message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.
It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).
Consider the following action:
* Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.
How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.
Constraints
* 1 \leq N, M \leq 200000
* 1 \leq K \leq 10^9
* 1 \leq A_i, B_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M K
A_1 A_2 \ldots A_N
B_1 B_2 \ldots B_M
Output
Print an integer representing the maximum number of books that can be read.
Examples
Input
3 4 240
60 90 120
80 150 80 150
Output
3
Input
3 4 730
60 90 120
80 150 80 150
Output
7
Input
5 4 1
1000000000 1000000000 1000000000 1000000000 1000000000
1000000000 1000000000 1000000000 1000000000
Output
0
Submitted Solution:
```
import itertools
import bisect
N, M, K = map(int, input().split())
A = tuple(map(int, input().split()))
B = tuple(map(int, input().split()))
A_sum = tuple(itertools.accumulate(A))
B_sum = tuple(itertools.accumulate(B))
ans = 0
for i, num in enumerate(A_sum):
target = K - num
if target < 0:
target, i = 0, -1
if target in set(B_sum):
j = bisect.bisect_left(B_sum, target)
ans = max(ans, (i+1)+(j+1))
else:
j = bisect.bisect_left(B_sum, target)
ans = max(ans, (i+1)+j)
print(ans)
``` | instruction | 0 | 28,857 | 8 | 57,714 |
No | output | 1 | 28,857 | 8 | 57,715 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.
It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).
Consider the following action:
* Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.
How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.
Constraints
* 1 \leq N, M \leq 200000
* 1 \leq K \leq 10^9
* 1 \leq A_i, B_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M K
A_1 A_2 \ldots A_N
B_1 B_2 \ldots B_M
Output
Print an integer representing the maximum number of books that can be read.
Examples
Input
3 4 240
60 90 120
80 150 80 150
Output
3
Input
3 4 730
60 90 120
80 150 80 150
Output
7
Input
5 4 1
1000000000 1000000000 1000000000 1000000000 1000000000
1000000000 1000000000 1000000000 1000000000
Output
0
Submitted Solution:
```
import math
import bisect
#map(int,input().split())
#int(x) for x in input().split()
#for _ in range(int(input())):
n,m,k=map(int,input().split())
a=[int(x) for x in input().split()]
b=[int(x) for x in input().split()]
pa=[0]*n
pb=[0]*m
pa[0]=a[0]
pb[0]=b[0]
for i in range(1,n):
pa[i]=pa[i-1]+a[i]
for i in range(1,m):
pb[i]=pb[i-1]+b[i]
ans=0
for i in range(n):
if pa[i]<=k:
p=bisect.bisect_right(pb,k-pa[i])
ans=max(ans,i+1+p)
print(ans)
``` | instruction | 0 | 28,858 | 8 | 57,716 |
No | output | 1 | 28,858 | 8 | 57,717 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have two desks: A and B. Desk A has a vertical stack of N books on it, and Desk B similarly has M books on it.
It takes us A_i minutes to read the i-th book from the top on Desk A (1 \leq i \leq N), and B_i minutes to read the i-th book from the top on Desk B (1 \leq i \leq M).
Consider the following action:
* Choose a desk with a book remaining, read the topmost book on that desk, and remove it from the desk.
How many books can we read at most by repeating this action so that it takes us at most K minutes in total? We ignore the time it takes to do anything other than reading.
Constraints
* 1 \leq N, M \leq 200000
* 1 \leq K \leq 10^9
* 1 \leq A_i, B_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M K
A_1 A_2 \ldots A_N
B_1 B_2 \ldots B_M
Output
Print an integer representing the maximum number of books that can be read.
Examples
Input
3 4 240
60 90 120
80 150 80 150
Output
3
Input
3 4 730
60 90 120
80 150 80 150
Output
7
Input
5 4 1
1000000000 1000000000 1000000000 1000000000 1000000000
1000000000 1000000000 1000000000 1000000000
Output
0
Submitted Solution:
```
n,m,k = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
o = n + m
if sum(a+b) <= k:
print(n+m)
exit()
a_l = int(len(a))
b_l = int(len(b))
if a_l < b_l:
c = b_l - a_l
a = a + [10 ** 10 for i in range(c)]
elif b_l < a_l:
c = a_l - b_l
b = b + [10 ** 10 for i in range(c)]
t = 0
#print(a,b)
for i in range(o):
#print(a)
if a[0] < b[0]:
t += a[0]
a = a[1:]
else:
t += b[0]
b = b[1:]
#print(a[0],b[0])
#print(t)
if t > k:
print(i)
break
``` | instruction | 0 | 28,859 | 8 | 57,718 |
No | output | 1 | 28,859 | 8 | 57,719 |
Provide a correct Python 3 solution for this coding contest problem.
Daruma Otoshi
You are playing a variant of a game called "Daruma Otoshi (Dharma Block Striking)".
At the start of a game, several wooden blocks of the same size but with varying weights are stacked on top of each other, forming a tower. Another block symbolizing Dharma is placed atop. You have a wooden hammer with its head thicker than the height of a block, but not twice that.
You can choose any two adjacent blocks, except Dharma on the top, differing at most 1 in their weight, and push both of them out of the stack with a single blow of your hammer. The blocks above the removed ones then fall straight down, without collapsing the tower. You cannot hit a block pair with weight difference of 2 or more, for that makes too hard to push out blocks while keeping the balance of the tower. There is no chance in hitting three blocks out at a time, for that would require superhuman accuracy.
The goal of the game is to remove as many blocks as you can. Your task is to decide the number of blocks that can be removed by repeating the blows in an optimal order.
<image>
Figure D1. Striking out two blocks at a time
In the above figure, with a stack of four blocks weighing 1, 2, 3, and 1, in this order from the bottom, you can hit middle two blocks, weighing 2 and 3, out from the stack. The blocks above will then fall down, and two blocks weighing 1 and the Dharma block will remain. You can then push out the remaining pair of weight-1 blocks after that.
Input
The input consists of multiple datasets. The number of datasets is at most 50. Each dataset is in the following format.
n
w1 w2 … wn
n is the number of blocks, except Dharma on the top. n is a positive integer not exceeding 300. wi gives the weight of the i-th block counted from the bottom. wi is an integer between 1 and 1000, inclusive.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output in a line the maximum number of blocks you can remove.
Sample Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output for the Sample Input
4
4
2
12
0
Example
Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output
4
4
2
12
0 | instruction | 0 | 29,040 | 8 | 58,080 |
"Correct Solution:
```
from functools import lru_cache
while True:
N = int(input())
if N==0:
break
W = list(map(int, input().split()))
assert len(W) == N
@lru_cache(maxsize=None)
def solve(l, r):
assert l<=r
if (r-l)%2==1:
return False
if r-l==2:
return abs(W[l]-W[l+1])<=1
if solve(l+1, r-1) and abs(W[l]-W[r-1])<=1:
return True
for i in range(l+2, r, 2):
if solve(l, i) and solve(i, r):
return True
return False
# return (solve(l, r-2) and solve(r-2, r)) \
# or (solve(l, l+2) and solve(l+2, r)) \
# or
dp = [0]*(N+1)
for i in range(1, N+1):
ma = 0
for j in range(i-1, -1, -1):
ma = max(ma, dp[j]+solve(j, i)*(i-j))
#if (j-i)%2==0:
# print(f"j={j}, i={i}, dp={solve(j,i)}")
dp[i] = ma
#print(dp)
print(dp[-1])
``` | output | 1 | 29,040 | 8 | 58,081 |
Provide a correct Python 3 solution for this coding contest problem.
Daruma Otoshi
You are playing a variant of a game called "Daruma Otoshi (Dharma Block Striking)".
At the start of a game, several wooden blocks of the same size but with varying weights are stacked on top of each other, forming a tower. Another block symbolizing Dharma is placed atop. You have a wooden hammer with its head thicker than the height of a block, but not twice that.
You can choose any two adjacent blocks, except Dharma on the top, differing at most 1 in their weight, and push both of them out of the stack with a single blow of your hammer. The blocks above the removed ones then fall straight down, without collapsing the tower. You cannot hit a block pair with weight difference of 2 or more, for that makes too hard to push out blocks while keeping the balance of the tower. There is no chance in hitting three blocks out at a time, for that would require superhuman accuracy.
The goal of the game is to remove as many blocks as you can. Your task is to decide the number of blocks that can be removed by repeating the blows in an optimal order.
<image>
Figure D1. Striking out two blocks at a time
In the above figure, with a stack of four blocks weighing 1, 2, 3, and 1, in this order from the bottom, you can hit middle two blocks, weighing 2 and 3, out from the stack. The blocks above will then fall down, and two blocks weighing 1 and the Dharma block will remain. You can then push out the remaining pair of weight-1 blocks after that.
Input
The input consists of multiple datasets. The number of datasets is at most 50. Each dataset is in the following format.
n
w1 w2 … wn
n is the number of blocks, except Dharma on the top. n is a positive integer not exceeding 300. wi gives the weight of the i-th block counted from the bottom. wi is an integer between 1 and 1000, inclusive.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output in a line the maximum number of blocks you can remove.
Sample Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output for the Sample Input
4
4
2
12
0
Example
Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output
4
4
2
12
0 | instruction | 0 | 29,041 | 8 | 58,082 |
"Correct Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
def main():
while True:
n = int(input())
if(n == 0):
break
w = list(map(int, input().split()))
dp = [[0] * (n+1) for _ in range(n+2)]
for gap in range(2,n+1):
for l in range(n-gap+1):
r = l+gap
if(dp[l+1][r-1] == gap-2):
if(abs(w[l]-w[r-1]) <= 1):
dp[l][r] = gap
continue
for k in range(l, r):
if(dp[l][r] < dp[l][k]+dp[k][r]):
dp[l][r] = dp[l][k]+dp[k][r]
print(dp[0][n])
if __name__ == '__main__':
main()
``` | output | 1 | 29,041 | 8 | 58,083 |
Provide a correct Python 3 solution for this coding contest problem.
Daruma Otoshi
You are playing a variant of a game called "Daruma Otoshi (Dharma Block Striking)".
At the start of a game, several wooden blocks of the same size but with varying weights are stacked on top of each other, forming a tower. Another block symbolizing Dharma is placed atop. You have a wooden hammer with its head thicker than the height of a block, but not twice that.
You can choose any two adjacent blocks, except Dharma on the top, differing at most 1 in their weight, and push both of them out of the stack with a single blow of your hammer. The blocks above the removed ones then fall straight down, without collapsing the tower. You cannot hit a block pair with weight difference of 2 or more, for that makes too hard to push out blocks while keeping the balance of the tower. There is no chance in hitting three blocks out at a time, for that would require superhuman accuracy.
The goal of the game is to remove as many blocks as you can. Your task is to decide the number of blocks that can be removed by repeating the blows in an optimal order.
<image>
Figure D1. Striking out two blocks at a time
In the above figure, with a stack of four blocks weighing 1, 2, 3, and 1, in this order from the bottom, you can hit middle two blocks, weighing 2 and 3, out from the stack. The blocks above will then fall down, and two blocks weighing 1 and the Dharma block will remain. You can then push out the remaining pair of weight-1 blocks after that.
Input
The input consists of multiple datasets. The number of datasets is at most 50. Each dataset is in the following format.
n
w1 w2 … wn
n is the number of blocks, except Dharma on the top. n is a positive integer not exceeding 300. wi gives the weight of the i-th block counted from the bottom. wi is an integer between 1 and 1000, inclusive.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output in a line the maximum number of blocks you can remove.
Sample Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output for the Sample Input
4
4
2
12
0
Example
Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output
4
4
2
12
0 | instruction | 0 | 29,042 | 8 | 58,084 |
"Correct Solution:
```
def solve(n,w):
dp=[[0]*(n+1) for _ in range(n+1)]
for kukan in range(2,n+1):
for l in range(0,n+1-kukan):
r=l+kukan
#decide dp[l][l+kukan]
if (kukan-2)%2==0:
if abs(w[l]-w[r-1])<=1 and dp[l+1][r-1]==r-l-2:
dp[l][r]=dp[l+1][r-1]+2
continue
for i in range(l+1,r):
if dp[l][r]<(dp[l][i]+dp[i][r]):
dp[l][r]=dp[l][i]+dp[i][r]
return dp[0][n]
n=int(input())
while(n!=0):
w=list(map(int,input().split()))
print(solve(n,w))
n=int(input())
``` | output | 1 | 29,042 | 8 | 58,085 |
Provide a correct Python 3 solution for this coding contest problem.
