue
else:
for j in "abcdefghijklmnopqrstuvwxyz":
if n[i] == j:
flag = True
break
if m[i] == j:
flag = False
break
break
else:
flag = False
elif len(n) == len(m):
for i in range(len(n)):
if n[i] == m[i]:
continue
else:
for j in "abcdefghijklmnopqrstuvwxyz":
if n[i] == j:
flag = True
break
if m[i] == j:
flag = False
break
break
else:
flag = False
else:
for i in range(len(m)):
if n[i] == m[i]:
continue
else:
for j in "abcdefghijklmnopqrstuvwxyz":
if n[i] == j:
flag = True
break
if m[i] == j:
flag = False
break
break
else:
flag = False
if flag:
print("Yes")
else:
print("No")
|
s309061909
|
Accepted
| 17 | 3,064 | 1,283 |
n = input()
m = input()
n = sorted(n)
m = sorted(m, reverse = True)
flag = True
if len(n) < len(m):
for i in range(len(n)):
if n[i] == m[i]:
continue
else:
for j in "abcdefghijklmnopqrstuvwxyz":
if n[i] == j:
flag = True
break
if m[i] == j:
flag = False
break
break
else:
flag = True
elif len(n) == len(m):
for i in range(len(n)):
if n[i] == m[i]:
continue
else:
for j in "abcdefghijklmnopqrstuvwxyz":
if n[i] == j:
flag = True
break
if m[i] == j:
flag = False
break
break
else:
flag = False
else:
for i in range(len(m)):
if n[i] == m[i]:
continue
else:
for j in "abcdefghijklmnopqrstuvwxyz":
if n[i] == j:
flag = True
break
if m[i] == j:
flag = False
break
break
else:
flag = False
if flag:
print("Yes")
else:
print("No")
|
s195190172
|
p02277
|
u530663965
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,604 | 932 |
Let's arrange a deck of cards. Your task is to sort totally n cards. A card consists of a part of a suit (S, H, C or D) and an number. Write a program which sorts such cards based on the following pseudocode: Partition(A, p, r) 1 x = A[r] 2 i = p-1 3 for j = p to r-1 4 do if A[j] <= x 5 then i = i+1 6 exchange A[i] and A[j] 7 exchange A[i+1] and A[r] 8 return i+1 Quicksort(A, p, r) 1 if p < r 2 then q = Partition(A, p, r) 3 run Quicksort(A, p, q-1) 4 run Quicksort(A, q+1, r) Here, A is an array which represents a deck of cards and comparison operations are performed based on the numbers. Your program should also report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
def main():
cards = []
for _ in range(int(input())):
cards.append(Card(input()))
base_cards = cards.copy()
quickSort(cards, 0, len(cards) - 1)
for c in cards:
print(c.cha, c.num)
print(check_stable(base_cards, cards))
def quickSort(A, p, r):
if p < r:
q = partiton(A, p, r)
quickSort(A, p, q - 1)
quickSort(A, q + 1, r)
def partiton(A, p, r):
x = A[-1]
i = p - 1
for j in range(p, r):
if A[j].num <= x.num:
i = i + 1
A[i], A[j] = A[j], A[i]
A[i + 1], A[r] = A[r], A[i + 1]
return i + 1
def check_stable(base, card):
for i in range(0, len(card) - 1):
if base.index(card[i]) > base.index(card[i + 1]):
return 'Not stable'
return 'Stable'
class Card:
def __init__(self, imp):
cha, num = imp.split(' ')
self.cha = cha
self.num = int(num)
main()
|
s640450554
|
Accepted
| 850 | 24,168 | 1,619 |
"""Quick Sort."""
def partition(A, p, r):
"""Divide list A into A[p:q] whose elements aren't greater than A[q] and
A[q+1:r] whose elements are greater than A[q].
Each element of A is a list [x, y].
x is a character of S, H, C or D (trump suits).
y is a natural number.
Sorting is done based on y numbers.
Default value of p is 0.
The number as a refernce of the division is A[r].
Return the index q.
"""
x = A[r][1]
i = p - 1
for j in range(p, r):
if A[j][1] <= x:
i += 1
A[i], A[j] = A[j], A[i]
A[i + 1], A[r] = A[r], A[i + 1]
return i + 1
def quickSort(A, p, r):
"""Sort list A in ascending order with quick sort algorithm.
Default value of p is 0 (the start index).
r is 1 less than the length of A (the last index of A).
"""
if p < r:
q = partition(A, p, r)
quickSort(A, p, q - 1)
quickSort(A, q + 1, r)
def is_stable(A, B):
"""Check the stability of sorted list A.
A is a sorted list.
B is a original list.
Return True or False.
