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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s527892710 | p03433 | u601082779 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 56 | E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins. | a=int(input());b=int(input());print("Yneos"[a%500<b::2]) | s414519637 | Accepted | 17 | 2,940 | 56 | a=int(input());b=int(input());print("YNeos"[a%500>b::2]) |
s980977572 | p03494 | u884959062 | 2,000 | 262,144 | Wrong Answer | 20 | 2,940 | 307 | There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform. | e = [[int(i) for i in input().split()] for i in range(2)]
max_count = 0
for i in range(e[0][0]):
tmp = e[1][i]
count = 0
while True:
tmp, m = divmod(tmp, 2)
if m == 1:
break
count += 1
if count < max_count:
max_count = count
print(max_count)
| s216774715 | Accepted | 19 | 3,060 | 312 | e = [[int(i) for i in input().split()] for i in range(2)]
max_count = 999999
for i in range(e[0][0]):
tmp = e[1][i]
count = 0
while True:
tmp, m = divmod(tmp, 2)
if m == 1:
break
count += 1
if count < max_count:
max_count = count
print(max_count)
|
s269136561 | p04044 | u577504524 | 2,000 | 262,144 | Wrong Answer | 25 | 9,108 | 164 | Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m. | #B
n,l = map(int,input().split())
word_list=[]
for i in range(n):
word = str(input())
word_list.append(word)
x = sorted(word_list)
print(''.join(word_list)) | s812948162 | Accepted | 27 | 9,172 | 156 | #B
n,l = map(int,input().split())
word_list=[]
for i in range(n):
word = str(input())
word_list.append(word)
x = sorted(word_list)
print(''.join(x)) |
s506298878 | p02255 | u843404779 | 1,000 | 131,072 | Wrong Answer | 20 | 7,680 | 457 | Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step. | def print_list(ele_list):
print(" ".join(map(str, ele_list)))
def insertion_sort(ele_list):
len_ele_list = len(ele_list)
for i in range(1, len_ele_list):
key = ele_list[i]
j = i - 1
while j >= 0 and ele_list[j] > key:
ele_list[j+1] = ele_list[j]
j -= 1
ele_list[j+1] = key
print_list(ele_list)
N = int(input())
ele_list = list(map(int, input().split()))
insertion_sort(ele_list) | s280184181 | Accepted | 20 | 7,672 | 482 | def print_list(ele_list):
print(" ".join(map(str, ele_list)))
def insertion_sort(ele_list):
len_ele_list = len(ele_list)
print_list(ele_list)
for i in range(1, len_ele_list):
key = ele_list[i]
j = i - 1
while j >= 0 and ele_list[j] > key:
ele_list[j+1] = ele_list[j]
j -= 1
ele_list[j+1] = key
print_list(ele_list)
N = int(input())
ele_list = list(map(int, input().split()))
insertion_sort(ele_list) |
s054842854 | p03680 | u936985471 | 2,000 | 262,144 | Wrong Answer | 195 | 8,948 | 163 | Takahashi wants to gain muscle, and decides to work out at AtCoder Gym. The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it. Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up. Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons. | n=int(input())
a=[None]*n
for i in range(len(a)):
a[i]=int(input())-1
cur=0
ok=False
for i in range(n+1):
cur=a[cur]
if cur==1:
break
print((-1,i)[ok]) | s268573231 | Accepted | 219 | 7,668 | 240 | n=int(input())
a=[-1]*n
for i in range(n):
a[i]=int(input())-1
reached=[False]*n
tar=0
cnt=0
suc=False
while reached[tar]==False:
reached[tar]=True
cnt+=1
tar=a[tar]
if tar==1:
suc=True
break
print((-1,cnt)[suc])
|
s149297287 | p03160 | u043236471 | 2,000 | 1,048,576 | Wrong Answer | 156 | 13,976 | 309 | 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 = [int(x) for x in input().split()]
def chmin(a, b):
return a if a <= b else b
dp = [float('inf')] * N
dp[0] = 0
for i in range(1, N):
dp[i] = chmin(dp[i], dp[i-1] + abs(h[i] - h[i-1]))
if i > 1:
dp[i] = chmin(dp[i], dp[i-2] + abs(h[i] - h[i-2]))
print(dp[1]) | s178854027 | Accepted | 160 | 14,052 | 310 | N = int(input())
h = [int(x) for x in input().split()]
def chmin(a, b):
return a if a <= b else b
dp = [float('inf')] * N
dp[0] = 0
for i in range(1, N):
dp[i] = chmin(dp[i], dp[i-1] + abs(h[i] - h[i-1]))
if i > 1:
dp[i] = chmin(dp[i], dp[i-2] + abs(h[i] - h[i-2]))
print(dp[-1]) |
s533837426 | p03455 | u026406748 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 121 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | a = input().split()
b = [int(i) for i in a]
c = b[0] * b[1]
if c%2 == 0:
print("偶数")
else:
print("奇数") | s984083086 | Accepted | 17 | 3,064 | 116 | a = input().split()
b = [int(i) for i in a]
c = b[0] * b[1]
if c%2 == 0:
print("Even")
else:
print("Odd") |
s064156669 | p02865 | u538276565 | 2,000 | 1,048,576 | Wrong Answer | 21 | 3,316 | 82 | How many ways are there to choose two distinct positive integers totaling N, disregarding the order? | def solve():
print(int(input()) // 2)
if __name__ == "__main__":
solve() | s226582364 | Accepted | 17 | 2,940 | 121 | def solve():
i = int(input())
print(i // 2 -1 if i % 2 == 0 else i // 2)
if __name__ == "__main__":
solve() |
s124622292 | p03352 | u020091453 | 2,000 | 1,048,576 | Wrong Answer | 19 | 3,188 | 271 | You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2. | import math
from bisect import bisect_right
X = int(input())
s = list([1])
for i in range(2,math.ceil(1000**0.5)):
for j in range(0,9):
if i**j > 1000:
break
else:
s.append(i**j)
s.sort()
print(s[bisect_right(s,X) -1])
| s470123917 | Accepted | 18 | 3,188 | 268 | import math
from bisect import bisect_right
X = int(input())
s = list([1])
for i in range(2,math.ceil(1000**0.5)):
for j in range(2,9):
if i**j > 1000:
break
else:
s.append(i**j)
s.sort()
print(s[bisect_right(s,X) -1]) |
s276986789 | p03067 | u239528020 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 138 | There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise. | a, b, c = map(int, input().split())
if a >= b:
tmp = a
b = a
a = tmp
if a<=c and c<=b:
print("yes")
else:
print("no") | s481828602 | Accepted | 17 | 2,940 | 148 |
a, b, c = map(int, input().split())
if a >= b:
tmp = b
b = a
a = tmp
if a<=c and c<=b:
print("Yes")
else:
print("No") |
s080263303 | p02383 | u299798926 | 1,000 | 131,072 | Wrong Answer | 20 | 7,812 | 1,016 | Write a program to simulate rolling a dice, which can be constructed by the following net. As shown in the figures, each face is identified by a different label from 1 to 6. Write a program which reads integers assigned to each face identified by the label and a sequence of commands to roll the dice, and prints the integer on the top face. At the initial state, the dice is located as shown in the above figures. | x=[int(i)for i in input().split()]
y=[[int(0)for i in range(4)]for i in range(3)]
y[0][1]=x[0]
y[1][1]=x[1]
y[1][2]=x[2]
y[1][0]=x[3]
y[1][3]=x[4]
y[2][1]=x[5]
m=[i for i in input()]
z=y.copy()
for i in range(len(m)):
if m[i]=='N':
y[0][1],z[1][1]=z[1][1],y[0][1]
y[1][1],z[2][1]=z[2][1],y[1][1]
y[1][3],z[0][1]=z[0][1],y[1][3]
y[2][1],z[1][3]=z[1][3],y[2][1]
z=y.copy()
elif m[i]=='S':
y[0][1],z[1][3]=z[1][3],y[0][1]
y[1][1],z[0][1]=z[0][1],y[1][1]
y[1][3],z[2][1]=z[2][1],y[1][3]
y[2][1],z[1][1]=z[1][1],y[2][1]
z=y.copy()
elif m[i]=='W':
y[0][1],z[1][2]=z[1][2],y[0][1]
y[1][0],z[0][1]=z[0][1],y[1][0]
y[1][2],z[2][1]=z[2][1],y[1][2]
y[2][1],z[1][0]=z[1][0],y[2][1]
z=y.copy()
else:
y[0][1],z[1][0]=z[1][0],y[0][1]
y[1][0],z[2][1]=z[2][1],y[1][0]
y[1][2],z[0][1]=z[0][1],y[1][2]
y[2][1],z[1][2]=z[1][2],y[2][1]
z=y.copy()
print(y[0][1]) | s212919276 | Accepted | 10 | 7,944 | 788 | x=[int(i)for i in input().split()]
y=[[int(0)for i in range(4)]for i in range(3)]
y[0][1]=x[0]
y[1][1]=x[1]
y[1][2]=x[2]
y[1][0]=x[3]
y[1][3]=x[4]
y[2][1]=x[5]
m=[i for i in input()]
for i in range(len(m)):
if m[i]=='N':
y[0][1],y[1][1]=y[1][1],y[0][1]
y[1][1],y[2][1]=y[2][1],y[1][1]
y[2][1],y[1][3]=y[1][3],y[2][1]
elif m[i]=='S':
y[0][1],y[1][3]=y[1][3],y[0][1]
y[1][3],y[2][1]=y[2][1],y[1][3]
y[2][1],y[1][1]=y[1][1],y[2][1]
elif m[i]=='W':
y[0][1],y[1][2]=y[1][2],y[0][1]
y[1][2],y[2][1]=y[2][1],y[1][2]
y[2][1],y[1][0]=y[1][0],y[2][1]
elif m[i]=='E':
y[0][1],y[1][0]=y[1][0],y[0][1]
y[1][0],y[2][1]=y[2][1],y[1][0]
y[2][1],y[1][2]=y[1][2],y[2][1]
print(y[0][1])
|
s903259883 | p02865 | u269778596 | 2,000 | 1,048,576 | Wrong Answer | 2,104 | 3,932 | 191 | How many ways are there to choose two distinct positive integers totaling N, disregarding the order? | n = int(input())
count = 0
seen = []
for i in range(1,n+1):
if i!=(n-i):
if i and n-i not in seen:
seen.append(i)
seen.append(n-i)
count += 1
print(count)
| s072836244 | Accepted | 17 | 2,940 | 32 | n = int(input())
print((n-1)//2) |
s990840260 | p02284 | u912143677 | 2,000 | 131,072 | Wrong Answer | 20 | 5,608 | 1,529 | Write a program which performs the following operations to a binary search tree $T$ by adding the find operation to A: Binary Search Tree I. * insert $k$: Insert a node containing $k$ as key into $T$. * find $k$: Report whether $T$ has a node containing $k$. * print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively. |
class Tree():
def __init__(self):
self.root = None
def insert(self, key):
z = Node(key)
y = None
x = self.root
while x:
y = x
if z.key < x.key:
x = x.left
else:
x = x.right
z.parent = y
if y == None:
self.root = z
elif z.key < y.key:
y.left = z
else:
y.right = z
def print(self):