Daruma Otoshi
You are playing a variant of a game called "Daruma Otoshi (Dharma Block Striking)".
At the start of a game, several wooden blocks of the same size but with varying weights are stacked on top of each other, forming a tower. Another block symbolizing Dharma is placed atop. You have a wooden hammer with its head thicker than the height of a block, but not twice that.
You can choose any two adjacent blocks, except Dharma on the top, differing at most 1 in their weight, and push both of them out of the stack with a single blow of your hammer. The blocks above the removed ones then fall straight down, without collapsing the tower. You cannot hit a block pair with weight difference of 2 or more, for that makes too hard to push out blocks while keeping the balance of the tower. There is no chance in hitting three blocks out at a time, for that would require superhuman accuracy.
The goal of the game is to remove as many blocks as you can. Your task is to decide the number of blocks that can be removed by repeating the blows in an optimal order.
<image>
Figure D1. Striking out two blocks at a time
In the above figure, with a stack of four blocks weighing 1, 2, 3, and 1, in this order from the bottom, you can hit middle two blocks, weighing 2 and 3, out from the stack. The blocks above will then fall down, and two blocks weighing 1 and the Dharma block will remain. You can then push out the remaining pair of weight-1 blocks after that.
Input
The input consists of multiple datasets. The number of datasets is at most 50. Each dataset is in the following format.
n
w1 w2 … wn
n is the number of blocks, except Dharma on the top. n is a positive integer not exceeding 300. wi gives the weight of the i-th block counted from the bottom. wi is an integer between 1 and 1000, inclusive.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output in a line the maximum number of blocks you can remove.
Sample Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output for the Sample Input
4
4
2
12
0
Example
Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output
4
4
2
12
0 | instruction | 0 | 29,043 | 8 | 58,086 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
def f(n):
a = LI()
l = len(a)
s = set()
r = 0
for i in range(l-1):
if abs(a[i] - a[i+1]) < 2:
s.add((i,i+1))
for i in range(4, l+1):
for j in range(l-i+1):
k = j + i - 1
if (j+1, k-1) in s and abs(a[j] - a[k]) < 2:
s.add((j,k))
continue
for m in range(j+2,k,2):
if (j,m-1) in s and (m,k) in s:
s.add((j,k))
break
m = {}
m[l] = 0
def _f(i):
if i in m:
return m[i]
r = 0
for j in range(i,l+1):
t = _f(j+1)
if (i,j) in s:
t += j - i + 1
if r < t:
r = t
m[i] = r
return r
return _f(0)
while True:
n = I()
if n == 0:
break
rr.append(f(n))
return '\n'.join(map(str, rr))
print(main())
``` | output | 1 | 29,043 | 8 | 58,087 |
Provide a correct Python 3 solution for this coding contest problem.
Daruma Otoshi
You are playing a variant of a game called "Daruma Otoshi (Dharma Block Striking)".
At the start of a game, several wooden blocks of the same size but with varying weights are stacked on top of each other, forming a tower. Another block symbolizing Dharma is placed atop. You have a wooden hammer with its head thicker than the height of a block, but not twice that.
You can choose any two adjacent blocks, except Dharma on the top, differing at most 1 in their weight, and push both of them out of the stack with a single blow of your hammer. The blocks above the removed ones then fall straight down, without collapsing the tower. You cannot hit a block pair with weight difference of 2 or more, for that makes too hard to push out blocks while keeping the balance of the tower. There is no chance in hitting three blocks out at a time, for that would require superhuman accuracy.
The goal of the game is to remove as many blocks as you can. Your task is to decide the number of blocks that can be removed by repeating the blows in an optimal order.
<image>
Figure D1. Striking out two blocks at a time
In the above figure, with a stack of four blocks weighing 1, 2, 3, and 1, in this order from the bottom, you can hit middle two blocks, weighing 2 and 3, out from the stack. The blocks above will then fall down, and two blocks weighing 1 and the Dharma block will remain. You can then push out the remaining pair of weight-1 blocks after that.
Input
The input consists of multiple datasets. The number of datasets is at most 50. Each dataset is in the following format.
n
w1 w2 … wn
n is the number of blocks, except Dharma on the top. n is a positive integer not exceeding 300. wi gives the weight of the i-th block counted from the bottom. wi is an integer between 1 and 1000, inclusive.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output in a line the maximum number of blocks you can remove.
Sample Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output for the Sample Input
4
4
2
12
0
Example
Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output
4
4
2
12
0 | instruction | 0 | 29,044 | 8 | 58,088 |
"Correct Solution:
```
n = int(input())
while True:
l1 = list(map(int,input().split()))
c = [[False] * n for i in range(n)]
d = [0] * (n + 1)
for i in range(n - 1):
if abs(l1[i + 1] - l1[i]) <= 1:
c[i][i + 1] = True
for i in range(3, n, 2):
for j in range(n - i):
for k in range(j + 1, j + i):
if c[j][j + i] == False and c[j][k] and c[k + 1][j + i]:
c[j][j + i] = True
break
if c[j][j + i] == False and abs(l1[j] - l1[j + i]) <= 1 and c[j + 1][j + i - 1]:
c[j][j + i] = True
for i in range(n):
for j in range(i):
if c[j][i]:
d[i] = max(d[i], d[j - 1] + i - j + 1)
d[i] = max(d[i], d[i - 1])
print(d[n - 1])
n = int(input())
if n == 0:
break
``` | output | 1 | 29,044 | 8 | 58,089 |
Provide a correct Python 3 solution for this coding contest problem.
Daruma Otoshi
You are playing a variant of a game called "Daruma Otoshi (Dharma Block Striking)".
At the start of a game, several wooden blocks of the same size but with varying weights are stacked on top of each other, forming a tower. Another block symbolizing Dharma is placed atop. You have a wooden hammer with its head thicker than the height of a block, but not twice that.
You can choose any two adjacent blocks, except Dharma on the top, differing at most 1 in their weight, and push both of them out of the stack with a single blow of your hammer. The blocks above the removed ones then fall straight down, without collapsing the tower. You cannot hit a block pair with weight difference of 2 or more, for that makes too hard to push out blocks while keeping the balance of the tower. There is no chance in hitting three blocks out at a time, for that would require superhuman accuracy.
The goal of the game is to remove as many blocks as you can. Your task is to decide the number of blocks that can be removed by repeating the blows in an optimal order.
<image>
Figure D1. Striking out two blocks at a time
In the above figure, with a stack of four blocks weighing 1, 2, 3, and 1, in this order from the bottom, you can hit middle two blocks, weighing 2 and 3, out from the stack. The blocks above will then fall down, and two blocks weighing 1 and the Dharma block will remain. You can then push out the remaining pair of weight-1 blocks after that.
Input
The input consists of multiple datasets. The number of datasets is at most 50. Each dataset is in the following format.
n
w1 w2 … wn
n is the number of blocks, except Dharma on the top. n is a positive integer not exceeding 300. wi gives the weight of the i-th block counted from the bottom. wi is an integer between 1 and 1000, inclusive.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output in a line the maximum number of blocks you can remove.
Sample Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output for the Sample Input
4
4
2
12
0
Example
Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output
4
4
2
12
0 | instruction | 0 | 29,045 | 8 | 58,090 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
ans = []
while True:
N = int(input())
if N == 0: break
W = list(map(int, input().split()))
# dp[i][j]: iからjのブロックを全て除外可能か
dp = [[False]*N for _ in range(N)]
for i in range(N-1):
if abs(W[i]-W[i+1]) <= 1:
dp[i][i+1] = True
for d in range(4, N+1):
for i in range(N-d+1):
j = i+d-1
for k in range(i, j):
if dp[i][k] and dp[k+1][j]:
dp[i][j] = True
if abs(W[i]-W[j]) <= 1 and dp[i+1][j-1]:
dp[i][j] = True
# dp2[i]: iまでで除外可能なブロックの最大値(1-indexed)
dp2 = [0]*(N+1)
for i in range(N):
for j in range(i+1, N):
if dp[i][j]:
dp2[j+1] = max(dp2[j+1], dp2[i]+j-i+1)
dp2[i+1] = max(dp2[i+1], dp2[i])
ans.append(dp2[N])
print(*ans, sep="\n")
``` | output | 1 | 29,045 | 8 | 58,091 |
Provide a correct Python 3 solution for this coding contest problem.
Daruma Otoshi
You are playing a variant of a game called "Daruma Otoshi (Dharma Block Striking)".
At the start of a game, several wooden blocks of the same size but with varying weights are stacked on top of each other, forming a tower. Another block symbolizing Dharma is placed atop. You have a wooden hammer with its head thicker than the height of a block, but not twice that.
You can choose any two adjacent blocks, except Dharma on the top, differing at most 1 in their weight, and push both of them out of the stack with a single blow of your hammer. The blocks above the removed ones then fall straight down, without collapsing the tower. You cannot hit a block pair with weight difference of 2 or more, for that makes too hard to push out blocks while keeping the balance of the tower. There is no chance in hitting three blocks out at a time, for that would require superhuman accuracy.
The goal of the game is to remove as many blocks as you can. Your task is to decide the number of blocks that can be removed by repeating the blows in an optimal order.
<image>
Figure D1. Striking out two blocks at a time
In the above figure, with a stack of four blocks weighing 1, 2, 3, and 1, in this order from the bottom, you can hit middle two blocks, weighing 2 and 3, out from the stack. The blocks above will then fall down, and two blocks weighing 1 and the Dharma block will remain. You can then push out the remaining pair of weight-1 blocks after that.
Input
The input consists of multiple datasets. The number of datasets is at most 50. Each dataset is in the following format.
n
w1 w2 … wn
n is the number of blocks, except Dharma on the top. n is a positive integer not exceeding 300. wi gives the weight of the i-th block counted from the bottom. wi is an integer between 1 and 1000, inclusive.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output in a line the maximum number of blocks you can remove.
Sample Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output for the Sample Input
4
4
2
12
0
Example
Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output
4
4
2
12
0 | instruction | 0 | 29,046 | 8 | 58,092 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(10**8)
def ii(): return int(sys.stdin.readline())
def mi(): return map(int, sys.stdin.readline().split())
def ti(): return tuple(map(int, sys.stdin.readline().split()))
def li2(N): return [list(map(int, sys.stdin.readline().split())) for _ in range(N)]
def dp2(ini, i, j): return [[ini]*i for _ in range(j)]
def dp3(ini, i, j, k): return [[[ini]*i for _ in range(j)] for _ in range(k)]
#import bisect #bisect.bisect_left(B, a)
#from collections import defaultdict #d = defaultdict(int) d[key] += value
#from itertools import accumulate #list(accumulate(A))
## DP
def solve(N):
A = ti()
dp = dp2(0, N, N)
for i in range(N-1):
if abs(A[i]-A[i+1]) < 2:
dp[i][i+1] = 2
for h in range(2, N): #hは(区間の長さ)-1
for i in range(N):
if i+h >= N:
break
j = i+h
# 区間の長さが偶数の場合
if h % 2:
if abs(A[i]-A[j]) < 2 and dp[i+1][j-1] == h-1:
dp[i][j] = h + 1
continue
else:
dp[i][j] = max(tuple(dp[i][k]+dp[k+1][j] for k in range(i, j)))
# 区間の長さが奇数の場合
else:
dp[i][j] = max(dp[i+1][j], dp[i][j-1])
else:
continue
print(dp[0][N-1])
while True:
N = ii()
if N == 0:
exit()
solve(N)
``` | output | 1 | 29,046 | 8 | 58,093 |
Provide a correct Python 3 solution for this coding contest problem.
Daruma Otoshi
You are playing a variant of a game called "Daruma Otoshi (Dharma Block Striking)".
At the start of a game, several wooden blocks of the same size but with varying weights are stacked on top of each other, forming a tower. Another block symbolizing Dharma is placed atop. You have a wooden hammer with its head thicker than the height of a block, but not twice that.
You can choose any two adjacent blocks, except Dharma on the top, differing at most 1 in their weight, and push both of them out of the stack with a single blow of your hammer. The blocks above the removed ones then fall straight down, without collapsing the tower. You cannot hit a block pair with weight difference of 2 or more, for that makes too hard to push out blocks while keeping the balance of the tower. There is no chance in hitting three blocks out at a time, for that would require superhuman accuracy.
The goal of the game is to remove as many blocks as you can. Your task is to decide the number of blocks that can be removed by repeating the blows in an optimal order.