"""
cA = list(A)
for x in B:
i = cA.index(x)
if i == 0:
pass
elif cA[i - 1][1] == x[1]:
return False
del cA[i]
return True
import sys
r = int(sys.stdin.readline()) - 1
A = []
for x in sys.stdin.readlines():
suit, num = x.split()
num = int(num)
A.append([suit, num])
B = list (A)
quickSort(A, 0, r)
if is_stable(A, B):
print('Stable')
else:
print('Not stable')
for x in A:
print(*x)
|
s061529991
|
p03401
|
u802963389
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 13,920 | 280 |
There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis. You planned a trip along the axis. In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0. However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i. You will visit the remaining spots as planned in the order they are numbered. You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned. For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.
|
N = int(input())
a = list(map(int, input().split()))
a.insert(0, 0)
a.append(0)
print(a)
for i in range(1, N + 1):
s = 0
for j in range(N + 1):
if j + 1 == i:
s += abs(a[j] - a[j+2])
elif j != i:
s += abs(a[j] - a[j+1])
print(s)
|
s816982503
|
Accepted
| 184 | 14,048 | 257 |
N = int(input())
a = list(map(int, input().split()))
a.insert(0, 0)
a.append(0)
# print(a)
li = [abs(a[i] - a[i+1]) for i in range(N + 1)]
sumli = sum(li)
for i in range(1, N + 1):
print(sumli - (li[i-1] + li[i]) + abs(a[i-1]-a[i+1]))
|
s599314038
|
p02741
|
u671211357
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 118 |
Print the K-th element of the following sequence of length 32: 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51
|
ans=[1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51]
print(ans[31])
|
s199852853
|
Accepted
| 17 | 2,940 | 135 |
ans=[1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51]
K=int(input())
print(ans[K-1])
|
s458828491
|
p03712
|
u818050295
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 326 |
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
a = input().split(" ")
tmp_out=[[""] for i in range(int(a[0])+2)]
for j in range(int(a[1])+2):
tmp_out[0]=["".join(tmp_out[0]+["#"])]
for i in range(1, int(a[0])+1):
tmp=input().split(" ")
tmp_out[i]=["".join(["#"]+tmp+["#"])]
tmp_out[int(a[0])+1]=tmp_out[0]
for i in range(int(a[0])+2):
print(tmp_out[i])
|
s460244882
|
Accepted
| 18 | 3,064 | 323 |
a = input().split(" ")
tmp_out=[[""] for i in range(int(a[0])+2)]
for j in range(int(a[1])+2):
tmp_out[0]=["".join(tmp_out[0]+["#"])]
for i in range(1, int(a[0])+1):
tmp_out[i]=["".join(["#"]+input().split(" ")+["#"])]
tmp_out[int(a[0])+1]=tmp_out[0]
for i in range(int(a[0])+2):
print("".join(tmp_out[i]))
|
s906197948
|
p03610
|
u427344224
| 2,000 | 262,144 |
Wrong Answer
| 29 | 3,572 | 76 |
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
s = input()
s = [h for i, h in enumerate(s) if i % 2 != 0]
print("".join(s))
|
s764378743
|
Accepted
| 30 | 3,572 | 79 |
s = input()
s = [h for i, h in enumerate(s, 1) if i % 2 != 0]
print("".join(s))
|
s095143989
|
p03696
|
u940102677
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 203 |
You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_. Here, a correct bracket sequence is defined as follows: * `()` is a correct bracket sequence. * If X is a correct bracket sequence, the concatenation of `(`, X and `)` in this order is also a correct bracket sequence. * If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence. * Every correct bracket sequence can be derived from the rules above. Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.
|
n = int(input())
s = input()
i = 0
k = 0
while i < len(s):
if s[i] == "(":
k += 1
i += 1
else:
if k > 0:
k -= 1
i += 1
else:
s = s[:i] + "(" + s[i:]
print(s+")"*k)
|
s377267633
|
Accepted
| 17 | 3,060 | 189 |
n = int(input())
s = input()
i = 0
k = 0
while i < len(s):
if s[i] == "(":
k += 1
else:
if k > 0:
k -= 1
else:
s = "(" + s
i += 1
i += 1
print(s+")"*k)
|
s795610184
|
p04043
|
u884323674
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 192 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
A, B, C = map(int, input().split())
if A == 5 and B == 5 and C == 7: print("YES")
if A == 5 and B == 7 and C == 5: print("YES")
if A == 7 and B == 5 and C == 5: print("YES")
else: print("NO")
|
s312691216
|
Accepted
| 17 | 2,940 | 117 |
length = input().split()
if length.count("5") == 2 and length.count("7") == 1:
print("YES")
else:
print("NO")
|
s393140219
|
p04000
|
u760794812
| 3,000 | 262,144 |
Wrong Answer
| 1,977 | 168,524 | 326 |
We have a grid with H rows and W columns. At first, all cells were painted white. Snuke painted N of these cells. The i-th ( 1 \leq i \leq N ) cell he painted is the cell at the a_i-th row and b_i-th column. Compute the following: * For each integer j ( 0 \leq j \leq 9 ), how many subrectangles of size 3×3 of the grid contains exactly j black cells, after Snuke painted N cells?