print('', ' '.join(map(str, self.root.inwalk())))
print('', ' '.join(map(str, self.root.prewalk())))
def find(self, x, k):
while x != None and k != x.key:
if k < x.key:
x = x.left
else:
x = x.right
return x
class Node():
def __init__(self, key):
self.key = key
self.parent = self.left = self.right = None
def prewalk(self):
ret = [self.key]
if self.left:
ret += self.left.prewalk()
if self.right:
ret += self.right.prewalk()
return ret
def inwalk(self):
ret = []
if self.left:
ret += self.left.inwalk()
ret += [self.key]
if self.right:
ret += self.right.inwalk()
return ret
tree = Tree()
n = int(input())
for i in range(n):
com = input().split()
if com[0] == "insert":
tree.insert(int(com[1]))
elif com[0] == "find":
tree.find(tree.root, int(com[1]))
else:
tree.print()
| s063320387 | Accepted | 6,900 | 116,656 | 1,619 |
class Tree():
def __init__(self):
self.root = None
def insert(self, key):
z = Node(key)
y = None
x = self.root
while x:
y = x
if z.key < x.key:
x = x.left
else:
x = x.right
z.parent = y
if y == None:
self.root = z
elif z.key < y.key:
y.left = z
else:
y.right = z
def print(self):
print('', ' '.join(map(str, self.root.inwalk())))
print('', ' '.join(map(str, self.root.prewalk())))
def find(self, x, k):
while x != None and k != x.key:
if k < x.key:
x = x.left
else:
x = x.right
return x
class Node():
def __init__(self, key):
self.key = key
self.parent = self.left = self.right = None
def prewalk(self):
ret = [self.key]
if self.left:
ret += self.left.prewalk()
if self.right:
ret += self.right.prewalk()
return ret
def inwalk(self):
ret = []
if self.left:
ret += self.left.inwalk()
ret += [self.key]
if self.right:
ret += self.right.inwalk()
return ret
tree = Tree()
n = int(input())
for i in range(n):
com = input().split()
if com[0] == "insert":
tree.insert(int(com[1]))
elif com[0] == "find":
ans = tree.find(tree.root, int(com[1]))
if ans != None:
print("yes")
else:
print("no")
else:
tree.print()
|
s953673935 | p02257 | u196653484 | 1,000 | 131,072 | Wrong Answer | 20 | 5,672 | 700 | A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7. Write a program which reads a list of _N_ integers and prints the number of prime numbers in the list. |
import math
def check(B,i):
sum=0
for j in range(len(B)):
if B[j]%10 in [1,2,4,5,6,8]:
del B[j]
check(B,i)
if B[j]%i==0:
sum+=1
del B[j]
return sum
def Prime_Numbers(A):
mx=max(A)
sum=0
if 2 in A:
sum+=1
if 3 in A:
sum+=1
if 5 in A:
sum+=1
if 7 in A:
sum+=1
i=10
while i <= math.sqrt(mx):
mx=max[A]
sum+=check(A,i)
i+=1
print("A={}".format(A))
return sum
if __name__ == "__main__":
A=[]
n=int(input())
for i in range(n):
A.append(int(input()))
print(Prime_Numbers(A))
| s057474588 | Accepted | 1,150 | 6,008 | 497 |
import math
def Prime_Numbers_Check(x):
i=2
flag=True
while i <= math.sqrt(x):
if x%i==0:
flag=False
break
i+=1
return flag
def Prime_Numbers_Count(A):
sum=0
for i in A:
if Prime_Numbers_Check(i):
sum+=1
return sum
if __name__ == "__main__":
A=[]
n=int(input())
for i in range(n):
A.append(int(input()))
sum=Prime_Numbers_Count(A)
print(sum)
|
s852935007 | p03574 | u824734140 | 2,000 | 262,144 | Wrong Answer | 34 | 9,176 | 521 | You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process. | h, w = map(int, input().split())
s = [list(input()) for _ in range(0, h)]
dx = [1, 0, -1, 0, 1, -1, -1, -1]
dy = [0, 1, 0, -1, 1, 1, -1, -1]
for i in range(0, h):
for j in range(0, w):
if s[i][j] != '.':
continue
cnt = 0
for k in range(0, 8):
ni = i + dy[k]
nj = j + dx[k]
if not (ni < 0 or h <= ni) and not (nj < 0 or w <= nj) and s[ni][nj] == '#':
cnt += 1
s[i][j] = str(cnt)
[print(s[ii]) for ii in range(0, h)]
| s449846540 | Accepted | 38 | 9,136 | 529 | h, w = map(int, input().split())
s = [list(input()) for _ in range(0, h)]
dx = [1, 1, 1, 0, 0, -1, -1, -1]
dy = [1, 0, -1, 1, -1, 1, 0, -1]
for i in range(0, h):
for j in range(0, w):
if s[i][j] != '.':
continue
cnt = 0
for k in range(0, 8):
ni = i + dy[k]
nj = j + dx[k]
if not (ni < 0 or h <= ni) and not (nj < 0 or w <= nj) and s[ni][nj] == '#':
cnt += 1
s[i][j] = str(cnt)
[print(''.join(s[ii])) for ii in range(0, h)]
|
s293369405 | p03493 | u300651214 | 2,000 | 262,144 | Wrong Answer | 25 | 8,856 | 32 | Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble. | s = list(input())
s.count("1")
| s259842484 | Accepted | 27 | 9,080 | 32 | a = input()
print(a.count('1'))
|
s484834080 | p03759 | u160659351 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 104 | 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. | #58
a, b, c = map(int, input().rstrip().split())
if b-a == c-a:
print("YES")
else:
print("NO") | s161513461 | Accepted | 17 | 2,940 | 104 | #58
a, b, c = map(int, input().rstrip().split())
if b-a == c-b:
print("YES")
else:
print("NO") |
s975871076 | p02388 | u335511832 | 1,000 | 131,072 | Wrong Answer | 30 | 6,724 | 17 | Write a program which calculates the cube of a given integer x. | x = 3
print(x**3) | s224024076 | Accepted | 30 | 6,724 | 28 | x = int(input())
print(x**3) |
s148622904 | p03543 | u778700306 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 130 | We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**? |
a = list(input())
if a[0] == a[1] and a[1] == a[2] or a[1] == a[2] and a[2] == a[3]:
print("YES")
else:
print("NO")
| s133782016 | Accepted | 18 | 2,940 | 130 |
a = list(input())
if a[0] == a[1] and a[1] == a[2] or a[1] == a[2] and a[2] == a[3]:
print("Yes")
else:
print("No")
|
s913153871 | p03303 | u674185143 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 86 | You are given a string S consisting of lowercase English letters. We will write down this string, starting a new line after every w letters. Print the string obtained by concatenating the letters at the beginnings of these lines from top to bottom. | s=input()
w=int(input())
ans=s[0]
for i in range(1,len(s)//w):
ans+=s[i*w]
print(ans) | s692651247 | Accepted | 17 | 2,940 | 39 | s=input()
w=int(input())
print(s[0::w]) |
s582124330 | p04031 | u901307908 | 2,000 | 262,144 | Wrong Answer | 23 | 3,064 | 163 | Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective. | n = int(input())
a = input()
a = list(map(int, a.split()))
#print(a)
x_min = round(sum(a) / 3 )
cost = sum([(x_min - i) * (x_min - i) for i in a])
print(cost) | s377213309 | Accepted | 24 | 3,188 | 163 | n = int(input())
a = input()
a = list(map(int, a.split()))
#print(a)
x_min = round(sum(a) / n )
cost = sum([(x_min - i) * (x_min - i) for i in a])
print(cost) |
s236310590 | p03436 | u742897895 | 2,000 | 262,144 | Wrong Answer | 32 | 3,700 | 819 | We have an H \times W grid whose squares are painted black or white. The square at the i-th row from the top and the j-th column from the left is denoted as (i, j). Snuke would like to play the following game on this grid. At the beginning of the game, there is a character called Kenus at square (1, 1). The player repeatedly moves Kenus up, down, left or right by one square. The game is completed when Kenus reaches square (H, W) passing only white squares. Before Snuke starts the game, he can change the color of some of the white squares to black. However, he cannot change the color of square (1, 1) and (H, W). Also, changes of color must all be carried out before the beginning of the game. When the game is completed, Snuke's score will be the number of times he changed the color of a square before the beginning of the game. Find the maximum possible score that Snuke can achieve, or print -1 if the game cannot be completed, that is, Kenus can never reach square (H, W) regardless of how Snuke changes the color of the squares. The color of the squares are given to you as characters s_{i, j}. If square (i, j) is initially painted by white, s_{i, j} is `.`; if square (i, j) is initially painted by black, s_{i, j} is `#`. | from collections import deque
h, w = map(int, input().split())
s = [list(input()) for j in range(h)]
dx = [-1, 1, 0, 0]
dy = [ 0, 0,-1, 1]
q = deque([(0, 0, 0)])
used = {(0, 0)}
res = None
while q:
i, j, cost = q.popleft()
if i == w - 1 and j == h - 1:
res = cost
break
for k in range(4):
if not (0 <= i + dx[k] < w):
continue
if not (0 <= j + dy[k] < h):
continue
if s[j + dy[k]][i + dx[k]] == '#':
continue
if (i + dx[k], j + dy[k]) in used:
continue
used.add((i + dx[k], j + dy[k]))
q.append((i + dx[k], j + dy[k], cost + 1))
if res:
ans = h * w - cost - 1
for j in range(h):
print(s[j][:].count('#'))
ans -= s[j][:].count('#')
print(ans)
else:
print(-1) | s812892276 | Accepted | 33 | 3,700 | 785 | from collections import deque
h, w = map(int, input().split())
s = [list(input()) for j in range(h)]
dx = [-1, 1, 0, 0]
dy = [ 0, 0,-1, 1]
q = deque([(0, 0, 0)])
used = {(0, 0)}
res = None
while q:
i, j, cost = q.popleft()
if i == w - 1 and j == h - 1:
res = cost
break
for k in range(4):
if not (0 <= i + dx[k] < w):
continue
if not (0 <= j + dy[k] < h):
continue
if s[j + dy[k]][i + dx[k]] == '#':
continue
if (i + dx[k], j + dy[k]) in used:
continue
used.add((i + dx[k], j + dy[k]))
q.append((i + dx[k], j + dy[k], cost + 1))
if res:
ans = h * w - cost - 1
for j in range(h):
ans -= s[j][:].count('#')
print(ans)
else:
print(-1) |