<image>
Figure D1. Striking out two blocks at a time
In the above figure, with a stack of four blocks weighing 1, 2, 3, and 1, in this order from the bottom, you can hit middle two blocks, weighing 2 and 3, out from the stack. The blocks above will then fall down, and two blocks weighing 1 and the Dharma block will remain. You can then push out the remaining pair of weight-1 blocks after that.
Input
The input consists of multiple datasets. The number of datasets is at most 50. Each dataset is in the following format.
n
w1 w2 … wn
n is the number of blocks, except Dharma on the top. n is a positive integer not exceeding 300. wi gives the weight of the i-th block counted from the bottom. wi is an integer between 1 and 1000, inclusive.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output in a line the maximum number of blocks you can remove.
Sample Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output for the Sample Input
4
4
2
12
0
Example
Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output
4
4
2
12
0 | instruction | 0 | 29,047 | 8 | 58,094 |
"Correct Solution:
```
import sys
while True:
n=int(input())
if n==0:
break
w = [int(x) for x in sys.stdin.readline().split()]
dp=[[0]*n for i in range(n)]
for l in range(n-1):
r=l+1
if abs(w[l]-w[r])<2:
dp[l][r]=2
for i in range(2,n):
if i%2==1:
for l in range(n-i):
r=l+i
if dp[l+1][r-1]==r-l-1 and abs(w[l]-w[r])<2:
dp[l][r]=r-l+1
else:
for lr in range(l+1,r):
if dp[l][r]< dp[l][lr]+dp[lr+1][r]:
dp[l][r]=dp[l][lr]+dp[lr+1][r]
else:
for l in range(n-i):
r=l+i
# dp[l][r]=max(dp[l][r],dp[l+1][r],dp[l][r-1])
if dp[l][r]<dp[l+1][r]:
dp[l][r]=dp[l+1][r]
if dp[l][r]<dp[l][r-1]:
dp[l][r]=dp[l][r-1]
print(dp[0][n-1])
``` | output | 1 | 29,047 | 8 | 58,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Daruma Otoshi
You are playing a variant of a game called "Daruma Otoshi (Dharma Block Striking)".
At the start of a game, several wooden blocks of the same size but with varying weights are stacked on top of each other, forming a tower. Another block symbolizing Dharma is placed atop. You have a wooden hammer with its head thicker than the height of a block, but not twice that.
You can choose any two adjacent blocks, except Dharma on the top, differing at most 1 in their weight, and push both of them out of the stack with a single blow of your hammer. The blocks above the removed ones then fall straight down, without collapsing the tower. You cannot hit a block pair with weight difference of 2 or more, for that makes too hard to push out blocks while keeping the balance of the tower. There is no chance in hitting three blocks out at a time, for that would require superhuman accuracy.
The goal of the game is to remove as many blocks as you can. Your task is to decide the number of blocks that can be removed by repeating the blows in an optimal order.
<image>
Figure D1. Striking out two blocks at a time
In the above figure, with a stack of four blocks weighing 1, 2, 3, and 1, in this order from the bottom, you can hit middle two blocks, weighing 2 and 3, out from the stack. The blocks above will then fall down, and two blocks weighing 1 and the Dharma block will remain. You can then push out the remaining pair of weight-1 blocks after that.
Input
The input consists of multiple datasets. The number of datasets is at most 50. Each dataset is in the following format.
n
w1 w2 … wn
n is the number of blocks, except Dharma on the top. n is a positive integer not exceeding 300. wi gives the weight of the i-th block counted from the bottom. wi is an integer between 1 and 1000, inclusive.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output in a line the maximum number of blocks you can remove.
Sample Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output for the Sample Input
4
4
2
12
0
Example
Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output
4
4
2
12
0
Submitted Solution:
```
# AOJ 1611 ダルマ落とし
# 貰うdp
import sys
def main():
while True:
N = int(input())
if N == 0:
break
w = list(map(int, sys.stdin.readline().strip().split()))
# dp 初期化 -------------------------------------------------
dp = [[0] * N for _ in range(N)]
for i in range(N - 1):
if abs(w[i] - w[i + 1]) <= 1:
dp[i][i + 1] = 2
# dp 更新 ------------------------------------------------------
for l in range(2, N):
for i in range(N - l):
j = i + l
if l % 2 == 1:
if dp[i + 1][j - 1] == l - 1 and (abs(w[i] - w[j]) <= 1):
dp[i][j] = l + 1
continue
for k in range(i, j):
if dp[i][j] < dp[i][k] + dp[k + 1][j]:
dp[i][j] = dp[i][k] + dp[k + 1][j]
if dp[i][j] == l:
break
# for i in range(len(dp)):
# print(dp[i])
print(dp[0][-1])
if __name__ == '__main__':
main()
``` | instruction | 0 | 29,048 | 8 | 58,096 |
Yes | output | 1 | 29,048 | 8 | 58,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Daruma Otoshi
You are playing a variant of a game called "Daruma Otoshi (Dharma Block Striking)".
At the start of a game, several wooden blocks of the same size but with varying weights are stacked on top of each other, forming a tower. Another block symbolizing Dharma is placed atop. You have a wooden hammer with its head thicker than the height of a block, but not twice that.
You can choose any two adjacent blocks, except Dharma on the top, differing at most 1 in their weight, and push both of them out of the stack with a single blow of your hammer. The blocks above the removed ones then fall straight down, without collapsing the tower. You cannot hit a block pair with weight difference of 2 or more, for that makes too hard to push out blocks while keeping the balance of the tower. There is no chance in hitting three blocks out at a time, for that would require superhuman accuracy.
The goal of the game is to remove as many blocks as you can. Your task is to decide the number of blocks that can be removed by repeating the blows in an optimal order.
<image>
Figure D1. Striking out two blocks at a time
In the above figure, with a stack of four blocks weighing 1, 2, 3, and 1, in this order from the bottom, you can hit middle two blocks, weighing 2 and 3, out from the stack. The blocks above will then fall down, and two blocks weighing 1 and the Dharma block will remain. You can then push out the remaining pair of weight-1 blocks after that.
Input
The input consists of multiple datasets. The number of datasets is at most 50. Each dataset is in the following format.
n
w1 w2 … wn
n is the number of blocks, except Dharma on the top. n is a positive integer not exceeding 300. wi gives the weight of the i-th block counted from the bottom. wi is an integer between 1 and 1000, inclusive.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output in a line the maximum number of blocks you can remove.
Sample Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output for the Sample Input
4
4
2
12
0
Example
Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output
4
4
2
12
0
Submitted Solution:
```
n=int(input())
while n!=0:
w=list(map(int,input().split()))
check=[[False]*n for i in range(n)]
for i in range(n-1):
if abs(w[i+1]-w[i])<=1:
check[i][i+1]=True
for i in range(3,n,2):
for j in range(n-i):
for k in range(j+1,j+i):
if check[j][j+i]==False and check[j][k] and check[k+1][j+i]:
check[j][j+i]=True
break
if check[j][j+i]==False and abs(w[j]-w[j+i])<=1 and check[j+1][j+i-1]:
check[j][j+i]=True
dp=[0]*(n+1)
for k in range(n):
for m in range(k):
if check[m][k]:
dp[k]=max(dp[k],dp[m-1]+k-m+1)
dp[k]=max(dp[k],dp[k-1])
print(dp[n-1])
n=int(input())
``` | instruction | 0 | 29,049 | 8 | 58,098 |
Yes | output | 1 | 29,049 | 8 | 58,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Daruma Otoshi
You are playing a variant of a game called "Daruma Otoshi (Dharma Block Striking)".
At the start of a game, several wooden blocks of the same size but with varying weights are stacked on top of each other, forming a tower. Another block symbolizing Dharma is placed atop. You have a wooden hammer with its head thicker than the height of a block, but not twice that.
You can choose any two adjacent blocks, except Dharma on the top, differing at most 1 in their weight, and push both of them out of the stack with a single blow of your hammer. The blocks above the removed ones then fall straight down, without collapsing the tower. You cannot hit a block pair with weight difference of 2 or more, for that makes too hard to push out blocks while keeping the balance of the tower. There is no chance in hitting three blocks out at a time, for that would require superhuman accuracy.
The goal of the game is to remove as many blocks as you can. Your task is to decide the number of blocks that can be removed by repeating the blows in an optimal order.
<image>
Figure D1. Striking out two blocks at a time
In the above figure, with a stack of four blocks weighing 1, 2, 3, and 1, in this order from the bottom, you can hit middle two blocks, weighing 2 and 3, out from the stack. The blocks above will then fall down, and two blocks weighing 1 and the Dharma block will remain. You can then push out the remaining pair of weight-1 blocks after that.
Input
The input consists of multiple datasets. The number of datasets is at most 50. Each dataset is in the following format.
n
w1 w2 … wn
n is the number of blocks, except Dharma on the top. n is a positive integer not exceeding 300. wi gives the weight of the i-th block counted from the bottom. wi is an integer between 1 and 1000, inclusive.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output in a line the maximum number of blocks you can remove.
Sample Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output for the Sample Input
4
4
2
12
0
Example
Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output
4
4
2
12
0
Submitted Solution:
```
def main(n):
a = list(map(int, input().split()))
dp = [[0] * n for i in range(n)]
for i in range(n - 1):
if abs(a[i] - a[i + 1]) <= 1:
dp[i][i + 1] = 2
for i in range(2, n):
for l in range(n - i):
r = i + l
res = 0
if abs(a[l] - a[r]) <= 1:
if dp[l + 1][r - 1] == r - l - 1:
dp[l][r] = r - l + 1
continue
res = dp[l + 1][r - 1]
for li in range(l + 1, r):
tmp = dp[l][li] + dp[li + 1][r]
if res < tmp:
res = tmp
dp[l][r] = res
print(dp[0][n-1])
def dfs(l, r):
if l >= r: return 0
if dp[l][r] != -1: return dp[l][r]
if abs(a[l] - a[r]) <= 1:
if dfs(l + 1, r - 1) == r - l - 1:
dp[l][r] = r - l + 1
return r - l + 1
res = 0
for li in range(l + 1, r):
tmp = 0
f = 0
if dp[l][li] != -1:
tmp += dp[l][li]
else:
tmp += dfs(l, li)
if dp[li + 1][r] != -1:
tmp += dp[li + 1][r]
else:
tmp += dfs(li + 1, r)
if res < tmp:
res = tmp
dp[l][r] = res
return res
# print(dfs(0, n - 1))
if __name__ == "__main__":
while 1:
n = int(input())
if n:
main(n)
else:
break
``` | instruction | 0 | 29,050 | 8 | 58,100 |
Yes | output | 1 | 29,050 | 8 | 58,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Daruma Otoshi
You are playing a variant of a game called "Daruma Otoshi (Dharma Block Striking)".
At the start of a game, several wooden blocks of the same size but with varying weights are stacked on top of each other, forming a tower. Another block symbolizing Dharma is placed atop. You have a wooden hammer with its head thicker than the height of a block, but not twice that.
You can choose any two adjacent blocks, except Dharma on the top, differing at most 1 in their weight, and push both of them out of the stack with a single blow of your hammer. The blocks above the removed ones then fall straight down, without collapsing the tower. You cannot hit a block pair with weight difference of 2 or more, for that makes too hard to push out blocks while keeping the balance of the tower. There is no chance in hitting three blocks out at a time, for that would require superhuman accuracy.
The goal of the game is to remove as many blocks as you can. Your task is to decide the number of blocks that can be removed by repeating the blows in an optimal order.
<image>
Figure D1. Striking out two blocks at a time
In the above figure, with a stack of four blocks weighing 1, 2, 3, and 1, in this order from the bottom, you can hit middle two blocks, weighing 2 and 3, out from the stack. The blocks above will then fall down, and two blocks weighing 1 and the Dharma block will remain. You can then push out the remaining pair of weight-1 blocks after that.
Input
The input consists of multiple datasets. The number of datasets is at most 50. Each dataset is in the following format.
n
w1 w2 … wn
n is the number of blocks, except Dharma on the top. n is a positive integer not exceeding 300. wi gives the weight of the i-th block counted from the bottom. wi is an integer between 1 and 1000, inclusive.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output in a line the maximum number of blocks you can remove.
Sample Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output for the Sample Input
4
4
2
12
0
Example
Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output
4
4
2
12
0
Submitted Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
def resolve():
while True:
n = int(input())
w = list(map(int, input().split()))
if n==0:
break
else:
dp = [[0]*(n+1) for _ in range(n+2)]
# 幅
for i in range(2, n+1):
for l in range(n-i+1):
r = l+i
if dp[l+1][r-1] == i-2:
if abs(w[l]-w[r-1])<=1:
dp[l][r] = i
else:
dp[l][r] = i-2
for j in range(l, r):
if dp[l][j] + dp[j][r] > dp[l][r]:
dp[l][r] = dp[l][j] + dp[j][r]
print(dp[0][-1])
if __name__ == '__main__':
resolve()
``` | instruction | 0 | 29,051 | 8 | 58,102 |
Yes | output | 1 | 29,051 | 8 | 58,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Daruma Otoshi
You are playing a variant of a game called "Daruma Otoshi (Dharma Block Striking)".