|
from collections import *
h,w,n = map(int,input().split())
dic = defaultdict(int)
for _ in range(n):
a,b = map(int,input().split())
for i in range(9):
dic[a - i //3,b - i % 3] += 1
ans =[0]*(n+1)
for i, j in dic:
ans[dic[i,j]] += h-1 > i > 0 < j < w -1
ans[0] = (h -2)*(w-2) - sum(ans)
for item in ans:
print(item)
|
s993974679
|
Accepted
| 2,121 | 166,872 | 323 |
from collections import *
h,w,n = map(int,input().split())
dic = defaultdict(int)
for _ in range(n):
a,b = map(int,input().split())
for i in range(9):
dic[a - i //3,b - i % 3] += 1
ans =[0]*10
for i, j in dic:
ans[dic[i,j]] += h-1 > i > 0 < j < w -1
ans[0] = (h -2)*(w-2) - sum(ans)
for item in ans:
print(item)
|
s292636472
|
p04043
|
u467831546
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 96 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a, b, c = map(int, input().split())
if a * b * c == 175:
print("yes")
else:
print("no")
|
s697561452
|
Accepted
| 18 | 2,940 | 96 |
a, b, c = map(int, input().split())
if a * b * c == 175:
print("YES")
else:
print("NO")
|
s202062308
|
p02396
|
u884445603
| 1,000 | 131,072 |
Wrong Answer
| 40 | 8,016 | 101 |
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.
|
import sys
l = sys.stdin.read().split()
for k,v in enumerate(l):
print("Case {}: {}".format(k,v))
|
s231034148
|
Accepted
| 150 | 7,380 | 121 |
cnt = 1
while True:
i = input()
if i == '0':
break
print("Case {}: {}".format(cnt,i))
cnt = cnt+1
|
s682579907
|
p03860
|
u663710122
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 46 |
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
A, S, C = input().split()
print(A + S[0] + C)
|
s779184432
|
Accepted
| 21 | 2,940 | 29 |
print('A' + input()[8] + 'C')
|
s813157513
|
p02610
|
u792078574
| 2,000 | 1,048,576 |
Wrong Answer
| 789 | 42,516 | 686 |
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row. The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise. Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel. Solve this problem for each of the T test cases given.
|
from heapq import heappush, heappop
T = int(input())
def process(arr):
arr.sort()
baseS = 0
diffS = 0
q = []
length = 0
for K, L, R in arr1:
baseS += R
diff = L - R
heappush(q, diff)
diffS += diff
length += 1
while length > K:
d = heappop(q)
diffS -= d
length -= 1
return baseS + diffS
for _ in range(T):
N = int(input())
arr1 = []
arr2 = []
for _ in range(N):
K, L, R = list(map(int, input().split()))
if L > R:
arr1.append((K, L, R))
else:
arr2.append((N-K, R, L))
print(process(arr1) + process(arr2))
|
s251575173
|
Accepted
| 727 | 42,532 | 685 |
from heapq import heappush, heappop
T = int(input())
def process(arr):
arr.sort()
baseS = 0
diffS = 0
q = []
length = 0
for K, L, R in arr:
baseS += R
diff = L - R
heappush(q, diff)
diffS += diff
length += 1
while length > K:
d = heappop(q)
diffS -= d
length -= 1
return baseS + diffS
for _ in range(T):
N = int(input())
arr1 = []
arr2 = []
for _ in range(N):
K, L, R = list(map(int, input().split()))
if L > R:
arr1.append((K, L, R))
else:
arr2.append((N-K, R, L))
print(process(arr1) + process(arr2))
|
s191106969
|
p03160
|
u141574039
| 2,000 | 1,048,576 |
Wrong Answer
| 156 | 13,928 | 393 |
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
N=int(input())
H=list(map(int,input().split()))
cost=0
t1,t2=0,0
A=[0]*(N)
i=1
while i<N:
if i==1:
A[i]=abs(H[i]-H[i-1])
#print(i,A[i])
elif i==2:
t1=abs(H[i]-H[i-2])
t2=abs(H[i]-H[i-1])+abs(H[i-2]-H[i-1])
A[i]=min(t1,t2)
#print(i,A[i],t1,t2)
else:
t1=abs(H[i]-H[i-1])+A[i-1]
t2=abs(H[i]-H[i-2])+A[i-2]
A[i]=min(t1,t2)
#print(i,A[i],t1,t2)
i=i+1
|
s658571883
|
Accepted
| 129 | 13,980 | 188 |
N=int(input())
H=list(map(int,input().split()))
DP=[0]*N
DP[0]=0
DP[1]=abs(H[1]-H[0])
for i in range(2,N):
DP[i]=min((abs(H[i]-H[i-1])+DP[i-1]),(abs(H[i]-H[i-2])+DP[i-2]))
print(DP[N-1])
|
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