s848276033 | p03438 | u692632484 | 2,000 | 262,144 | Wrong Answer | 29 | 4,600 | 345 | You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N. Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal. Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions **simultaneously** : * Add 2 to a_i. * Add 1 to b_j. | N=int(input())
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
total_do=sum(a)-sum(b)
must_do_a=0
must_do_b=0
for i in range(N):
dist=a[i]-b[i]
if dist<0:
dist=-dist
must_do_a+=int(dist/2)
if dist%2!=0:
must_do_b+=1
elif dist>0:
must_do_b+=dist
if must_do_a>=must_do_b:
print("YES")
else:
print("NO")
| s861995510 | Accepted | 28 | 4,600 | 370 | N=int(input())
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
can = sum(b)-sum(a)
actA,actB = 0,0
for i in range(N):
diff = b[i]-a[i]
if diff==0:continue
if diff>0:
actA+=(diff+1)//2
actB+=diff%2
else:
actB+=(-diff)
leftA = can-actA
leftB = can-actB
print("Yes" if leftA>=0 and leftA*2==leftB else "No")
|
s695990924 | p03251 | u175204984 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,064 | 606 | Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out. | l = input().split()
N = int(l[0])
M = int(l[1])
X = int(l[2])
Y = int(l[3])
if X < -100:
X = -100
elif X > 100:
X = 100
if Y < -100:
Y = -100
elif Y > 100:
Y = 100
lx = input().split()
ly = input().split()
lx = [int(z) for z in lx]
ly = [int(z) for z in ly]
lx = sorted(lx)
ly = sorted(ly)
flag = True
if int(ly[0]) - int(lx[N-1]) >= 1:
for z in range(int(lx[N-1]) + 1, int(ly[0]) + 1):
if z > X and z <= Y:
flag = True
else:
flag = False
break
else:
flag = False
if flag:
print("No war")
else:
print("War")
| s444526876 | Accepted | 17 | 3,064 | 554 | l = input().split()
N = int(l[0])
M = int(l[1])
X = int(l[2])
Y = int(l[3])
"""
if X < -100:
X = -100
elif X > 100:
X = 100
if Y < -100:
Y = -100
elif Y > 100:
Y = 100
"""
lx = input().split()
ly = input().split()
lx = [int(z) for z in lx]
ly = [int(z) for z in ly]
lx = sorted(lx)
ly = sorted(ly)
flag = True
if int(ly[0]) - int(lx[N-1]) >= 1:
for z in range(int(lx[N-1]) + 1, int(ly[0]) + 1):
if z > X and z <= Y:
flag = False
break
if flag:
print("War")
else:
print("No War")
|
s550172635 | p03543 | u274045692 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 125 | We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**? | N = input()
if (N[0] == N[1] and N[1] == N[2]) or (N[1] == N[2] and N[2] == N[3]):
print("true")
else:
print("false") | s927355379 | Accepted | 18 | 2,940 | 121 | N = input()
if (N[0] == N[1] and N[1] == N[2]) or (N[1] == N[2] and N[2] == N[3]):
print("Yes")
else:
print("No") |
s834322207 | p03337 | u883792993 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 54 | You are given two integers A and B. Find the largest value among A+B, A-B and A \times B. | A,B=list(map(int, input().split()))
max(A+B, A-B, A*B) | s485367244 | Accepted | 17 | 2,940 | 61 | A,B=list(map(int, input().split()))
print(max(A+B, A-B, A*B)) |
s542836844 | p03379 | u057993957 | 2,000 | 262,144 | Wrong Answer | 287 | 25,620 | 155 | When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N. | n = int(input())
x = list(map(int, input().split()))
sort_x = sorted(x)
l = sort_x[n//2 - 1]
r = sort_x[n//2]
for xi in x:
print(l if xi <= l else r)
| s683743771 | Accepted | 282 | 25,224 | 155 | n = int(input())
x = list(map(int, input().split()))
sort_x = sorted(x)
l = sort_x[n//2 - 1]
r = sort_x[n//2]
for xi in x:
print(r if xi <= l else l)
|
s125406347 | p03878 | u945181840 | 2,000 | 262,144 | Wrong Answer | 236 | 34,096 | 224 | There are N computers and N sockets in a one-dimensional world. The coordinate of the i-th computer is a_i, and the coordinate of the i-th socket is b_i. It is guaranteed that these 2N coordinates are pairwise distinct. Snuke wants to connect each computer to a socket using a cable. Each socket can be connected to only one computer. In how many ways can he minimize the total length of the cables? Compute the answer modulo 10^9+7. | import sys
import numpy as np
read = sys.stdin.read
N, *ab = map(int, read().split())
mod = 10 ** 9 + 7
a = np.array(ab[:N], np.int64)
b = np.array(ab[N:], np.int64)
a.sort()
b.sort()
idx = np.searchsorted(b, a)
print(0)
| s304073541 | Accepted | 278 | 27,484 | 566 | import sys
from operator import itemgetter
read = sys.stdin.read
N, *ab = map(int, read().split())
mod = 10 ** 9 + 7
ab = list(zip(ab, [0] * N + [1] * N))
ab.sort(key=itemgetter(0))
remain_pc = 0
remain_power = 0
answer = 1
for x, p in ab:
if p == 0:
if remain_power == 0:
remain_pc += 1
continue
answer *= remain_power
remain_power -= 1
else:
if remain_pc == 0:
remain_power += 1
continue
answer *= remain_pc
remain_pc -= 1
answer %= mod
print(answer)
|
s212244612 | p02414 | u012062731 | 1,000 | 131,072 | Wrong Answer | 20 | 7,616 | 539 | Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula: \\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\] where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively. | n, m, l = map(int, input().split()) # n = 3, c = 2, l = 3
a_matrix = [list(map(int, input().split())) for row in range(n)] # [[1, 2], [0, 3], [4, 5]]
b_matrix = [list(map(int, input().split())) for row in range(m)] # [[1, 2, 1], [0, 3, 2]]
result_matrix = [[0 for x in range(l)] for y in range(n)] # [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
print(result_matrix)
for i in range(n):
for j in range(l):
for k in range(m):
result_matrix[i][j] += a_matrix[i][k] * b_matrix[k][j]
for i in range(n):
print(' '.join(map(str, result_matrix[i]))) | s176105612 | Accepted | 510 | 8,896 | 518 | n, m, l = map(int, input().split()) # n = 3, c = 2, l = 3
a_matrix = [list(map(int, input().split())) for row in range(n)] # [[1, 2], [0, 3], [4, 5]]
b_matrix = [list(map(int, input().split())) for row in range(m)] # [[1, 2, 1], [0, 3, 2]]
result_matrix = [[0 for x in range(l)] for y in range(n)] # [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
for i in range(n):
for j in range(l):
for k in range(m):
result_matrix[i][j] += a_matrix[i][k] * b_matrix[k][j]
for i in range(n):
print(' '.join(map(str, result_matrix[i]))) |
s152367606 | p04043 | u629540524 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 86 | 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 = input().split()
print('Yes' if a.count('5') == 2 and a.count('7') == 1 else 'No') | s540025696 | Accepted | 17 | 2,940 | 78 | a = input()
print('YES' if a.count('5') == 2 and a.count('7') == 1 else 'NO') |
s641948024 | p02866 | u571969099 | 2,000 | 1,048,576 | Wrong Answer | 115 | 14,396 | 281 | Given is an integer sequence D_1,...,D_N of N elements. Find the number, modulo 998244353, of trees with N vertices numbered 1 to N that satisfy the following condition: * For every integer i from 1 to N, the distance between Vertex 1 and Vertex i is D_i. | n = int(input())
d = [int(i) for i in input().split()]
a = [0] * n
for i in d:
a[i] += 1
if d[0] != 0:
print(0)
else:
j = 1
for i in range(n - 1):
if a[i] == 0:
break
j *= a[i] ** a[i + 1]
j %= 998244353
print(a)
print(j) | s153777833 | Accepted | 126 | 14,396 | 275 | n = int(input())
d = [int(i) for i in input().split()]
a = [0] * n
k = 0
for i in d:
a[i] += 1
k=max(k,i)
if d[0] != 0:
print(0)
elif a[0] != 1:
print(0)
else:
j = 1
for i in range(k):
j *= a[i] ** a[i + 1]
j %= 998244353
print(j)
|
s062215171 | p03433 | u190079347 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 85 | E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins. | n = int(input())
a = int(input())
if n % 500 <= a:
print("YES")
else:
print("NO") | s997369149 | Accepted | 17 | 2,940 | 85 | n = int(input())
a = int(input())
if n % 500 <= a:
print("Yes")
else:
print("No") |
s395633984 | p03377 | u294385082 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 87 | There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals. | A,B,X = map(int,input().split())
if A <= X <= A+B :
print("Yes")
else :
print("No") | s985859041 | Accepted | 17 | 2,940 | 87 | A,B,X = map(int,input().split())
if A <= X <= A+B :
print("YES")
else :
print("NO") |
s194382911 | p03338 | u265281278 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 128 | 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 = input()
ans = 0
for i in range(1,N):
x = set(S[:i]) and set(S[i:])
ans = max(ans,len(x))
print(ans) | s527369613 | Accepted | 17 | 2,940 | 126 | N = int(input())
S = input()
ans = 0
for i in range(1,N):
x = set(S[:i]) & set(S[i:])
ans = max(ans,len(x))
print(ans) |
s159462724 | p04029 | u451017206 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 37 | 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) | s188441235 | Accepted | 18 | 2,940 | 38 | N = int(input())
print(N * (N + 1)//2) |
s459283168 | p02260 | u456917014 | 1,000 | 131,072 | Wrong Answer | 20 | 5,592 | 207 | Write a program of the Selection Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: SelectionSort(A) 1 for i = 0 to A.length-1 2 mini = i 3 for j = i to A.length-1 4 if A[j] < A[mini] 5 mini = j 6 swap A[i] and A[mini] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 6 of the pseudocode in the case where i ≠ mini. | n=int(input())
l=list(map(int,input().split()))
count=0
for i in range(n):
m=i
for j in range(i,n):