At the start of a game, several wooden blocks of the same size but with varying weights are stacked on top of each other, forming a tower. Another block symbolizing Dharma is placed atop. You have a wooden hammer with its head thicker than the height of a block, but not twice that.
You can choose any two adjacent blocks, except Dharma on the top, differing at most 1 in their weight, and push both of them out of the stack with a single blow of your hammer. The blocks above the removed ones then fall straight down, without collapsing the tower. You cannot hit a block pair with weight difference of 2 or more, for that makes too hard to push out blocks while keeping the balance of the tower. There is no chance in hitting three blocks out at a time, for that would require superhuman accuracy.
The goal of the game is to remove as many blocks as you can. Your task is to decide the number of blocks that can be removed by repeating the blows in an optimal order.
<image>
Figure D1. Striking out two blocks at a time
In the above figure, with a stack of four blocks weighing 1, 2, 3, and 1, in this order from the bottom, you can hit middle two blocks, weighing 2 and 3, out from the stack. The blocks above will then fall down, and two blocks weighing 1 and the Dharma block will remain. You can then push out the remaining pair of weight-1 blocks after that.
Input
The input consists of multiple datasets. The number of datasets is at most 50. Each dataset is in the following format.
n
w1 w2 … wn
n is the number of blocks, except Dharma on the top. n is a positive integer not exceeding 300. wi gives the weight of the i-th block counted from the bottom. wi is an integer between 1 and 1000, inclusive.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output in a line the maximum number of blocks you can remove.
Sample Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output for the Sample Input
4
4
2
12
0
Example
Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output
4
4
2
12
0
Submitted Solution:
```
while True:
n = int(input())
if n == 0:
break
w = list(map(int, input().split()))
dp = [[-1 for j in range(n + 1)] for i in range(n + 1)]
def rec(l, r):
# 既に計算済み？
if dp[l][r] != -1:
return dp[l][r]
# これ以上取り除けない？
if abs(l - r) <= 1:
return 0
res = 0
# パターン1.
if abs(w[l] - w[r - 1]) <= 1 and rec(l + 1, r - 1) == r - l - 2:
res = r - l
# パターン２.区間を分ける
for mid in range(l + 1, r - 1):
res = max(res, rec(l, mid) + rec(mid, r))
dp[l][r] = res
return res
print(rec(0, n))
``` | instruction | 0 | 29,052 | 8 | 58,104 |
No | output | 1 | 29,052 | 8 | 58,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Daruma Otoshi
You are playing a variant of a game called "Daruma Otoshi (Dharma Block Striking)".
At the start of a game, several wooden blocks of the same size but with varying weights are stacked on top of each other, forming a tower. Another block symbolizing Dharma is placed atop. You have a wooden hammer with its head thicker than the height of a block, but not twice that.
You can choose any two adjacent blocks, except Dharma on the top, differing at most 1 in their weight, and push both of them out of the stack with a single blow of your hammer. The blocks above the removed ones then fall straight down, without collapsing the tower. You cannot hit a block pair with weight difference of 2 or more, for that makes too hard to push out blocks while keeping the balance of the tower. There is no chance in hitting three blocks out at a time, for that would require superhuman accuracy.
The goal of the game is to remove as many blocks as you can. Your task is to decide the number of blocks that can be removed by repeating the blows in an optimal order.
<image>
Figure D1. Striking out two blocks at a time
In the above figure, with a stack of four blocks weighing 1, 2, 3, and 1, in this order from the bottom, you can hit middle two blocks, weighing 2 and 3, out from the stack. The blocks above will then fall down, and two blocks weighing 1 and the Dharma block will remain. You can then push out the remaining pair of weight-1 blocks after that.
Input
The input consists of multiple datasets. The number of datasets is at most 50. Each dataset is in the following format.
n
w1 w2 … wn
n is the number of blocks, except Dharma on the top. n is a positive integer not exceeding 300. wi gives the weight of the i-th block counted from the bottom. wi is an integer between 1 and 1000, inclusive.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output in a line the maximum number of blocks you can remove.
Sample Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output for the Sample Input
4
4
2
12
0
Example
Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output
4
4
2
12
0
Submitted Solution:
```
n = int(input())
w = list(map(int, input().split()))
dp = [[-1 for j in range(n + 1)] for i in range(n + 1)]
def rec(l, r):
if dp[l][r] != -1:
return dp[l][r]
if abs(l - r) <= 1:
return 0
res = 0
if abs(w[l] - w[r - 1]) <= 1 and rec(l + 1, r - 1) == r - l - 2:
res = r - l
for mid in range(l + 1, r - 1):
res = max(res, rec(l, mid) + rec(mid, r))
dp[l][r] = res
return res
print(rec(0, n))
``` | instruction | 0 | 29,053 | 8 | 58,106 |
No | output | 1 | 29,053 | 8 | 58,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Daruma Otoshi
You are playing a variant of a game called "Daruma Otoshi (Dharma Block Striking)".
At the start of a game, several wooden blocks of the same size but with varying weights are stacked on top of each other, forming a tower. Another block symbolizing Dharma is placed atop. You have a wooden hammer with its head thicker than the height of a block, but not twice that.
You can choose any two adjacent blocks, except Dharma on the top, differing at most 1 in their weight, and push both of them out of the stack with a single blow of your hammer. The blocks above the removed ones then fall straight down, without collapsing the tower. You cannot hit a block pair with weight difference of 2 or more, for that makes too hard to push out blocks while keeping the balance of the tower. There is no chance in hitting three blocks out at a time, for that would require superhuman accuracy.
The goal of the game is to remove as many blocks as you can. Your task is to decide the number of blocks that can be removed by repeating the blows in an optimal order.
<image>
Figure D1. Striking out two blocks at a time
In the above figure, with a stack of four blocks weighing 1, 2, 3, and 1, in this order from the bottom, you can hit middle two blocks, weighing 2 and 3, out from the stack. The blocks above will then fall down, and two blocks weighing 1 and the Dharma block will remain. You can then push out the remaining pair of weight-1 blocks after that.
Input
The input consists of multiple datasets. The number of datasets is at most 50. Each dataset is in the following format.
n
w1 w2 … wn
n is the number of blocks, except Dharma on the top. n is a positive integer not exceeding 300. wi gives the weight of the i-th block counted from the bottom. wi is an integer between 1 and 1000, inclusive.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output in a line the maximum number of blocks you can remove.
Sample Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output for the Sample Input
4
4
2
12
0
Example
Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output
4
4
2
12
0
Submitted Solution:
```
from functools import lru_cache
while True:
N = int(input())
if N==0:
break
W = list(map(int, input().split()))
@lru_cache(maxsize=None)
def solve(l, r):
assert l<=r
if (r-l)%2==1:
return False
if r-l==2:
return abs(W[l]-W[l+1])<=1
return (solve(l, r-2) and solve(r-2, r)) \
or (solve(l, l+2) and solve(l+2, r)) \
or (solve(l+1, r-1) and abs(W[l]-W[r-1])<=1)
dp = [0]*(N+1)
for i in range(N+1):
ma = 0
for j in range(i-1, -1, -1):
ma = max(ma, dp[j]+solve(j, i)*(i-j))
dp[i] = ma
print(dp[-1])
``` | instruction | 0 | 29,054 | 8 | 58,108 |
No | output | 1 | 29,054 | 8 | 58,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Daruma Otoshi
You are playing a variant of a game called "Daruma Otoshi (Dharma Block Striking)".
At the start of a game, several wooden blocks of the same size but with varying weights are stacked on top of each other, forming a tower. Another block symbolizing Dharma is placed atop. You have a wooden hammer with its head thicker than the height of a block, but not twice that.
You can choose any two adjacent blocks, except Dharma on the top, differing at most 1 in their weight, and push both of them out of the stack with a single blow of your hammer. The blocks above the removed ones then fall straight down, without collapsing the tower. You cannot hit a block pair with weight difference of 2 or more, for that makes too hard to push out blocks while keeping the balance of the tower. There is no chance in hitting three blocks out at a time, for that would require superhuman accuracy.
The goal of the game is to remove as many blocks as you can. Your task is to decide the number of blocks that can be removed by repeating the blows in an optimal order.
<image>
Figure D1. Striking out two blocks at a time
In the above figure, with a stack of four blocks weighing 1, 2, 3, and 1, in this order from the bottom, you can hit middle two blocks, weighing 2 and 3, out from the stack. The blocks above will then fall down, and two blocks weighing 1 and the Dharma block will remain. You can then push out the remaining pair of weight-1 blocks after that.
Input
The input consists of multiple datasets. The number of datasets is at most 50. Each dataset is in the following format.
n
w1 w2 … wn
n is the number of blocks, except Dharma on the top. n is a positive integer not exceeding 300. wi gives the weight of the i-th block counted from the bottom. wi is an integer between 1 and 1000, inclusive.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output in a line the maximum number of blocks you can remove.
Sample Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output for the Sample Input
4
4
2
12
0
Example
Input
4
1 2 3 4
4
1 2 3 1
5
5 1 2 3 6
14
8 7 1 4 3 5 4 1 6 8 10 4 6 5
5
1 3 5 1 3
0
Output
4
4
2
12
0
Submitted Solution:
```
while True:
N = int(input())
if not N:
break
W = [int(x) for x in input().split()]
dp = [[0] * N for _ in range(N)]
for w in range(2, N + 1):
for left in range(N - w + 1):
# [left, right]
right = left + w - 1
for mid in range(left + 1, right):
dp[left][right] = max(dp[left][right], dp[left][mid] + dp[mid + 1][right])
if right - left - 1 == dp[left + 1][right - 1] and abs(W[left] - W[right]) <= 1:
dp[left][right] = right - left + 1
print(dp[0][N-1])
``` | instruction | 0 | 29,055 | 8 | 58,110 |
No | output | 1 | 29,055 | 8 | 58,111 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a beautiful garden of stones in Innopolis.
Its most beautiful place is the n piles with stones numbered from 1 to n.
EJOI participants have visited this place twice.
When they first visited it, the number of stones in piles was x_1, x_2, …, x_n, correspondingly. One of the participants wrote down this sequence in a notebook.
They visited it again the following day, and the number of stones in piles was equal to y_1, y_2, …, y_n. One of the participants also wrote it down in a notebook.
It is well known that every member of the EJOI jury during the night either sits in the room 108 or comes to the place with stones. Each jury member who comes there either takes one stone for himself or moves one stone from one pile to another. We can assume that there is an unlimited number of jury members. No one except the jury goes to the place with stones at night.
Participants want to know whether their notes can be correct or they are sure to have made a mistake.
Input
The first line of the input file contains a single integer n, the number of piles with stones in the garden (1 ≤ n ≤ 50).
The second line contains n integers separated by spaces x_1, x_2, …, x_n, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the first time (0 ≤ x_i ≤ 1000).
The third line contains n integers separated by spaces y_1, y_2, …, y_n, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the second time (0 ≤ y_i ≤ 1000).
Output
If the records can be consistent output "Yes", otherwise output "No" (quotes for clarity).
Examples
Input
5
1 2 3 4 5
2 1 4 3 5
Output
Yes
Input
5
1 1 1 1 1
1 0 1 0 1
Output
Yes
Input
3
2 3 9
1 7 9
Output
No
Note
In the first example, the following could have happened during the night: one of the jury members moved one stone from the second pile to the first pile, and the other jury member moved one stone from the fourth pile to the third pile.
In the second example, the jury took stones from the second and fourth piles.
It can be proved that it is impossible for the jury members to move and took stones to convert the first array into the second array. | instruction | 0 | 29,099 | 8 | 58,198 |
Tags: math
Correct Solution:
```
n=input()
R=lambda:sum(list(map(int,input().split())))
print("Yes") if R()>=R() else print("No")
``` | output | 1 | 29,099 | 8 | 58,199 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a beautiful garden of stones in Innopolis.
Its most beautiful place is the n piles with stones numbered from 1 to n.
EJOI participants have visited this place twice.
When they first visited it, the number of stones in piles was x_1, x_2, …, x_n, correspondingly. One of the participants wrote down this sequence in a notebook.