if l[j]<l[m]:
m=j
l[i],l[m]=l[m],l[i]
count+=1
print(*l)
print(count)
| s508557251 | Accepted | 20 | 5,608 | 230 | n=int(input())
l=list(map(int,input().split()))
count=0
for i in range(n):
m=i
for j in range(i,n):
if l[j]<l[m]:
m=j
l[i],l[m]=l[m],l[i]
if l[i]!=l[m]:
count+=1
print(*l)
print(count)
|
s015970135 | p02259 | u865220118 | 1,000 | 131,072 | Wrong Answer | 20 | 5,592 | 684 | Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode. |
A = []
LENGTH =int(input())
A = input().split()
N = 0
M = LENGTH - 1
CHANGE = 0
while N <= LENGTH -1 :
M = LENGTH - 1
print ('N=',N)
# for j = A.length-1 downto i+1
while M >= N + 1:
print ('-M=',M)
# if A[j] < A[j-1]
if A[M] < A[M-1] :
tmp = A[M-1]
A[M-1] = A[M]
A[M] = tmp
CHANGE += 1
M -= 1
N += 1
print(" ".join(map(str,A)))
print (CHANGE)
| s553320550 | Accepted | 20 | 5,600 | 695 |
A = []
LENGTH =int(input())
A = input().split()
N = 0
M = LENGTH - 1
CHANGE = 0
while N <= LENGTH -1 :
M = LENGTH - 1
# print ('N=',N)
# for j = A.length-1 downto i+1
while M >= N + 1:
# print ('-M=',M)
# if A[j] < A[j-1]
if int(A[M]) < int(A[M-1]) :
tmp = A[M-1]
A[M-1] = A[M]
A[M] = tmp
CHANGE += 1
M -= 1
N += 1
print(" ".join(map(str,A)))
print (CHANGE)
|
s317598015 | p03796 | u703890795 | 2,000 | 262,144 | Wrong Answer | 51 | 2,940 | 82 | 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())
s = 1
for i in range(1, N+1):
s *= i
s %= (7 + 1E+9)
print(s) | s215632958 | Accepted | 50 | 2,940 | 87 | N = int(input())
s = 1
for i in range(1, N+1):
s *= i
s %= (7 + 1E+9)
print(int(s)) |
s544975059 | p03673 | u385309449 | 2,000 | 262,144 | Wrong Answer | 2,104 | 25,156 | 107 | You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations. | n = int(input())
a = list(map(int,input().split()))
b = []
for i in a:
b.append(i)
b = b[::-1]
print(b) | s203713851 | Accepted | 68 | 27,204 | 180 | n = int(input())
a = list(map(str,input().split()))
b = a[::-1]
b = b[::2]
if n%2 == 0:
c = a[::2]
else:
c = a[1::2]
d = b+c
if n >= 2:
print(' '.join(d))
else:
print(a[0]) |
s968350468 | p02613 | u674190122 | 2,000 | 1,048,576 | Wrong Answer | 152 | 16,296 | 252 | 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. | result = []
for _ in range(int(input())):
result.append(input())
print("AC X {}".format(result.count("AC")))
print("WA X {}".format(result.count("WA")))
print("TLE X {}".format(result.count("TLE")))
print("RE X {}".format(result.count("RE")))
| s614507616 | Accepted | 147 | 16,196 | 245 | result = []
for _ in range(int(input())):
result.append(input())
print("AC x {}".format(result.count("AC")))
print("WA x {}".format(result.count("WA")))
print("TLE x {}".format(result.count("TLE")))
print("RE x {}".format(result.count("RE")))
|
s750473663 | p03729 | u840570107 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 185 | 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`. | lis = input().split()
lis_a = lis[0][len(lis[0])-1]
lis_b = lis[1][len(lis[1])-1]
lis_c = lis[2][len(lis[2])-1]
if lis_a == lis_b and lis_b == lis_c:
print("YES")
else:
print("NO") | s305700403 | Accepted | 17 | 3,060 | 195 | lis = input().split()
lis_a = lis[0][len(lis[0])-1]
lis_bt = lis[1][0]
lis_bf = lis[1][len(lis[1])-1]
lis_c = lis[2][0]
if lis_a == lis_bt and lis_bf == lis_c:
print("YES")
else:
print("NO") |
s326874377 | p03813 | u863442865 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 79 | Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. | x = int(input())
s = x // 11
a = x % 11
print(s*2+2) if a > 6 else print(s*2+1) | s664918802 | Accepted | 17 | 2,940 | 53 | print('ABC') if int(input()) < 1200 else print('ARC') |
s609316362 | p03054 | u052499405 | 2,000 | 1,048,576 | Wrong Answer | 393 | 3,784 | 664 | We have a rectangular grid of squares with H horizontal rows and W vertical columns. Let (i,j) denote the square at the i-th row from the top and the j-th column from the left. On this grid, there is a piece, which is initially placed at square (s_r,s_c). Takahashi and Aoki will play a game, where each player has a string of length N. Takahashi's string is S, and Aoki's string is T. S and T both consist of four kinds of letters: `L`, `R`, `U` and `D`. The game consists of N steps. The i-th step proceeds as follows: * First, Takahashi performs a move. He either moves the piece in the direction of S_i, or does not move the piece. * Second, Aoki performs a move. He either moves the piece in the direction of T_i, or does not move the piece. Here, to move the piece in the direction of `L`, `R`, `U` and `D`, is to move the piece from square (r,c) to square (r,c-1), (r,c+1), (r-1,c) and (r+1,c), respectively. If the destination square does not exist, the piece is removed from the grid, and the game ends, even if less than N steps are done. Takahashi wants to remove the piece from the grid in one of the N steps. Aoki, on the other hand, wants to finish the N steps with the piece remaining on the grid. Determine if the piece will remain on the grid at the end of the game when both players play optimally. | h, w, n = [int(item) for item in input().split()]
sr, sc = [int(item) for item in input().split()]
s = input().rstrip()
t = input().rstrip()
# l, r, u, d
move_limit = [-(w-sr+1), -sr, -(h-sc+1), -sc]
max_move = [0] * 4
move = [0] * 4
direction = "LRUD"
reverse = "RLDU"
for i in range(n):
move[direction.index(s[i])] += 1
max_move[direction.index(s[i])] = max(move[direction.index(s[i])], max_move[direction.index(s[i])])
if move[reverse.index(t[i])] < move_limit[reverse.index(t[i])]:
move[reverse.index(t[i])] -= 1
if max_move[0] >= sr or max_move[1] > w-sr or max_move[2] >= sc or max_move[3] > h-sc:
print("NO")
else:
print("YES") | s626117054 | Accepted | 247 | 3,888 | 857 | h, w, n = [int(item) for item in input().split()]
sr, sc = [int(item) for item in input().split()]
sr -= 1
sc -= 1
s = input().rstrip()
t = input().rstrip()
x = sc
# Check left
for i in range(n):
if s[i] == "L":
x -= 1
if x < 0:
print("NO")
exit()
if t[i] == "R" and x+1 < w-1:
x += 1
x = sc
# Check Right
for i in range(n):
if s[i] == "R":
x += 1
if x > w-1:
print("NO")
exit()
if t[i] == "L" and x > 0:
x -= 1
y = sr
# Check Down
for i in range(n):
if s[i] == "D":
y += 1
if y > h-1:
print("NO")
exit()
if t[i] == "U" and y > 0:
y -= 1
y = sr
# Check up
for i in range(n):
if s[i] == "U":
y -= 1
if y < 0:
print("NO")
exit()
if t[i] == "D" and y+1 < h-1:
y += 1
print("YES") |
s064914879 | p02399 | u731710433 | 1,000 | 131,072 | Wrong Answer | 30 | 7,480 | 72 | 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(float, input().split())
print(int(a // b), int(a % b), a / b) | s456933563 | Accepted | 20 | 7,728 | 94 | a, b = [int(x) for x in input().split()]
print("{0} {1} {2:.5f}".format(a // b, a % b, a / b)) |
s672416299 | p02408 | u635647915 | 1,000 | 131,072 | Wrong Answer | 30 | 7,780 | 457 | Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers). The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond. | #coding:utf-8
def ch2int(a):
s=["S","H","C","D"]
for i,b in enumerate(s):
if a==b:
return i
def int2ch(i):
s=["S","H","C","D"]
return s[i]
n=int(input())
card=[[0 for i in range(0,4)]for j in range(0,13)]
for i in range(n):
c,x=list(map(str,input().split()))
card[int(x)-1][ch2int(c)]=1
for i,x in enumerate(card):
for j,y in enumerate(x):
if y==0:
print(int2ch(j)+" "+str(i+1))
| s912385780 | Accepted | 20 | 7,712 | 457 | #coding:utf-8
def ch2int(a):
s=["S","H","C","D"]
for i,b in enumerate(s):
if a==b:
return i
def int2ch(i):
s=["S","H","C","D"]
return s[i]
n=int(input())
card=[[0 for i in range(0,13)]for j in range(0,4)]
for i in range(n):
c,x=list(map(str,input().split()))
card[ch2int(c)][int(x)-1]=1
for i,x in enumerate(card):
for j,y in enumerate(x):
if y==0:
print(int2ch(i)+" "+str(j+1))
|
s542307176 | p03578 | u966601619 | 2,000 | 262,144 | Wrong Answer | 2,105 | 35,420 | 312 | Rng is preparing a problem set for a qualification round of CODEFESTIVAL. He has N candidates of problems. The difficulty of the i-th candidate is D_i. There must be M problems in the problem set, and the difficulty of the i-th problem must be T_i. Here, one candidate of a problem cannot be used as multiple problems. Determine whether Rng can complete the problem set without creating new candidates of problems. | q=input()
x=[int(i) for i in input().split()]
cnt=input()
j=0
y=[int(i) for i in input().split()]
for k in range(len(y)) :
for i in range(len(x)) :
if x[i] == y[k] :
del x[i]
break
if i+1 == len(x) :
print('NO')
j=1
break
if j==0 :
print('YES') | s103935280 | Accepted | 305 | 41,816 | 443 |
q=input()
x=[int(i) for i in input().split()]
cnt=input()
y=[int(i) for i in input().split()]
if max(x) >= max(y):
MAX=max(x)
else:
MAX=max(y)
dic = {}
for k in x :
if k in dic.keys() :
dic[k] += 1
else :
dic[k] = 1
for i in y :
if i in dic.keys() :
dic[i] -= 1
if dic[i] == -1 :
print('NO')
exit()
else :
print('NO')
exit()
print('YES')
|
s770316780 | p03470 | u401077816 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 80 | An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have? | N = int(input())
A = list(map(int, input().split()))
print(len(set(sorted(A)))) | s741839572 | Accepted | 17 | 2,940 | 76 | N = int(input())