They visited it again the following day, and the number of stones in piles was equal to y_1, y_2, …, y_n. One of the participants also wrote it down in a notebook.
It is well known that every member of the EJOI jury during the night either sits in the room 108 or comes to the place with stones. Each jury member who comes there either takes one stone for himself or moves one stone from one pile to another. We can assume that there is an unlimited number of jury members. No one except the jury goes to the place with stones at night.
Participants want to know whether their notes can be correct or they are sure to have made a mistake.
Input
The first line of the input file contains a single integer n, the number of piles with stones in the garden (1 ≤ n ≤ 50).
The second line contains n integers separated by spaces x_1, x_2, …, x_n, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the first time (0 ≤ x_i ≤ 1000).
The third line contains n integers separated by spaces y_1, y_2, …, y_n, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the second time (0 ≤ y_i ≤ 1000).
Output
If the records can be consistent output "Yes", otherwise output "No" (quotes for clarity).
Examples
Input
5
1 2 3 4 5
2 1 4 3 5
Output
Yes
Input
5
1 1 1 1 1
1 0 1 0 1
Output
Yes
Input
3
2 3 9
1 7 9
Output
No
Note
In the first example, the following could have happened during the night: one of the jury members moved one stone from the second pile to the first pile, and the other jury member moved one stone from the fourth pile to the third pile.
In the second example, the jury took stones from the second and fourth piles.
It can be proved that it is impossible for the jury members to move and took stones to convert the first array into the second array. | instruction | 0 | 29,100 | 8 | 58,200 |
Tags: math
Correct Solution:
```
input()
if sum(list(map(int,input().split())))<sum(list(map(int,input().split()))):
print("No")
else:
print("Yes")
``` | output | 1 | 29,100 | 8 | 58,201 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a beautiful garden of stones in Innopolis.
Its most beautiful place is the n piles with stones numbered from 1 to n.
EJOI participants have visited this place twice.
When they first visited it, the number of stones in piles was x_1, x_2, …, x_n, correspondingly. One of the participants wrote down this sequence in a notebook.
They visited it again the following day, and the number of stones in piles was equal to y_1, y_2, …, y_n. One of the participants also wrote it down in a notebook.
It is well known that every member of the EJOI jury during the night either sits in the room 108 or comes to the place with stones. Each jury member who comes there either takes one stone for himself or moves one stone from one pile to another. We can assume that there is an unlimited number of jury members. No one except the jury goes to the place with stones at night.
Participants want to know whether their notes can be correct or they are sure to have made a mistake.
Input
The first line of the input file contains a single integer n, the number of piles with stones in the garden (1 ≤ n ≤ 50).
The second line contains n integers separated by spaces x_1, x_2, …, x_n, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the first time (0 ≤ x_i ≤ 1000).
The third line contains n integers separated by spaces y_1, y_2, …, y_n, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the second time (0 ≤ y_i ≤ 1000).
Output
If the records can be consistent output "Yes", otherwise output "No" (quotes for clarity).
Examples
Input
5
1 2 3 4 5
2 1 4 3 5
Output
Yes
Input
5
1 1 1 1 1
1 0 1 0 1
Output
Yes
Input
3
2 3 9
1 7 9
Output
No
Note
In the first example, the following could have happened during the night: one of the jury members moved one stone from the second pile to the first pile, and the other jury member moved one stone from the fourth pile to the third pile.
In the second example, the jury took stones from the second and fourth piles.
It can be proved that it is impossible for the jury members to move and took stones to convert the first array into the second array. | instruction | 0 | 29,101 | 8 | 58,202 |
Tags: math
Correct Solution:
```
n = int(input())
a = input().split()
b = input().split()
a = [int(x) for x in a]
b = [int(x) for x in b]
s1 = i = s2 = 0
while i < n :
s1 += a[i]
s2 += b[i]
i+=1
if s2>s1 :
print("No")
else :
print("Yes")
``` | output | 1 | 29,101 | 8 | 58,203 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a beautiful garden of stones in Innopolis.
Its most beautiful place is the n piles with stones numbered from 1 to n.
EJOI participants have visited this place twice.
When they first visited it, the number of stones in piles was x_1, x_2, …, x_n, correspondingly. One of the participants wrote down this sequence in a notebook.
They visited it again the following day, and the number of stones in piles was equal to y_1, y_2, …, y_n. One of the participants also wrote it down in a notebook.
It is well known that every member of the EJOI jury during the night either sits in the room 108 or comes to the place with stones. Each jury member who comes there either takes one stone for himself or moves one stone from one pile to another. We can assume that there is an unlimited number of jury members. No one except the jury goes to the place with stones at night.
Participants want to know whether their notes can be correct or they are sure to have made a mistake.
Input
The first line of the input file contains a single integer n, the number of piles with stones in the garden (1 ≤ n ≤ 50).
The second line contains n integers separated by spaces x_1, x_2, …, x_n, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the first time (0 ≤ x_i ≤ 1000).
The third line contains n integers separated by spaces y_1, y_2, …, y_n, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the second time (0 ≤ y_i ≤ 1000).
Output
If the records can be consistent output "Yes", otherwise output "No" (quotes for clarity).
Examples
Input
5
1 2 3 4 5
2 1 4 3 5
Output
Yes
Input
5
1 1 1 1 1
1 0 1 0 1
Output
Yes
Input
3
2 3 9
1 7 9
Output
No
Note
In the first example, the following could have happened during the night: one of the jury members moved one stone from the second pile to the first pile, and the other jury member moved one stone from the fourth pile to the third pile.
In the second example, the jury took stones from the second and fourth piles.
It can be proved that it is impossible for the jury members to move and took stones to convert the first array into the second array. | instruction | 0 | 29,102 | 8 | 58,204 |
Tags: math
Correct Solution:
```
n = int(input())
a_1 = list(map(int,input().strip().split(' ')))
a_2 = list(map(int,input().strip().split(' ')))
print("Yes" if sum(a_1) >= sum(a_2) else "No")
``` | output | 1 | 29,102 | 8 | 58,205 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a beautiful garden of stones in Innopolis.
Its most beautiful place is the n piles with stones numbered from 1 to n.
EJOI participants have visited this place twice.
When they first visited it, the number of stones in piles was x_1, x_2, …, x_n, correspondingly. One of the participants wrote down this sequence in a notebook.
They visited it again the following day, and the number of stones in piles was equal to y_1, y_2, …, y_n. One of the participants also wrote it down in a notebook.
It is well known that every member of the EJOI jury during the night either sits in the room 108 or comes to the place with stones. Each jury member who comes there either takes one stone for himself or moves one stone from one pile to another. We can assume that there is an unlimited number of jury members. No one except the jury goes to the place with stones at night.
Participants want to know whether their notes can be correct or they are sure to have made a mistake.
Input
The first line of the input file contains a single integer n, the number of piles with stones in the garden (1 ≤ n ≤ 50).
The second line contains n integers separated by spaces x_1, x_2, …, x_n, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the first time (0 ≤ x_i ≤ 1000).
The third line contains n integers separated by spaces y_1, y_2, …, y_n, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the second time (0 ≤ y_i ≤ 1000).
Output
If the records can be consistent output "Yes", otherwise output "No" (quotes for clarity).
Examples
Input
5
1 2 3 4 5
2 1 4 3 5
Output
Yes
Input
5
1 1 1 1 1
1 0 1 0 1
Output
Yes
Input
3
2 3 9
1 7 9
Output
No
Note
In the first example, the following could have happened during the night: one of the jury members moved one stone from the second pile to the first pile, and the other jury member moved one stone from the fourth pile to the third pile.
In the second example, the jury took stones from the second and fourth piles.
It can be proved that it is impossible for the jury members to move and took stones to convert the first array into the second array. | instruction | 0 | 29,103 | 8 | 58,206 |
Tags: math
Correct Solution:
```
n=int(input())
l=list(map(int,input().split()))
a=list(map(int,input().split()))
if(sum(a)<=sum(l)):
print('Yes')
else:
print('No')
``` | output | 1 | 29,103 | 8 | 58,207 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a beautiful garden of stones in Innopolis.
Its most beautiful place is the n piles with stones numbered from 1 to n.
EJOI participants have visited this place twice.
When they first visited it, the number of stones in piles was x_1, x_2, …, x_n, correspondingly. One of the participants wrote down this sequence in a notebook.
They visited it again the following day, and the number of stones in piles was equal to y_1, y_2, …, y_n. One of the participants also wrote it down in a notebook.
It is well known that every member of the EJOI jury during the night either sits in the room 108 or comes to the place with stones. Each jury member who comes there either takes one stone for himself or moves one stone from one pile to another. We can assume that there is an unlimited number of jury members. No one except the jury goes to the place with stones at night.
Participants want to know whether their notes can be correct or they are sure to have made a mistake.
Input
The first line of the input file contains a single integer n, the number of piles with stones in the garden (1 ≤ n ≤ 50).
The second line contains n integers separated by spaces x_1, x_2, …, x_n, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the first time (0 ≤ x_i ≤ 1000).
The third line contains n integers separated by spaces y_1, y_2, …, y_n, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the second time (0 ≤ y_i ≤ 1000).
Output
If the records can be consistent output "Yes", otherwise output "No" (quotes for clarity).
Examples
Input
5
1 2 3 4 5
2 1 4 3 5
Output
Yes
Input
5
1 1 1 1 1
1 0 1 0 1
Output
Yes
Input
3
2 3 9
1 7 9
Output
No
Note
In the first example, the following could have happened during the night: one of the jury members moved one stone from the second pile to the first pile, and the other jury member moved one stone from the fourth pile to the third pile.
In the second example, the jury took stones from the second and fourth piles.
It can be proved that it is impossible for the jury members to move and took stones to convert the first array into the second array. | instruction | 0 | 29,104 | 8 | 58,208 |
Tags: math
Correct Solution:
```
n=int(input())
x=list(map(int,input().split()))
y=list(map(int,input().split()))
c=0
e=0
for i in range (n):
c=c+x[i]
e=e+y[i]
if e>c:
print("No")
else :
print("Yes")
``` | output | 1 | 29,104 | 8 | 58,209 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a beautiful garden of stones in Innopolis.
Its most beautiful place is the n piles with stones numbered from 1 to n.
EJOI participants have visited this place twice.
When they first visited it, the number of stones in piles was x_1, x_2, …, x_n, correspondingly. One of the participants wrote down this sequence in a notebook.
They visited it again the following day, and the number of stones in piles was equal to y_1, y_2, …, y_n. One of the participants also wrote it down in a notebook.
It is well known that every member of the EJOI jury during the night either sits in the room 108 or comes to the place with stones. Each jury member who comes there either takes one stone for himself or moves one stone from one pile to another. We can assume that there is an unlimited number of jury members. No one except the jury goes to the place with stones at night.
Participants want to know whether their notes can be correct or they are sure to have made a mistake.
Input
The first line of the input file contains a single integer n, the number of piles with stones in the garden (1 ≤ n ≤ 50).
The second line contains n integers separated by spaces x_1, x_2, …, x_n, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the first time (0 ≤ x_i ≤ 1000).
The third line contains n integers separated by spaces y_1, y_2, …, y_n, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the second time (0 ≤ y_i ≤ 1000).
Output
If the records can be consistent output "Yes", otherwise output "No" (quotes for clarity).
Examples
Input
5
1 2 3 4 5
2 1 4 3 5
Output
Yes
Input
5
1 1 1 1 1
1 0 1 0 1
Output
Yes
Input
3
2 3 9
1 7 9
Output
No
Note
In the first example, the following could have happened during the night: one of the jury members moved one stone from the second pile to the first pile, and the other jury member moved one stone from the fourth pile to the third pile.
In the second example, the jury took stones from the second and fourth piles.
It can be proved that it is impossible for the jury members to move and took stones to convert the first array into the second array. | instruction | 0 | 29,105 | 8 | 58,210 |
Tags: math
Correct Solution:
```
a=int(input())
b=sum([int(i) for i in input().split()])
c=sum([int(j) for j in input().split()])
if(c>b):
print("No")
else:
print("Yes")
``` | output | 1 | 29,105 | 8 | 58,211 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a beautiful garden of stones in Innopolis.
Its most beautiful place is the n piles with stones numbered from 1 to n.
EJOI participants have visited this place twice.
When they first visited it, the number of stones in piles was x_1, x_2, …, x_n, correspondingly. One of the participants wrote down this sequence in a notebook.
They visited it again the following day, and the number of stones in piles was equal to y_1, y_2, …, y_n. One of the participants also wrote it down in a notebook.
It is well known that every member of the EJOI jury during the night either sits in the room 108 or comes to the place with stones. Each jury member who comes there either takes one stone for himself or moves one stone from one pile to another. We can assume that there is an unlimited number of jury members. No one except the jury goes to the place with stones at night.
Participants want to know whether their notes can be correct or they are sure to have made a mistake.
Input
The first line of the input file contains a single integer n, the number of piles with stones in the garden (1 ≤ n ≤ 50).
The second line contains n integers separated by spaces x_1, x_2, …, x_n, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the first time (0 ≤ x_i ≤ 1000).
The third line contains n integers separated by spaces y_1, y_2, …, y_n, the number of stones in piles recorded in the notebook when the participants came to the place with stones for the second time (0 ≤ y_i ≤ 1000).
Output
If the records can be consistent output "Yes", otherwise output "No" (quotes for clarity).
Examples
Input
5
1 2 3 4 5
2 1 4 3 5
Output
Yes
Input
5
1 1 1 1 1
1 0 1 0 1
Output
Yes
Input
3
2 3 9
1 7 9
Output
No
Note
In the first example, the following could have happened during the night: one of the jury members moved one stone from the second pile to the first pile, and the other jury member moved one stone from the fourth pile to the third pile.
In the second example, the jury took stones from the second and fourth piles.
It can be proved that it is impossible for the jury members to move and took stones to convert the first array into the second array. | instruction | 0 | 29,106 | 8 | 58,212 |
Tags: math
Correct Solution:
```
n= int(input())
arr1=input()
lis1=list(map(int,arr1.split()))
arr2=input()
lis2=list(map(int,arr2.split()))
sum1=sum(lis1)
sum2=sum(lis2)
if (sum2>sum1):
print("No")
else :
print("Yes")
``` | output | 1 | 29,106 | 8 | 58,213 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored A·B·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) × (B - 2) × (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 × 1 × 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B и C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 ≤ n ≤ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 × 4 × 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) × (4 - 2) × (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 × 3 × 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) × (3 - 2) × (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks. | instruction | 0 | 29,335 | 8 | 58,670 |
Tags: brute force, math
Correct Solution:
```
from math import sqrt
p, n = [], int(input())
def f(x, y): return (x + 2) * (y + 2) + (2 * (x + y + 2) * n) // (x * y)
for x in range(2, int(sqrt(n)) + 1):
if n % x == 0: p.append(x)
p += [n // x for x in reversed(p)]
p.append(n)
u = v = f(1, 1)
for m in p:
for x in range(1, int(sqrt(m)) + 1):
if m % x == 0: u = min(u, f(x, m // x))
print(u, v)
# Made By Mostafa_Khaled
``` | output | 1 | 29,335 | 8 | 58,671 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored A·B·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) × (B - 2) × (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 × 1 × 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B и C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 ≤ n ≤ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 × 4 × 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) × (4 - 2) × (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 × 3 × 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) × (3 - 2) × (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks. | instruction | 0 | 29,336 | 8 | 58,672 |
Tags: brute force, math
Correct Solution:
```
n=int(input())
a,mi=1,999999999999
while a**3<=n:
if n%a==0:
b=1
while b**2<=(n//a):
if (n//a)%b==0:
c=n//a//b
mi=min(mi,(a+1)*(b+2)*(c+2))
b+=1
a+=1
print(mi-n, 9*n+9-n)
``` | output | 1 | 29,336 | 8 | 58,673 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored A·B·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) × (B - 2) × (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 × 1 × 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B и C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 ≤ n ≤ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 × 4 × 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) × (4 - 2) × (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 × 3 × 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) × (3 - 2) × (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks. | instruction | 0 | 29,337 | 8 | 58,674 |
Tags: brute force, math
Correct Solution:
```
n = int(input())
mn = 999999999999
for i in range(1, int(n**(1/3))+1):
if n%i==0:
for j in range(1, int((n//i)**(1/2))+1):
if (n/i) % j == 0:
k = (n//i)//j
mn = min(mn, (i+1)*(k+2)*(j+2))
print(mn-n, (9*n+9-n))
``` | output | 1 | 29,337 | 8 | 58,675 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored A·B·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) × (B - 2) × (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 × 1 × 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B и C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 ≤ n ≤ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 × 4 × 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) × (4 - 2) × (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 × 3 × 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) × (3 - 2) × (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks. | instruction | 0 | 29,338 | 8 | 58,676 |
Tags: brute force, math
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import defaultdict
from math import ceil,floor,sqrt,log2,gcd,pi
from heapq import heappush,heappop
from bisect import bisect_left,bisect
import sys
abc='abcdefghijklmnopqrstuvwxyz'
ABC="ABCDEFGHIJKLMNOPQRSTUVWXYZ"
n=int(input())
arr=[]
for i in range(1,min(n,100000)+1):
if n%i==0:
for j in range(1,min(n,1000)+1):
if n%(i*j)==0:
k=n//(i*j)
# print(i,j,k)
arr.append((k+1)*(i+2)*(j+2))
arr.append((i+1)*(k+2)*(j+2))
arr.append((j+1)*(k+2)*(i+2))
print(min(arr)-n,max(arr)-n)
``` | output | 1 | 29,338 | 8 | 58,677 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored A·B·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) × (B - 2) × (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 × 1 × 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B и C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 ≤ n ≤ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 × 4 × 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) × (4 - 2) × (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 × 3 × 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) × (3 - 2) × (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks. | instruction | 0 | 29,339 | 8 | 58,678 |
Tags: brute force, math
Correct Solution:
```
import itertools as it
from functools import reduce
def factor(n):
"""
>>> factor(2)
[(2, 1)]
>>> factor(6)
[(3, 1), (2, 1)]
>>> factor(98)
[(7, 2), (2, 1)]
>>> factor(1)
[]
"""
result = []
i = 2
while i * i <= n:
j = 0
while n % i == 0:
j += 1
n //= i
if j > 0:
result += [(i, j)]
i += 1
if n > 1:
result += [(n, 1)]
return result[::-1]
def all_divisors(prime_factorization):
"""
>>> all_divisors(factor(6))
[1, 2, 3, 6]
>>> all_divisors(factor(12))
[1, 2, 4, 3, 6, 12]
>>> all_divisors(factor(7))
[1, 7]
>>> all_divisors(factor(1))
[1]
"""
if len(prime_factorization) == 0:
return [1]
result = []
factor_with_mult = []
for pfactor in prime_factorization:
factor_with_mult += [[1]]
for _ in range(pfactor[1]):
factor_with_mult[-1] += [factor_with_mult[-1][-1] * pfactor[0]]
for divisor in it.product(*factor_with_mult):
result += [reduce(lambda x, y: x * y, divisor)]
return result
if __name__ == '__main__':
n = int(input())
min_result = 10**20
max_result = 0
for divisor in all_divisors(factor(n)):
for div2 in all_divisors(factor(divisor)):
a = n // divisor + 1
b = divisor // div2 + 2
c = div2 + 2
min_result = min([min_result,
c * b + (a - 1) * 2 *
(b + c - 2)])
max_result = max([max_result,
c * b + (a - 1) * 2 *
(b + c - 2)])
print(str(min_result) + " " + str(max_result))
``` | output | 1 | 29,339 | 8 | 58,679 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored A·B·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) × (B - 2) × (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 × 1 × 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B и C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 ≤ n ≤ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 × 4 × 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) × (4 - 2) × (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 × 3 × 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) × (3 - 2) × (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks. | instruction | 0 | 29,340 | 8 | 58,680 |
Tags: brute force, math
Correct Solution:
```
n=int(input())
min1 = 10000000000000000000000
max1 = 0
i=1
while i*i*i <= n:
if n%i==0:
j=1
while j*j <= int(n/i):
if n % (i*j) ==0:
a=i+1
b=j+2
c=int(n/(i*j)) + 2
min1=min(min1,(a*b*c)-n)
max1=max(max1,(a*b*c)-n)
a=i+2
b=j+1
c=int(n/(i*j) )+ 2
min1=min(min1,(a*b*c)-n)
max1=max(max1,(a*b*c)-n)
a=i+2
b=j+2
c=int(n/(i*j)) + 1
min1=min(min1,(a*b*c)-n)
max1=max(max1,(a*b*c)-n)
j+=1
i+=1
print(min1,max1)
``` | output | 1 | 29,340 | 8 | 58,681 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored A·B·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) × (B - 2) × (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 × 1 × 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B и C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 ≤ n ≤ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 × 4 × 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) × (4 - 2) × (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 × 3 × 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) × (3 - 2) × (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks. | instruction | 0 | 29,341 | 8 | 58,682 |
Tags: brute force, math
Correct Solution:
```
# ========= /\ /| |====/|
# | / \ | | / |
# | /____\ | | / |
# | / \ | | / |
# ========= / \ ===== |/====|
# code
def main():
n = int(input())
m1 = float('inf')
m2 = float('-inf')
i = 1
while i*i*i <= n:
if n%i != 0:
i += 1
continue
j = 1
while j*j <= n//i:
if (n//i)%j != 0:
j += 1
continue
k = (n // i)//j
# print(i,j,k)
m1 = min(m1 , (i + 1)*(j + 2)*(k + 2) - n)
m1 = min(m1 , (j + 1)*(i + 2)*(k + 2) - n)
m1 = min(m1 , (k + 1)*(j + 2)*(i + 2) - n)
m2 = max(m2 , (i + 1)*(j + 2)*(k + 2) - n)
m2 = max(m2 , (j + 1)*(i + 2)*(k + 2) - n)
m2 = max(m2 , (k + 1)*(j + 2)*(i + 2) - n)
j += 1
i += 1
print(m1 , m2)
return
if __name__ == "__main__":
main()
``` | output | 1 | 29,341 | 8 | 58,683 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored A·B·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) × (B - 2) × (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 × 1 × 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B и C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 ≤ n ≤ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 × 4 × 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) × (4 - 2) × (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 × 3 × 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) × (3 - 2) × (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks. | instruction | 0 | 29,342 | 8 | 58,684 |
Tags: brute force, math
Correct Solution:
```
from math import sqrt
n = int(input())
def f(x, y):
return (x+2) * (y+2) + (2*n * (x+y+2)) // (x * y)
factors = []
for i in range(2, int(sqrt(n))+1):
if n % i == 0:
factors.append(i)
factors += [n // i for i in reversed(factors)] + [n]
res = f(1, 1)
for i in factors:
for j in range(1, int(sqrt(i))+1):
if i % j == 0:
res = min(res, f(j, i // j))
print(res, f(1, 1))
``` | output | 1 | 29,342 | 8 | 58,685 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored A·B·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) × (B - 2) × (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 × 1 × 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B и C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 ≤ n ≤ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 × 4 × 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) × (4 - 2) × (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 × 3 × 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) × (3 - 2) × (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Submitted Solution:
```
n=int(input())
L=[]
i=1
while(i*i*i<=n):
if(n%i!=0):
i+=1
continue
x=n//i
j=i
while(j*j<=x):
if(x%j!=0):
j+=1
continue
L.append((i,j,x//j))
j+=1
i+=1
maxx=0
minn=10**20
E=[[1,2,2],[2,1,2],[2,2,1]]
for item in L:
x=item[0]
y=item[1]
z=item[2]
A=[x,y,z]
for item in E:
m=(A[0]+item[0])*(A[1]+item[1])*(A[2]+item[2])
if(m<minn):
minn=m
if(m>maxx):
maxx=m
print(minn-n,maxx-n)
``` | instruction | 0 | 29,343 | 8 | 58,686 |
Yes | output | 1 | 29,343 | 8 | 58,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored A·B·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) × (B - 2) × (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 × 1 × 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B и C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 ≤ n ≤ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 × 4 × 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) × (4 - 2) × (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 × 3 × 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) × (3 - 2) × (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Submitted Solution:
```
import itertools as it
from functools import reduce
def factor(n):
"""
>>> factor(2)
[(2, 1)]
>>> factor(6)
[(3, 1), (2, 1)]
>>> factor(98)
[(7, 2), (2, 1)]
>>> factor(1)
[]
"""
result = []
i = 2
while i * i <= n:
j = 0
while n % i == 0:
j += 1
n //= i
if j > 0:
result += [(i, j)]
i += 1
if n > 1:
result += [(n, 1)]
return result[::-1]
def all_divisors(prime_factorization):
"""
>>> all_divisors(factor(6))
[1, 2, 3, 6]
>>> all_divisors(factor(12))
[1, 2, 4, 3, 6, 12]
>>> all_divisors(factor(7))
[1, 7]
>>> all_divisors(factor(1))
[1]
"""
if len(prime_factorization) == 0:
return [1]
result = []
factor_with_mult = []
for pfactor in prime_factorization:
factor_with_mult += [[1]]
for _ in range(pfactor[1]):
factor_with_mult[-1] += [factor_with_mult[-1][-1] * pfactor[0]]
for divisor in it.product(*factor_with_mult):
result += [reduce(lambda x, y: x * y, divisor)]
return result
if __name__ == '__main__':
n = int(input())
min_result = 10**9
max_result = 0
for divisor in all_divisors(factor(n)):
for div2 in all_divisors(factor(divisor)):
if div2 * div2 > divisor:
break
a = n // divisor + 1
b = divisor // div2 + 2
c = div2 + 2
min_result = min([min_result,
c * b + (a - 1) * 2 *
(b + c - 2)])
max_result = max([max_result,
c * b + (a - 1) * 2 *
(b + c - 2)])
print(str(min_result) + " " + str(max_result))
``` | instruction | 0 | 29,344 | 8 | 58,688 |
No | output | 1 | 29,344 | 8 | 58,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored A·B·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) × (B - 2) × (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 × 1 × 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B и C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 ≤ n ≤ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 × 4 × 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) × (4 - 2) × (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 × 3 × 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) × (3 - 2) × (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import defaultdict
from math import ceil,floor,sqrt,log2,gcd,pi
from heapq import heappush,heappop
from bisect import bisect_left,bisect
import sys
abc='abcdefghijklmnopqrstuvwxyz'
ABC="ABCDEFGHIJKLMNOPQRSTUVWXYZ"
n=int(input())
l=1
r=n
for i in range(2,ceil(sqrt(n))+1):
if n%i==0:
# print(n//i,i)
if ((n//i)-i)<(r-l):
r=n//i
l=i
# print(l,r)
print((2*(l+2)*(r+2))-n,(n+1)*9-n)
``` | instruction | 0 | 29,345 | 8 | 58,690 |
No | output | 1 | 29,345 | 8 | 58,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored A·B·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) × (B - 2) × (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 × 1 × 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B и C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 ≤ n ≤ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 × 4 × 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) × (4 - 2) × (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 × 3 × 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) × (3 - 2) × (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Submitted Solution:
```
n=int(input())
min1 = 10**9
max1 = -1
i=1
while i*i*i <= n:
if n%i==0:
j=1
while j*j <= int(n/i):
if n % i*j ==0:
a=i+1
b=j+2
c=int(n/(i*j) + 2 )
min1=min(min1,(a*b*c)-n)
max1=max(max1,(a*b*c)-n)
a=i+2
b=j+1
c=int(n/(i*j) + 2)
min1=min(min1,(a*b*c)-n)
max1=max(max1,(a*b*c)-n)
a=i+2
b=j+2
c=int(n/(i*j) + 1)
min1=min(min1,(a*b*c)-n)
max1=max(max1,(a*b*c)-n)
j+=1
i+=1
print(min1,max1)
``` | instruction | 0 | 29,346 | 8 | 58,692 |
No | output | 1 | 29,346 | 8 | 58,693 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time in the Kingdom of Far Far Away lived Sam the Farmer. Sam had a cow named Dawn and he was deeply attached to her. Sam would spend the whole summer stocking hay to feed Dawn in winter. Sam scythed hay and put it into haystack. As Sam was a bright farmer, he tried to make the process of storing hay simpler and more convenient to use. He collected the hay into cubical hay blocks of the same size. Then he stored the blocks in his barn. After a summer spent in hard toil Sam stored A·B·C hay blocks and stored them in a barn as a rectangular parallelepiped A layers high. Each layer had B rows and each row had C blocks.