l = sorted([input() for _ in range(N)])
print(len(set(l))) |
s018667884 | p03089 | u731368968 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,064 | 420 | Snuke has an empty sequence a. He will perform N operations on this sequence. In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1). You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it. | N=int(input())
b = list(map(int, input().split()))
ans=[]
for i in range(N):
print(b)
#j:index of delete + 1(ans base)
j = 1
if j < len(b):
while b[j] == j + 1:
j += 1
if j == len(b):break
if b[j - 1] == j:
del b[j-1]
ans.append(j)
else:
print(-1)
import sys
sys.exit()
for i in range(len(ans)-1,-1,-1):
print(ans[i]) | s040202527 | Accepted | 17 | 3,064 | 422 | N=int(input())
b = list(map(int, input().split()))
ans=[]
#delete N times
for i in range(N):
#j:index of delete + 1(ans base)
j = False
for k in range(len(b)-1,-1,-1):
if b[k] == k + 1:
j = True
del b[k]
ans.append(k+1)
break
if not j:
print(-1)
import sys
sys.exit()
for i in range(len(ans) - 1, -1, -1):
print(ans[i]) |
s822957986 | p03549 | u463655976 | 2,000 | 262,144 | Wrong Answer | 19 | 3,188 | 78 | Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`. When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases. Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds. Now, he goes through the following process: * Submit the code. * Wait until the code finishes execution on all the cases. * If the code fails to correctly solve some of the M cases, submit it again. * Repeat until the code correctly solve all the cases in one submission. Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer). | N, M = map(int, input().split())
print((1900 * M + 100 * (N-M)) / pow(1/2, M)) | s240115259 | Accepted | 18 | 2,940 | 96 | N, M = map(int, input().split())
print("{:.0f}".format((1900 * M + 100 * (N-M)) / pow(1/2, M)))
|
s345213768 | p03836 | u752513456 | 2,000 | 262,144 | Wrong Answer | 23 | 3,064 | 292 | 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())
s = ''
s += 'R' * (tx - sx) + 'U' * (ty - sy)
s += 'L' + 'D' * (ty - sy - 1) + 'L' * (tx - sx - 1) + 'D'
s += 'D' + 'R' * (tx - sx + 1) + 'U' * (ty - sy + 1) + 'L'
s+= 'U' + 'L' * 2 + 'D' * (ty - sy - 1) + 'L' * (tx - sx - 1) + 'D' + 'R'
print(s)
| s223291104 | Accepted | 22 | 3,064 | 256 | sx, sy, tx, ty = map(int, input().split())
s = ''
s += 'U' * (ty - sy) + 'R' * (tx - sx)
s += 'D' * (ty - sy) + 'L' * (tx - sx)
s += 'L' + 'U' * (ty - sy + 1) + 'R' * (tx - sx + 1) + 'D'
s += 'R' + 'D' * (ty - sy + 1) + 'L' * (tx -sx + 1) + 'U'
print(s)
|
s943553690 | p03192 | u858523893 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 18 | You are given an integer N that has exactly four digits in base ten. How many times does `2` occur in the base-ten representation of N? | input().count("2") | s831356119 | Accepted | 20 | 2,940 | 25 | print(input().count("2")) |
s873401721 | p03149 | u065793287 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 87 | You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974". | a = list(map(int,input().split()))
a.sort()
print( 'YES' if a == [1,9,7,4] else 'NO' )
| s063261084 | Accepted | 18 | 2,940 | 87 | a = list(map(int,input().split()))
a.sort()
print( 'YES' if a == [1,4,7,9] else 'NO' )
|
s355021238 | p02742 | u902380746 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,060 | 217 | 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 sys
import math
import bisect
def main():
n, m = map(int, input().split())
a = (m + 1) // 2
b = m // 2
ans = a * (n + 1) // 2 + b * n // 2
print(ans)
if __name__ == "__main__":
main()
| s677655376 | Accepted | 17 | 3,060 | 278 | import sys
import math
import bisect
def main():
n, m = map(int, input().split())
if n == 1 or m == 1:
print(1)
else:
ans = ((m + 1) // 2) * ((n + 1) // 2)
ans += (m // 2) * (n // 2)
print(ans)
if __name__ == "__main__":
main()
|
s419224621 | p03471 | u281303342 | 2,000 | 262,144 | Wrong Answer | 685 | 3,060 | 178 | The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough. | N,Y = map(int,input().split())
for i in range(N+1):
for j in range(N-i+1):
if i*10000 + j*5000 + (N-(i+j))*1000 == Y:
print(i,j,N-(i+j))
break | s269323620 | Accepted | 712 | 3,060 | 537 | # python 3.4.3
import sys
input = sys.stdin.readline
# -------------------------------------------------------------
# library
# -------------------------------------------------------------
# -------------------------------------------------------------
# main
# -------------------------------------------------------------
N,Y = map(int,input().split())
ans = [-1,-1,-1]
for i in range(N+1):
for j in range(N+1-i):
if Y == 10000*i + 5000*j + 1000*(N-i-j):
ans = [i,j,N-i-j]
print(" ".join(map(str,ans))) |
s143581906 | p02602 | u377834804 | 2,000 | 1,048,576 | Wrong Answer | 1,705 | 25,184 | 222 | M-kun is a student in Aoki High School, where a year is divided into N terms. There is an exam at the end of each term. According to the scores in those exams, a student is given a grade for each term, as follows: * For the first through (K-1)-th terms: not given. * For each of the K-th through N-th terms: the multiplication of the scores in the last K exams, including the exam in the graded term. M-kun scored A_i in the exam at the end of the i-th term. For each i such that K+1 \leq i \leq N, determine whether his grade for the i-th term is **strictly** greater than the grade for the (i-1)-th term. | N, K = map(int, input().split())
A = []
for i, a in enumerate(input().split()):
if i < K:
A.append(a)
continue
else:
A.append(a)
x = A.pop(0)
if x < a:
print('Yes')
else:
print('No') | s835318911 | Accepted | 1,654 | 31,544 | 238 | N, K = map(int, input().split())
A = []
for i, a in enumerate(list(map(int, input().split()))):
if i < K:
A.append(a)
continue
else:
A.append(a)
x = A.pop(0)
if x < a:
print('Yes')
else:
print('No') |
s897746303 | p02853 | u368796742 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,064 | 412 | We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned. | x,y = map(int, input().split())
if x > 3 and y > 3:
print("0")
elif x == 1:
if y == 1:
print("100000")
elif y == 2:
print("50000")
elif y == 3:
print("40000")
elif x == 2:
if y == 1:
print("500000")
elif y == 2:
print("40000")
elif y == 3:
print("30000")
elif x == 3:
if y == 1:
print("400000")
elif y == 2:
print("30000")
elif y == 3:
print("20000")
| s487102081 | Accepted | 17 | 3,064 | 250 | x,y = map(int,input().split())
def point(a):
if a == 1:
return 300000
elif a == 2:
return 200000
elif a == 3:
return 100000
else:
return 0
c = point(x)
b = point(y)
if x == 1 and y == 1:
print(1000000)
else:
print(c+b) |
s232051202 | p03573 | u117193815 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 68 | You are given three integers, A, B and C. Among them, two are the same, but the remaining one is different from the rest. For example, when A=5,B=7,C=5, A and C are the same, but B is different. Find the one that is different from the rest among the given three integers. | a,b,c=map(int, input().split())
if a==b:
print(c)
else:
print(a) | s550372279 | Accepted | 17 | 3,064 | 91 | a,b,c=map(int, input().split())
if a==b:
print(c)
elif a==c:
print(b)
else:
print(a)
|
s156151691 | p02669 | u923270446 | 2,000 | 1,048,576 | Wrong Answer | 23 | 9,136 | 458 | You start with the number 0 and you want to reach the number N. You can change the number, paying a certain amount of coins, with the following operations: * Multiply the number by 2, paying A coins. * Multiply the number by 3, paying B coins. * Multiply the number by 5, paying C coins. * Increase or decrease the number by 1, paying D coins. You can perform these operations in arbitrary order and an arbitrary number of times. What is the minimum number of coins you need to reach N? **You have to solve T testcases.** | t = int(input())
for i in range(t):
n, a, b, c, d = map(int, input().split())
ans = 0
for i in range(60):
if n == 1:
ans += d
n = 0
break
if n % 5 == 0:
ans += c
n //= 5
elif n % 3 == 0:
ans += b
n //= 3
elif n % 2 == 0:
ans += a
n //= 2
else:
ans += d
n += 1
print(ans) | s133993692 | Accepted | 286 | 11,004 | 509 | def dfs(n):
if n == 1:
return d
if n == 0:
return 0
if n in dic:
return dic[n]
ans = min(n * d, dfs(n // 2) + a + (n % 2) * d, dfs((n + 1) // 2) + a + ((-1 * n) % 2) * d, dfs(n // 3) + b + (n % 3) * d, dfs((n + 2) // 3) + b + ((-1 * n) % 3) * d, dfs(n // 5) + c + (n % 5) * d, dfs((n + 4) // 5) + c + ((-1 * n) % 5) * d)
dic[n] = ans
return ans
t = int(input())
for i in range(t):
n, a, b, c, d = map(int, input().split())
dic = dict()
print(dfs(n)) |
s024264750 | p03712 | u530383736 | 2,000 | 262,144 | Wrong Answer | 24 | 3,952 | 305 | 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. | # -*- coding: utf-8 -*-
H,W = list(map(int, input().rstrip().split()))
a_list = [list(input().strip()) for _ in range(H)]
#-----
b_list=[ ["#" for c in range(W+2)] for r in range(H+2) ]
for i in range(H):
for j in range(W):
b_list[i+1][j+1] = a_list[i][j]
for li in b_list:
print(*li)
| s414538360 | Accepted | 25 | 4,596 | 312 | # -*- coding: utf-8 -*-
H,W = list(map(int, input().rstrip().split()))
a_list = [list(input().strip()) for _ in range(H)]
#-----
b_list=[ ["#" for c in range(W+2)] for r in range(H+2) ]
for i in range(H):
for j in range(W):
b_list[i+1][j+1] = a_list[i][j]
for li in b_list:
print(*li,sep="")
|
s194254079 | p04029 | u777394984 | 2,000 | 262,144 | Wrong Answer | 21 | 3,316 | 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((1+n)*n/2) | s223641914 | Accepted | 23 | 3,316 | 37 | n=int(input())
print(int((1+n)*n/2))
|
s503464086 | p03657 | u457957084 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 122 | Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies. | a, b = map(int,input().split())
if (a % 3 != 0) and ( b % 3 != 0):
print("Impossible")
else:
print("Possible")
| s537994892 | Accepted | 17 | 2,940 | 141 | a, b = map(int,input().split())
if (a % 3 == 0) or( b % 3 == 0) or ((a + b) % 3 ==0):
print("Possible")
else:
print("Impossible")
|
s176564177 | p03501 | u344888046 | 2,000 | 262,144 | Wrong Answer | 26 | 9,100 | 53 | You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours. | N, A, B = map(int, input().split())
print(N * A + B) | s944688488 | Accepted | 28 | 9,092 | 80 | N, A, B = map(int, input().split())
ans = N * A if N * A < B else B
print(ans) |
s380315834 | p02615 | u608755339 | 2,000 | 1,048,576 | Wrong Answer | 238 | 31,432 | 170 | Quickly after finishing the tutorial of the online game _ATChat_ , you have decided to visit a particular place with N-1 players who happen to be there. These N players, including you, are numbered 1 through N, and the **friendliness** of Player i is A_i. The N players will arrive at the place one by one in some order. To make sure nobody gets lost, you have set the following rule: players who have already arrived there should form a circle, and a player who has just arrived there should cut into the circle somewhere. When each player, except the first one to arrive, arrives at the place, the player gets **comfort** equal to the smaller of the friendliness of the clockwise adjacent player and that of the counter-clockwise adjacent player. The first player to arrive there gets the comfort of 0. What is the maximum total comfort the N players can get by optimally choosing the order of arrivals and the positions in the circle to cut into? | N = int(input())
A = sorted([int(i) for i in input().split()], reverse=True)
S = A[0]
step = 2
print(S)
for i in range(1,N-1):
S += (A[(i+1)//2])
print(S)
print(S) | s707535539 | Accepted | 150 | 31,500 | 150 | N = int(input())
A = sorted([int(i) for i in input().split()], reverse=True)
S = A[0]
step = 2
for i in range(1,N-1):
S += (A[(i+1)//2])
print(S) |
s586676636 | p02850 | u602740328 | 2,000 | 1,048,576 | Wrong Answer | 2,105 | 29,184 | 581 | 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. | from functools import reduce
from operator import mul
N = int(input())
ab = [tuple(map(int, input().split())) for _ in range(N-1)]
print(N, ab)
E_n = {k: 0 for k in ab}
for i in range(1,N+1):
items = {k:v for k,v in E_n.items() if i in k}
c = 0
for k in items.keys():
if items[k] != 0: continue
while True:
c += 1
if c not in items.values():
items[k] = c
break
for k,v in items.items(): E_n[k] = v
if reduce(mul, E_n.values()) != 0: break
print(len(set(E_n.values())))
for k in sorted(E_n.keys()): print(E_n[k])
| s227730394 | Accepted | 605 | 49,600 | 417 | N = int(input())
edge = {}
tree = [[] for i in range(N)]
colors = [1]*(N-1)
c_p = [[] for i in range(N)]
for i in range(N-1):
a,b = map(int,input().split())
a,b = a-1,b-1
tree[a].append(b)
edge[(a,b)] = i
for i in range(N):
c = 1
for v_c in tree[i]:
while c in c_p[i]: c+=1
colors[edge[(i,v_c)]] = c
c_p[v_c].append(c)
c+=1
print(max(colors))
print(*colors, sep="\n") |
s027836696 | p02608 | u243748614 | 2,000 | 1,048,576 | Wrong Answer | 2,206 | 9,204 | 497 | Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N). | from math import trunc, sqrt
def p(x,y,z):
return x**2+y**2+z**2+x*y+y*z+z*x
def perm(x,y,z):
if x==y and y==z and z==x:
return 1
if x!=y and y!=z and z!=x:
return 6
return 3
def f(n):
mn=trunc(sqrt(n // 3))+1
k=0
for x in range(1,mn+1):
for y in range(x,mn+1):
for z in range(y,mn+1):
if p(x,y,z)==n:
k += perm(x,y,z)
return k
N=int(input())
for n in range(1,N):
print(n, f(n))
| s032108665 | Accepted | 235 | 9,252 | 245 | N=int(input())
a=[0]*N
x=1
while x*x<=N:
y=1
while x*x+y*y+x*y<=N:
z=1
while z*z+x*x+y*y+x*y+z*y+z*x<=N:
a[z*z+x*x+y*y+x*y+z*y+z*x-1]+=1
z+=1
y+=1
x+=1
for n in a:
print(n)
|
s498309868 | p03854 | u472534477 | 2,000 | 262,144 | Wrong Answer | 18 | 3,188 | 370 | You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`. | S= input()
i=0
while True:
flag = False
if S[i:i+7] == "dreamer":
i += 7
flag = True
if S[i:i+6] == "eraser":
i += 6
flag = True
if S[i:i+5] == "erase" or S[i:i+5] == "dream":
i += 5
flag = True
if not flag:
break
if i == len(S):
print("Yes")
else:
print("No")
| s642435279 | Accepted | 18 | 3,188 | 128 | s=input().replace("eraser","").replace("erase","").replace("dreamer","").replace("dream","")
print("YES" if len(s)==0 else "NO") |
s261435705 | p03944 | u589969467 | 2,000 | 262,144 | Wrong Answer | 29 | 9,076 | 454 | There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting. | w,h,n = map(int,input().split())
#print(x_list)
#print(y_list)
min_x = 0
max_x = w
min_y = 0
max_y = h
for i in range(n):
x,y,a = map(int,input().split())
if a==1:
min_x = x
elif a==2:
max_x = x
elif a==3:
min_y = y
else:
max_y = y
if min_x > max_x or min_y > max_y:
print(0)
else:
print((max_x-min_y)*(max_y-min_y)) | s836923603 | Accepted | 29 | 9,100 | 498 | w,h,n = map(int,input().split())
#print(x_list)
#print(y_list)
min_x = 0
max_x = w
min_y = 0
max_y = h
for i in range(n):
x,y,a = map(int,input().split())
if a==1:
min_x = max(min_x,x)
elif a==2:
max_x = min(max_x,x)
elif a==3:
min_y = max(min_y,y)
else:
max_y = min(max_y,y)
if min_x > max_x or min_y > max_y:
print(0)
else:
print((max_x-min_x)*(max_y-min_y)) |
s862226060 | p03360 | u936985471 | 2,000 | 262,144 | Wrong Answer | 19 | 2,940 | 80 | There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? | l=list(map(int,input().split()))
k=int(input())
print(sum(l)+max(l)**k-max(l))
| s048957645 | Accepted | 17 | 2,940 | 73 | A=list(map(int,input().split()))
print(sum(A)+max(A)*(2**int(input())-1)) |
s814248298 | p03573 | u126747509 | 2,000 | 262,144 | Wrong Answer | 22 | 3,316 | 229 | You are given three integers, A, B and C. Among them, two are the same, but the remaining one is different from the rest. For example, when A=5,B=7,C=5, A and C are the same, but B is different. Find the one that is different from the rest among the given three integers. | from collections import defaultdict
def main():
d = defaultdict(lambda: 0)
for i in input().split():
d[int(i)] += 1
d = {v:k for k, v in d.items()}
print(d[2])
if __name__ == "__main__":
main() | s970267132 | Accepted | 21 | 3,316 | 229 | from collections import defaultdict
def main():
d = defaultdict(lambda: 0)
for i in input().split():
d[int(i)] += 1
d = {v:k for k, v in d.items()}
print(d[1])
if __name__ == "__main__":
main() |
s729025168 | p03993 | u502175663 | 2,000 | 262,144 | Wrong Answer | 2,103 | 16,308 | 255 | There are N rabbits, numbered 1 through N. The i-th (1≤i≤N) rabbit likes rabbit a_i. Note that no rabbit can like itself, that is, a_i≠i. For a pair of rabbits i and j (i＜j), we call the pair (i，j) a _friendly pair_ if the following condition is met. * Rabbit i likes rabbit j and rabbit j likes rabbit i. Calculate the number of the friendly pairs. | n = int(input())
line = input().split(' ')
total_list =[]
for i in range(0,n):
total = i + int(line[i])
total_list.append(total)
print(total_list)
answer = 0
for i in range(n *2 -1):
if total_list.count(i) == 2:
answer += 1
print(answer) | s725586290 | Accepted | 113 | 10,740 | 158 | n = int(input())
a = input().split(' ')
t = 0
for o in range(n):
i = o + 1
b = int(a[o]) - i
if i == int(a[o + b]):
t += 1
print(int(t / 2)) |
s753410179 | p02233 | u216229195 | 1,000 | 131,072 | Wrong Answer | 20 | 7,604 | 152 | Write a program which prints $n$-th fibonacci number for a given integer $n$. The $n$-th fibonacci number is defined by the following recursive formula: \begin{equation*} fib(n)= \left \\{ \begin{array}{ll} 1 & (n = 0) \\\ 1 & (n = 1) \\\ fib(n - 1) + fib(n - 2) & \\\ \end{array} \right. \end{equation*} | def fib(n):
if n < 0:
print("error")
exit()
if n < 2:
result = 1
result = fib(n-1) + fib(n-2)
return result
n = int(input())
fib(n) | s052848630 | Accepted | 20 | 5,604 | 218 | def fibr(n):
if arr[n] != 0:
return arr[n]
if n < 2:
arr[n] = 1
else:
arr[n] = fibr(n-1) + fibr(n-2)
return arr[n]
n = int(input())
arr = [0 for i in range(n+1)]
print(fibr(n))
|
s800568439 | p02748 | u490489966 | 2,000 | 1,048,576 | Wrong Answer | 464 | 18,612 | 297 | You are visiting a large electronics store to buy a refrigerator and a microwave. The store sells A kinds of refrigerators and B kinds of microwaves. The i-th refrigerator ( 1 \le i \le A ) is sold at a_i yen (the currency of Japan), and the j-th microwave ( 1 \le j \le B ) is sold at b_j yen. You have M discount tickets. With the i-th ticket ( 1 \le i \le M ), you can get a discount of c_i yen from the total price when buying the x_i-th refrigerator and the y_i-th microwave together. Only one ticket can be used at a time. You are planning to buy one refrigerator and one microwave. Find the minimum amount of money required. | #B