At the end of the autumn Sam came into the barn to admire one more time the hay he'd been stacking during this hard summer. Unfortunately, Sam was horrified to see that the hay blocks had been carelessly scattered around the barn. The place was a complete mess. As it turned out, thieves had sneaked into the barn. They completely dissembled and took away a layer of blocks from the parallelepiped's front, back, top and sides. As a result, the barn only had a parallelepiped containing (A - 1) × (B - 2) × (C - 2) hay blocks. To hide the evidence of the crime, the thieves had dissembled the parallelepiped into single 1 × 1 × 1 blocks and scattered them around the barn. After the theft Sam counted n hay blocks in the barn but he forgot numbers A, B и C.
Given number n, find the minimally possible and maximally possible number of stolen hay blocks.
Input
The only line contains integer n from the problem's statement (1 ≤ n ≤ 109).
Output
Print space-separated minimum and maximum number of hay blocks that could have been stolen by the thieves.
Note that the answer to the problem can be large enough, so you must use the 64-bit integer type for calculations. Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
Examples
Input
4
Output
28 41
Input
7
Output
47 65
Input
12
Output
48 105
Note
Let's consider the first sample test. If initially Sam has a parallelepiped consisting of 32 = 2 × 4 × 4 hay blocks in his barn, then after the theft the barn has 4 = (2 - 1) × (4 - 2) × (4 - 2) hay blocks left. Thus, the thieves could have stolen 32 - 4 = 28 hay blocks. If Sam initially had a parallelepiped consisting of 45 = 5 × 3 × 3 hay blocks in his barn, then after the theft the barn has 4 = (5 - 1) × (3 - 2) × (3 - 2) hay blocks left. Thus, the thieves could have stolen 45 - 4 = 41 hay blocks. No other variants of the blocks' initial arrangement (that leave Sam with exactly 4 blocks after the theft) can permit the thieves to steal less than 28 or more than 41 blocks.
Submitted Solution:
```
import itertools as it
from functools import reduce
def factor(n):
"""
>>> factor(2)
[(2, 1)]
>>> factor(6)
[(3, 1), (2, 1)]
>>> factor(98)
[(7, 2), (2, 1)]
>>> factor(1)
[]
"""
prev_n = n
result = []
i = 2
while i * i <= prev_n:
j = 0
while n % i == 0:
j += 1
n //= i
if j > 0:
result += [(i, j)]
i += 1
if n > 1:
result += [(n, 1)]
return result[::-1]
def all_divisors(prime_factorization):
"""
>>> all_divisors(factor(6))
[1, 2, 3, 6]
>>> all_divisors(factor(12))
[1, 2, 4, 3, 6, 12]
>>> all_divisors(factor(7))
[1, 7]
>>> all_divisors(factor(1))
[1]
"""
if len(prime_factorization) == 0:
return [1]
result = []
factor_with_mult = []
for pfactor in prime_factorization:
factor_with_mult += [[1]]
for _ in range(pfactor[1]):
factor_with_mult[-1] += [factor_with_mult[-1][-1] * pfactor[0]]
for divisor in it.product(*factor_with_mult):
result += [reduce(lambda x, y: x * y, divisor)]
return result
if __name__ == '__main__':
n = int(input())
min_result = 10**9
max_result = 0
for divisor in all_divisors(factor(n)):
for div2 in all_divisors(factor(divisor)):
if div2 * div2 > divisor:
break
a = n // divisor + 1
b = divisor // div2 + 2
c = div2 + 2
min_result = min([min_result,
c * b + (a - 1) * 2 *
(b + c - 2)])
max_result = max([max_result,
c * b + (a - 1) * 2 *
(b + c - 2)])
print(str(min_result) + " " + str(max_result))
``` | instruction | 0 | 29,347 | 8 | 58,694 |
No | output | 1 | 29,347 | 8 | 58,695 |
Provide a correct Python 3 solution for this coding contest problem.
B-Mansion and courier
Problem Statement
Taro lives alone in a mansion. Taro, who loves studying, intends to study in his study in the house today. Taro can't concentrate outside the study, so he always studies in the study.
However, on this day, $ N $ of courier service to Taro arrived. $ i $ ($ 1 \ leq i \ leq N $) The arrival time of the third courier is $ a_i $. It is painful to have the delivery person wait at the front door, so Taro decided to be at the front door by the time the courier arrives. Due to the large size of the mansion, it takes $ M $ one way to move between the study and the entrance.
On the other hand, Taro wants to study for as long as possible. Find the maximum amount of time Taro can study in the study from time $ 0 $ to time $ T $.
Taro is in the study at time $ 0 $, and the courier does not arrive earlier than the time $ M $, and the courier does not arrive later than the time $ T $. Also, the time it takes for Taro to receive the courier can be ignored.
Input
Each dataset consists of two lines. The first line consists of three integers $ N, M, T $ separated by blanks. These integers satisfy $ 1 \ leq N \ leq 100 $, $ 1 \ leq M \ leq 10 {,} 000 $, $ 1 \ leq T \ leq 10 {,} 000 $. The second line consists of $ N $ integers $ a_1, a_2, \ dots, a_N $ separated by blanks. Each $ a_i $ fills $ M \ leq a_i \ leq T $ and is also $ a_i <a_ {i + 1} $ ($ 1 \ leq i <N $).
Output
Output an integer representing the maximum amount of time Taro can study on one line.
Sample Input 1
1 1 5
3
Output for the Sample Input 1
3
Sample Input 2
2 1 10
2 7
Output for the Sample Input 2
6
Sample Input 3
2 4 10
6 8
Output for the Sample Input 3
2
Example
Input
1 1 5
3
Output
3 | instruction | 0 | 30,003 | 8 | 60,006 |
"Correct Solution:
```
N, M, T = map(int, input().split())
A = list(map(int, input().split()))
prv = 0
cnt = 0
for a in A:
if prv == 0:
cnt += a - M
else:
if prv + M < a - M:
cnt += (a - M) - (prv + M)
prv = a
if prv + M < T:
cnt += T - (prv + M)
print(cnt)
``` | output | 1 | 30,003 | 8 | 60,007 |
Provide a correct Python 3 solution for this coding contest problem.
B-Mansion and courier
Problem Statement
Taro lives alone in a mansion. Taro, who loves studying, intends to study in his study in the house today. Taro can't concentrate outside the study, so he always studies in the study.
However, on this day, $ N $ of courier service to Taro arrived. $ i $ ($ 1 \ leq i \ leq N $) The arrival time of the third courier is $ a_i $. It is painful to have the delivery person wait at the front door, so Taro decided to be at the front door by the time the courier arrives. Due to the large size of the mansion, it takes $ M $ one way to move between the study and the entrance.
On the other hand, Taro wants to study for as long as possible. Find the maximum amount of time Taro can study in the study from time $ 0 $ to time $ T $.
Taro is in the study at time $ 0 $, and the courier does not arrive earlier than the time $ M $, and the courier does not arrive later than the time $ T $. Also, the time it takes for Taro to receive the courier can be ignored.
Input
Each dataset consists of two lines. The first line consists of three integers $ N, M, T $ separated by blanks. These integers satisfy $ 1 \ leq N \ leq 100 $, $ 1 \ leq M \ leq 10 {,} 000 $, $ 1 \ leq T \ leq 10 {,} 000 $. The second line consists of $ N $ integers $ a_1, a_2, \ dots, a_N $ separated by blanks. Each $ a_i $ fills $ M \ leq a_i \ leq T $ and is also $ a_i <a_ {i + 1} $ ($ 1 \ leq i <N $).
Output
Output an integer representing the maximum amount of time Taro can study on one line.
Sample Input 1
1 1 5
3
Output for the Sample Input 1
3
Sample Input 2
2 1 10
2 7
Output for the Sample Input 2
6
Sample Input 3
2 4 10
6 8
Output for the Sample Input 3
2
Example
Input
1 1 5
3
Output
3 | instruction | 0 | 30,004 | 8 | 60,008 |
"Correct Solution:
```
# coding: utf-8
n,m,t=map(int,input().split())
a=list(map(int,input().split()))
d=[0]
for time in a:
f=False
if d[-1]<time-m:
d.append(max(time-m,0))
else:
f=True
if f:
d[-1]=max(d[-1],time+m)
else:
d.append(min(time+m,t))
del d[0]
data=[False for i in range(t)]
i=0
while i<len(d)-1:
data[d[i]:d[i+1]]=[True for j in range(len(data[d[i]:d[i+1]]))]
i+=2
print(data.count(False))
``` | output | 1 | 30,004 | 8 | 60,009 |
Provide a correct Python 3 solution for this coding contest problem.
B-Mansion and courier
Problem Statement
Taro lives alone in a mansion. Taro, who loves studying, intends to study in his study in the house today. Taro can't concentrate outside the study, so he always studies in the study.