a,b,m=map(int,input().split())
al=list(map(int,input().split()))
bl=list(map(int,input().split()))
amin=min(al)
bmin=min(bl)
minp=amin+bmin
print(amin,bmin)
for i in range(m):
num=list(map(int,input().split()))
price=al[num[0]-1]+bl[num[1]-1]-num[2]
minp=min(price,minp)
print(minp) | s446408584 | Accepted | 458 | 18,612 | 280 | #B
a,b,m=map(int,input().split())
al=list(map(int,input().split()))
bl=list(map(int,input().split()))
amin=min(al)
bmin=min(bl)
minp=amin+bmin
for i in range(m):
num=list(map(int,input().split()))
price=al[num[0]-1]+bl[num[1]-1]-num[2]
minp=min(price,minp)
print(minp) |
s297675886 | p03448 | u698348858 | 2,000 | 262,144 | Wrong Answer | 30 | 3,064 | 268 | 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. | #coding:utf-8
a = int(input())
b = int(input())
c = int(input())
x = int(input())
cnt = 0
for i in range(a):
for j in range(b):
for k in range(c):
p = a * 500 + b * 100 + c * 50
if p == x:
cnt += 1
elif p > x:
break
print(cnt) | s213251245 | Accepted | 48 | 3,064 | 280 | #coding:utf-8
a = int(input())
b = int(input())
c = int(input())
x = int(input())
cnt = 0
for i in range(a + 1):
for j in range(b + 1):
for k in range(c + 1):
p = i * 500 + j * 100 + k * 50
if p > x:
break
if p == x:
cnt += 1
print(cnt) |
s126579976 | p02401 | u104931506 | 1,000 | 131,072 | Time Limit Exceeded | 40,000 | 7,864 | 179 | Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part. | a, op, b = input().split()
while True:
if op == '+' : print(a + b)
elif op == '-' : print(a - b)
elif op == '/' : print(a / b)
elif op == '*' : print(a * b)
else : break | s214950619 | Accepted | 40 | 7,656 | 203 | while True:
a, op, b = input().split()
a = int(a)
b = int(b)
if op == '+': print(a + b)
elif op == '-': print(a - b)
elif op == '/': print(a // b)
elif op == '*': print(a * b)
else: break |
s295979512 | p03455 | u323532272 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 82 | 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 == 0:
print('Even')
else:
print('Odd') | s764873628 | Accepted | 17 | 2,940 | 85 | a, b = map(int, input().split())
if a*b%2 == 0:
print('Even')
else:
print('Odd')
|
s788949008 | p03130 | u989623817 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,064 | 392 | There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads. Determine if we can visit all the towns by traversing each of the roads exactly once. | a1, b1 = map(int, input().split())
a2, b2 = map(int, input().split())
a3, b3 = map(int, input().split())
cnt = {1: 0, 2: 0, 3: 0, 4: 0}
for i in range(4):
if not i+1 in [a1, a2, a3, b1, b2, b3]:
print('NO')
exit()
lis = [a1, a2, a3, b1, b2, b3]
c = 0
for l in range(4):
cnt = lis.count(l+1)
if cnt > 1:
c += 1
if c <= 2:
print('NO')
print('YES')
| s987488845 | Accepted | 17 | 3,064 | 401 | a1, b1 = map(int, input().split())
a2, b2 = map(int, input().split())
a3, b3 = map(int, input().split())
cnt = {1: 0, 2: 0, 3: 0, 4: 0}
for i in range(4):
if not i+1 in [a1, a2, a3, b1, b2, b3]:
print('NO')
exit()
lis = [a1, a2, a3, b1, b2, b3]
c = 0
for l in range(4):
cnt = lis.count(l+1)
if cnt > 1:
c += 1
if c < 2:
print('NO')
exit()
print('YES')
|
s164764282 | p02742 | u571395477 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 208 | 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
def main():
H, W = [int(_) for _ in input().split()]
if H % 2 == 0 or W % 2 == 0:
m = (H * W) /2
else:
m = (H * W) /2
m = int(math.ceil(m))
print(m)
main() | s983001390 | Accepted | 18 | 3,060 | 294 | import sys
import math
def main():
H, W = [int(_) for _ in input().split()]
if H == 1 or W == 1:
m = 1
print(int(m))
sys.exit()
elif H % 2 == 1 and W % 2 == 1:
m = (H * W) // 2 + 1
else:
m = (H * W) // 2
print(int(m))
main() |
s803407351 | p03379 | u962127640 | 2,000 | 262,144 | Wrong Answer | 2,109 | 34,196 | 233 | When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N. | import numpy as np
N= int(input())
a = list(map(int, input().split(' ')))
a = np.array(a)
ret = []
for i in range(N):
ind = np.ones(N, dtype=bool)
ind[i] = False
tmp = a[ind]
ret.append(int(np.median(tmp)))
print(ret) | s525779032 | Accepted | 413 | 34,156 | 214 | import numpy as np
N= int(input())
a = list(map(int, input().split(' ')))
sort_a = sorted(a)
small = sort_a[N//2-1]
big = sort_a[N//2]
for aa in a:
if aa < big:
print(big)
else:
print(small) |
s284051930 | p03359 | u777846995 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 118 | In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi? | x, y = map(int, input().split())
cnt = 0
for i in range(1, 13):
if x == i and y == i:
cnt += 1
print(cnt) | s639572936 | Accepted | 17 | 2,940 | 74 | x, y = map(int, input().split())
if x <= y:
print(x)
else:
print(x-1) |
s798753036 | p03502 | u061545295 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 89 | An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number. | N = input()
f = [int(i) for i in N]
if N%f == 0:
print("Yes")
else:
print('No')
| s915782919 | Accepted | 17 | 2,940 | 110 | N = input()
num = [int(i) for i in N]
f = sum(num)
if int(N)%f == 0:
print("Yes")
else:
print('No')
|
s534572122 | p04029 | u701318346 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 40 | 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((1 + N) * N / 2) | s162981110 | Accepted | 17 | 2,940 | 45 | N = int(input())
print(int((1 + N) * N / 2)) |
s066801736 | p03644 | u527190736 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 188 | 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())
x = 1
cnt = 0
list = []
for i in range(n):
i = x
while i % 2 == 0:
i /= 2
cnt += 1
x += 1
list.append(cnt)
cnt = 0
print(max(list))
| s099147105 | Accepted | 17 | 2,940 | 196 | n = int(input())
x = 1
cnt = 0
list = []
for i in range(n):
i = x
while i % 2 == 0:
i = i//2
cnt += 1
x += 1
list.append(cnt)
cnt = 0
print(2 ** max(list))
|
s620950634 | p03371 | u859897687 | 2,000 | 262,144 | Wrong Answer | 47 | 3,060 | 185 | "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())
if a+b>=c+c:
print(a*x+b*y)
else:
ans=10000000000
for i in range(0,(max(x,y)+1)//2):
k=c*2*i+a*(x-i)+b*(y-i)
ans=min(ans,k)
print(ans) | s511777933 | Accepted | 121 | 3,060 | 150 | a,b,c,x,y=map(int,input().split())
ans=10000000000
for i in range(0,(max(x,y)+1)):
k=c*2*i+a*max(0,(x-i))+max(0,b*(y-i))
ans=min(ans,k)
print(ans) |
s987175021 | p03193 | u518987899 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,188 | 160 | There are N rectangular plate materials made of special metal called AtCoder Alloy. The dimensions of the i-th material are A_i \times B_i (A_i vertically and B_i horizontally). Takahashi wants a rectangular plate made of AtCoder Alloy whose dimensions are exactly H \times W. He is trying to obtain such a plate by choosing one of the N materials and cutting it if necessary. When cutting a material, the cuts must be parallel to one of the sides of the material. Also, the materials have fixed directions and cannot be rotated. For example, a 5 \times 3 material cannot be used as a 3 \times 5 plate. Out of the N materials, how many can produce an H \times W plate if properly cut? | import sys
lines = [list(map(int, line.strip().split(' '))) for line in sys.stdin]
print(sum([1 for h,w in lines[1:] if h <= lines[0][1] and w <= lines[0][2]])) | s210542012 | Accepted | 18 | 3,188 | 160 | import sys
lines = [list(map(int, line.strip().split(' '))) for line in sys.stdin]
print(sum([1 for h,w in lines[1:] if lines[0][1] <= h and lines[0][2] <= w])) |
s984572968 | p00067 | u847467233 | 1,000 | 131,072 | Wrong Answer | 30 | 5,620 | 1,085 | 地勢を示す縦 12, 横 12 のマスからなる平面図があります。おのおののマスは白か黒に塗られています。白は海を、黒は陸地を表します。二つの黒いマスが上下、あるいは左右に接しているとき、これらは地続きであるといいます。この平面図では、黒いマス一つのみ、あるいは地続きの黒いマスが作る領域を「島」といいます。例えば下図には、5 つの島があります。 ■■■■□□□□■■■■ ■■■□□□□□■■■■ ■■□□□□□□■■■■ ■□□□□□□□■■■■ □□□■□□□■□□□□ □□□□□□■■■□□□ □□□□□■■■■■□□ ■□□□■■■■■■■□ ■■□□□■■■■■□□ ■■■□□□■■■□□□ ■■■■□□□■□□□□ □□□□□□□□□□□□ マスのデータを読み込んで、島の数を出力するプログラムを作成してください。 | # AOJ 0067 The Number of Island
# Python3 2018.6.16 bal4u
# UNION-FIND library
MAX = 12*12
id, size = [0]*MAX, [0]*MAX
def init(n):
for i in range(n): id[i], size[i] = i, 1
def root(i):
while i != id[i]:
id[i] = id[id[i]]
i = id[i]
return i
def connected(p, q):
return root(p) == root(q)
def unite(p, q):
i, j = root(p), root(q)
if i == j: return
if size[i] < size[j]:
id[i] = j
size[j] += size[i]
else:
id[j] = i
size[i] += size[j]
# UNION-FIND library
arr = [[0 for r in range(12)] for c in range(12)]
dr = [-1,0,1, 0 ]
dc = [ 0,1,0,-1 ]
while True:
init(12*12)
for r in range(12): arr[r] = list(map(int, input()))
for r in range(12): print('r=', arr[r])
for r in range(12):
for c in range(12):
if arr[r][c] == 0: continue
for i in range(4):
nr, nc = r + dr[i], c + dc[i]
if nr >= 0 and nr < 12 and nc >= 0 and nc < 12:
if arr[nr][nc] == 1:
unite(r*12+c, nr*12+nc)
ans = 0
for r in range(12):
for c in range(12):
if arr[r][c] == 1 and root(r*12+c) == r*12+c: ans += 1
print(ans)
try: input()
except: break
| s824486759 | Accepted | 30 | 5,608 | 1,045 | # AOJ 0067 The Number of Island
# Python3 2018.6.16 bal4u
# UNION-FIND library
MAX = 12*12
id, size = [0]*MAX, [0]*MAX
def init(n):
for i in range(n): id[i], size[i] = i, 1
def root(i):
while i != id[i]:
id[i] = id[id[i]]
i = id[i]
return i
def connected(p, q):
return root(p) == root(q)
def unite(p, q):
i, j = root(p), root(q)
if i == j: return
if size[i] < size[j]:
id[i] = j
size[j] += size[i]
else:
id[j] = i
size[i] += size[j]
# UNION-FIND library
arr = [[0 for r in range(12)] for c in range(12)]
dr = [-1, 0, 1, 0]
dc = [ 0, 1, 0,-1]
while True:
init(12*12)
for r in range(12): arr[r] = list(map(int, input()))
for r in range(12):
for c in range(12):
if arr[r][c] == 0: continue
for i in range(4):
nr, nc = r + dr[i], c + dc[i]
if nr >= 0 and nr < 12 and nc >= 0 and nc < 12:
if arr[nr][nc] == 1:
unite(r*12+c, nr*12+nc)
ans = 0
for r in range(12):
for c in range(12):
if arr[r][c] == 1 and root(r*12+c) == r*12+c: ans += 1
print(ans)
try: input()
except: break
|
s436730464 | p03610 | u626468554 | 2,000 | 262,144 | Wrong Answer | 76 | 5,292 | 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 = list(input())
for i in range(1,len(s),2):
print(s[i],end="")
print() | s277091605 | Accepted | 68 | 5,292 | 76 | s = list(input())
for i in range(0,len(s),2):
print(s[i],end="")
print() |
s433879903 | p04045 | u488884575 | 2,000 | 262,144 | Wrong Answer | 35 | 3,064 | 528 | Iroha is very particular about numbers. There are K digits that she dislikes: D_1, D_2, ..., D_K. She is shopping, and now paying at the cashier. Her total is N yen (the currency of Japan), thus she has to hand at least N yen to the cashier (and possibly receive the change). However, as mentioned before, she is very particular about numbers. When she hands money to the cashier, the decimal notation of the amount must not contain any digits that she dislikes. Under this condition, she will hand the minimum amount of money. Find the amount of money that she will hand to the cashier. |
import itertools
N, K = input().split()
P = input().split()
a= sorted(set([str(i) for i in range(10)])-set(P))
b1 = itertools.product(a, repeat=len(N))
b2 = itertools.product(a, repeat=len(N)+1)
ans=str()
cnt=0
for i in b1:
#print(type(i))
ans = "".join(i)
if N <= ans:
print(ans)
cnt=1
break
if cnt==1:
for i in b2:
ans = "".join(i)
if N <= ans:
print(ans)
break | s421744213 | Accepted | 19 | 3,064 | 548 |
import itertools
N, K = input().split()
P = input().split()
a= sorted(set([str(i) for i in range(10)])-set(P))
b1 = itertools.product(a, repeat=len(N))
b2 = itertools.product(a, repeat=len(N)+1)
ans=str()
cnt=0
for i in b1:
#print(type(i))
ans = "".join(i)
if N <= ans:
print(ans)
cnt=1
break
if cnt==0:
for i in b2:
ans = "".join(i)
#print(ans)
if "0"+N <= ans:
print(ans)
break |
s942125315 | p03433 | u421499233 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 125 | E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins. | N = int(input())
A = int(input())
if N <= A:
print("Yes")
elif N - 500 <= A:
print ("Yes")
else:
print("No")
| s490810038 | Accepted | 17 | 2,940 | 93 | N = int(input())
A = int(input())
r = N%500
if A >= r:
print("Yes")
else:
print("No") |
s904587578 | p03067 | u559346857 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 71 | There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise. | a,b,c=map(int,input().split())
print("YES" if a<c<b or b<c<a else "NO") | s981580960 | Accepted | 18 | 2,940 | 71 | a,b,c=map(int,input().split())
print("Yes" if a<c<b or b<c<a else "No") |
s861338851 | p03680 | u716660050 | 2,000 | 262,144 | Wrong Answer | 162 | 12,988 | 178 | Takahashi wants to gain muscle, and decides to work out at AtCoder Gym. The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it. Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up. Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons. | N=int(input())
a=[int(input())for i in range(N)]
p=1
ok=False
ans=0
for i in range(N):
ans+=1
if p==2:
ok=True
break
p=a[p-1]
print(ans if ok else -1) | s946087932 | Accepted | 160 | 13,048 | 178 | N=int(input())
a=[int(input())for i in range(N)]
p=1
ok=False
ans=0
for i in range(N):
if p==2:
ok=True
break
p=a[p-1]
ans+=1
print(ans if ok else -1) |
s590092663 | p03573 | u798260206 | 2,000 | 262,144 | Wrong Answer | 25 | 9,020 | 66 | You are given three integers, A, B and C. Among them, two are the same, but the remaining one is different from the rest. For example, when A=5,B=7,C=5, A and C are the same, but B is different. Find the one that is different from the rest among the given three integers. | a = list(map(int,input().split()))
print(sum(a)-sum(list(set(a)))) | s459807971 | Accepted | 33 | 9,128 | 144 | a = list(map(int,input().split()))
if a[0]!=a[1]and a[0]!=a[2]:
print(a[0])
elif a[1]!=a[0]and a[1]!=a[2]:
print(a[1])
else:
print(a[2]) |
s074099561 | p03493 | u348868667 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 90 | Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble. | S = input()
count = 0
for i in range(3):
if S[i] == 1:
count += 1
print(count) | s059615988 | Accepted | 17 | 2,940 | 92 | S = input()
count = 0
for i in range(3):
if S[i] == "1":
count += 1
print(count) |
s191578545 | p04012 | u281303342 | 2,000 | 262,144 | Wrong Answer | 21 | 3,316 | 173 | Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful. | from collections import Counter
W = input()
Ans = "YES"
C = Counter(W).items()
for c in C:
_x,cnt = c
if cnt%2 != 0:
Ans = "NO"
break
print(Ans)
| s858613553 | Accepted | 20 | 3,316 | 526 | # python 3.4.3
import sys
input = sys.stdin.readline
# import numpy as np
from collections import Counter
# -------------------------------------------------------------
# function
# -------------------------------------------------------------
# -------------------------------------------------------------
# main
# -------------------------------------------------------------
W = input().rstrip()
C = Counter(W).most_common()
ans = "Yes"
for _,cnt in C:
if cnt%2 == 1:
ans = "No"
break
print(ans) |
s562527423 | p02406 | u682153677 | 1,000 | 131,072 | Wrong Answer | 20 | 5,592 | 167 | 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; } | # -*- coding: utf-8 -*-
n = int(input())
i = 1
while True:
if i % 3 == 0 | i % 10 == 3:
print(' {0}'.format(i))
i += 1
if i > n:
break
| s345331143 | Accepted | 70 | 6,532 | 411 | # -*- coding: utf-8 -*-
from functools import reduce
n = int(input())
i = 1
while True:
a = [int(x) for x in list(str(i))]
if 3 in a:
print(" {0}".format(int(reduce(lambda x, y: x + y, [str(x) for x in a]))), end='')
elif sum(a) % 3 == 0:
print(" {0}".format(int(reduce(lambda x, y: x + y, [str(x) for x in a]))), end='')
i += 1
if i > n:
print()
break
|
s535452182 | p02398 | u941509088 | 1,000 | 131,072 | Wrong Answer | 20 | 7,608 | 113 | Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b. | a, b, c = map(int, input().split())
numbers_list = [i for i in range(a,b+1) if i%c==0]
print(len(numbers_list)) | s787558495 | Accepted | 60 | 7,720 | 117 | a, b, c = map(int, input().split())
numbers_list = [i for i in range(a,b+1) if c % i == 0]
print(len(numbers_list)) |
s242061075 | p03719 | u646792990 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 102 | You are given three integers A, B and C. Determine whether C is not less than A and not greater than B. | a, b, c = input().split()
if a[-1] == b[0] and b[-1] == c[0]:
print('YES')
else:
print('NO')
| s971263887 | Accepted | 17 | 2,940 | 98 | a, b, c = map(int, input().split())
if a <= c and c <= b:
print('Yes')
else:
print('No')
|
s949098648 | p02390 | u120055135 | 1,000 | 131,072 | Wrong Answer | 20 | 5,580 | 106 | Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively. | S = int(input())
h = int(S/360)
m = int((S - h*360)/60)
s = int(S - h*360 -m*60)
print(h, ':', m, ':', s)
| s900517760 | Accepted | 20 | 5,588 | 118 | S = int(input())
h = int(S/3600)
m = int((S - h*3600)/60)
s = int(S - h*3600 - m*60)
print(h, ':', m, ':', s, sep='')
|
s905234598 | p03493 | u100652283 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 128 | Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble. | s = input()
print(s)
count = 0
if s[0] == '1':
count+=1
if s[1] == '1':
count+=1
if s[2] == '1':
count+=1
print(count) | s455106065 | Accepted | 17 | 2,940 | 119 | s = input()
count = 0
if s[0] == '1':
count+=1
if s[1] == '1':
count+=1
if s[2] == '1':
count+=1
print(count) |
s707111910 | p03556 | u441201379 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 72 | Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer. | from math import sqrt, floor
n = int(input())
print((floor(sqrt(n)))^2) | s387732979 | Accepted | 17 | 2,940 | 79 | from math import sqrt, floor
n = int(input())
rt = floor(sqrt(n))
print(rt**2) |
s095975976 | p02865 | u163320134 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 25 | How many ways are there to choose two distinct positive integers totaling N, disregarding the order? | n=int(input())
print(n-1) | s301581079 | Accepted | 17 | 2,940 | 61 | n=int(input())
if n%2==0:
print(n//2-1)
else:
print(n//2) |
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