However, on this day, $ N $ of courier service to Taro arrived. $ i $ ($ 1 \ leq i \ leq N $) The arrival time of the third courier is $ a_i $. It is painful to have the delivery person wait at the front door, so Taro decided to be at the front door by the time the courier arrives. Due to the large size of the mansion, it takes $ M $ one way to move between the study and the entrance.
On the other hand, Taro wants to study for as long as possible. Find the maximum amount of time Taro can study in the study from time $ 0 $ to time $ T $.
Taro is in the study at time $ 0 $, and the courier does not arrive earlier than the time $ M $, and the courier does not arrive later than the time $ T $. Also, the time it takes for Taro to receive the courier can be ignored.
Input
Each dataset consists of two lines. The first line consists of three integers $ N, M, T $ separated by blanks. These integers satisfy $ 1 \ leq N \ leq 100 $, $ 1 \ leq M \ leq 10 {,} 000 $, $ 1 \ leq T \ leq 10 {,} 000 $. The second line consists of $ N $ integers $ a_1, a_2, \ dots, a_N $ separated by blanks. Each $ a_i $ fills $ M \ leq a_i \ leq T $ and is also $ a_i <a_ {i + 1} $ ($ 1 \ leq i <N $).
Output
Output an integer representing the maximum amount of time Taro can study on one line.
Sample Input 1
1 1 5
3
Output for the Sample Input 1
3
Sample Input 2
2 1 10
2 7
Output for the Sample Input 2
6
Sample Input 3
2 4 10
6 8
Output for the Sample Input 3
2
Example
Input
1 1 5
3
Output
3 | instruction | 0 | 30,005 | 8 | 60,010 |
"Correct Solution:
```
N, M, T = map(int, input().split())
a = list(map(int, input().split()))
current = 0
ans = 0
ans += max(a[0] - M, 0)
current = a[0]
for i in range(1, N):
ans += max(a[i] - a[i - 1] - 2 * M, 0)
current = a[i]
ans += max(T - a[N - 1] - M, 0)
print(ans)
``` | output | 1 | 30,005 | 8 | 60,011 |
Provide a correct Python 3 solution for this coding contest problem.
B-Mansion and courier
Problem Statement
Taro lives alone in a mansion. Taro, who loves studying, intends to study in his study in the house today. Taro can't concentrate outside the study, so he always studies in the study.
However, on this day, $ N $ of courier service to Taro arrived. $ i $ ($ 1 \ leq i \ leq N $) The arrival time of the third courier is $ a_i $. It is painful to have the delivery person wait at the front door, so Taro decided to be at the front door by the time the courier arrives. Due to the large size of the mansion, it takes $ M $ one way to move between the study and the entrance.
On the other hand, Taro wants to study for as long as possible. Find the maximum amount of time Taro can study in the study from time $ 0 $ to time $ T $.
Taro is in the study at time $ 0 $, and the courier does not arrive earlier than the time $ M $, and the courier does not arrive later than the time $ T $. Also, the time it takes for Taro to receive the courier can be ignored.
Input
Each dataset consists of two lines. The first line consists of three integers $ N, M, T $ separated by blanks. These integers satisfy $ 1 \ leq N \ leq 100 $, $ 1 \ leq M \ leq 10 {,} 000 $, $ 1 \ leq T \ leq 10 {,} 000 $. The second line consists of $ N $ integers $ a_1, a_2, \ dots, a_N $ separated by blanks. Each $ a_i $ fills $ M \ leq a_i \ leq T $ and is also $ a_i <a_ {i + 1} $ ($ 1 \ leq i <N $).
Output
Output an integer representing the maximum amount of time Taro can study on one line.
Sample Input 1
1 1 5
3
Output for the Sample Input 1
3
Sample Input 2
2 1 10
2 7
Output for the Sample Input 2
6
Sample Input 3
2 4 10
6 8
Output for the Sample Input 3
2
Example
Input
1 1 5
3
Output
3 | instruction | 0 | 30,006 | 8 | 60,012 |
"Correct Solution:
```
l_raw = input().split()
l = [int(n) for n in l_raw]
a_raw = input().split()
a_ = [int(n) for n in a_raw]
study = 0
state = 0
now=0
for a in a_:
if state==0:
if l[1]<a-now:
study+=a-now-l[1]
now=a
state=1
elif state==1:
if 2*l[1]<a-now:
study+=a-now-2*l[1]
now=a
if l[2]-l[1]-a>0:
study+=l[2]-l[1]-a
print(study)
``` | output | 1 | 30,006 | 8 | 60,013 |
Provide a correct Python 3 solution for this coding contest problem.
B-Mansion and courier
Problem Statement
Taro lives alone in a mansion. Taro, who loves studying, intends to study in his study in the house today. Taro can't concentrate outside the study, so he always studies in the study.
However, on this day, $ N $ of courier service to Taro arrived. $ i $ ($ 1 \ leq i \ leq N $) The arrival time of the third courier is $ a_i $. It is painful to have the delivery person wait at the front door, so Taro decided to be at the front door by the time the courier arrives. Due to the large size of the mansion, it takes $ M $ one way to move between the study and the entrance.
On the other hand, Taro wants to study for as long as possible. Find the maximum amount of time Taro can study in the study from time $ 0 $ to time $ T $.
Taro is in the study at time $ 0 $, and the courier does not arrive earlier than the time $ M $, and the courier does not arrive later than the time $ T $. Also, the time it takes for Taro to receive the courier can be ignored.
Input
Each dataset consists of two lines. The first line consists of three integers $ N, M, T $ separated by blanks. These integers satisfy $ 1 \ leq N \ leq 100 $, $ 1 \ leq M \ leq 10 {,} 000 $, $ 1 \ leq T \ leq 10 {,} 000 $. The second line consists of $ N $ integers $ a_1, a_2, \ dots, a_N $ separated by blanks. Each $ a_i $ fills $ M \ leq a_i \ leq T $ and is also $ a_i <a_ {i + 1} $ ($ 1 \ leq i <N $).
Output
Output an integer representing the maximum amount of time Taro can study on one line.
Sample Input 1
1 1 5
3
Output for the Sample Input 1
3
Sample Input 2
2 1 10
2 7
Output for the Sample Input 2
6
Sample Input 3
2 4 10
6 8
Output for the Sample Input 3
2
Example
Input
1 1 5
3
Output
3 | instruction | 0 | 30,007 | 8 | 60,014 |
"Correct Solution:
```
n, m, t = map(int, input().split())
a = list(map(int, input().split()))
s = t
for i in range(n):
if i == 0:
s -= m
elif a[i] - a[i - 1] >= 2 * m:
s -= 2 * m
else:
s -= a[i] - a[i - 1]
if i == n - 1:
s -= min(t - a[-1], m)
print(s)
``` | output | 1 | 30,007 | 8 | 60,015 |
Provide a correct Python 3 solution for this coding contest problem.
B-Mansion and courier
Problem Statement
Taro lives alone in a mansion. Taro, who loves studying, intends to study in his study in the house today. Taro can't concentrate outside the study, so he always studies in the study.
However, on this day, $ N $ of courier service to Taro arrived. $ i $ ($ 1 \ leq i \ leq N $) The arrival time of the third courier is $ a_i $. It is painful to have the delivery person wait at the front door, so Taro decided to be at the front door by the time the courier arrives. Due to the large size of the mansion, it takes $ M $ one way to move between the study and the entrance.
On the other hand, Taro wants to study for as long as possible. Find the maximum amount of time Taro can study in the study from time $ 0 $ to time $ T $.
Taro is in the study at time $ 0 $, and the courier does not arrive earlier than the time $ M $, and the courier does not arrive later than the time $ T $. Also, the time it takes for Taro to receive the courier can be ignored.
Input
Each dataset consists of two lines. The first line consists of three integers $ N, M, T $ separated by blanks. These integers satisfy $ 1 \ leq N \ leq 100 $, $ 1 \ leq M \ leq 10 {,} 000 $, $ 1 \ leq T \ leq 10 {,} 000 $. The second line consists of $ N $ integers $ a_1, a_2, \ dots, a_N $ separated by blanks. Each $ a_i $ fills $ M \ leq a_i \ leq T $ and is also $ a_i <a_ {i + 1} $ ($ 1 \ leq i <N $).
Output
Output an integer representing the maximum amount of time Taro can study on one line.
Sample Input 1
1 1 5
3
Output for the Sample Input 1
3
Sample Input 2
2 1 10
2 7
Output for the Sample Input 2
6
Sample Input 3
2 4 10
6 8
Output for the Sample Input 3
2
Example
Input
1 1 5
3
Output
3 | instruction | 0 | 30,008 | 8 | 60,016 |
"Correct Solution:
```
# coding: utf-8
# Your code here!
a = list(map(int,input().split()))
b = list(map(int,input().split()))
k = [0]*a[2]
pp = 0
su = 0
for i in b:
for j in range(a[2]):
if j >= i - a[1] and j <= i + a[1] - 1:
k[j] += 1
for i in range(a[2]):
if k[i] == 0:
su += 1
print(su)
``` | output | 1 | 30,008 | 8 | 60,017 |
Provide a correct Python 3 solution for this coding contest problem.
B-Mansion and courier
Problem Statement
Taro lives alone in a mansion. Taro, who loves studying, intends to study in his study in the house today. Taro can't concentrate outside the study, so he always studies in the study.
However, on this day, $ N $ of courier service to Taro arrived. $ i $ ($ 1 \ leq i \ leq N $) The arrival time of the third courier is $ a_i $. It is painful to have the delivery person wait at the front door, so Taro decided to be at the front door by the time the courier arrives. Due to the large size of the mansion, it takes $ M $ one way to move between the study and the entrance.
On the other hand, Taro wants to study for as long as possible. Find the maximum amount of time Taro can study in the study from time $ 0 $ to time $ T $.
Taro is in the study at time $ 0 $, and the courier does not arrive earlier than the time $ M $, and the courier does not arrive later than the time $ T $. Also, the time it takes for Taro to receive the courier can be ignored.
Input
Each dataset consists of two lines. The first line consists of three integers $ N, M, T $ separated by blanks. These integers satisfy $ 1 \ leq N \ leq 100 $, $ 1 \ leq M \ leq 10 {,} 000 $, $ 1 \ leq T \ leq 10 {,} 000 $. The second line consists of $ N $ integers $ a_1, a_2, \ dots, a_N $ separated by blanks. Each $ a_i $ fills $ M \ leq a_i \ leq T $ and is also $ a_i <a_ {i + 1} $ ($ 1 \ leq i <N $).
Output
Output an integer representing the maximum amount of time Taro can study on one line.
Sample Input 1
1 1 5
3
Output for the Sample Input 1
3
Sample Input 2
2 1 10
2 7
Output for the Sample Input 2
6
Sample Input 3
2 4 10
6 8
Output for the Sample Input 3
2
Example
Input
1 1 5
3
Output
3 | instruction | 0 | 30,009 | 8 | 60,018 |
"Correct Solution:
```
n, m, t = map(int, input().split())
a_list = list(map(int, input().split()))
ans = 0
ans += max(0, a_list[0] - m)
before = a_list[0]
for i in range(1, n):
ans += max(0, a_list[i]-before-m*2)
before = a_list[i]
ans += max(0, t-before-m)
print(ans)
``` | output | 1 | 30,009 | 8 | 60,019 |
Provide a correct Python 3 solution for this coding contest problem.
B-Mansion and courier
Problem Statement
Taro lives alone in a mansion. Taro, who loves studying, intends to study in his study in the house today. Taro can't concentrate outside the study, so he always studies in the study.
However, on this day, $ N $ of courier service to Taro arrived. $ i $ ($ 1 \ leq i \ leq N $) The arrival time of the third courier is $ a_i $. It is painful to have the delivery person wait at the front door, so Taro decided to be at the front door by the time the courier arrives. Due to the large size of the mansion, it takes $ M $ one way to move between the study and the entrance.
On the other hand, Taro wants to study for as long as possible. Find the maximum amount of time Taro can study in the study from time $ 0 $ to time $ T $.
Taro is in the study at time $ 0 $, and the courier does not arrive earlier than the time $ M $, and the courier does not arrive later than the time $ T $. Also, the time it takes for Taro to receive the courier can be ignored.
Input
Each dataset consists of two lines. The first line consists of three integers $ N, M, T $ separated by blanks. These integers satisfy $ 1 \ leq N \ leq 100 $, $ 1 \ leq M \ leq 10 {,} 000 $, $ 1 \ leq T \ leq 10 {,} 000 $. The second line consists of $ N $ integers $ a_1, a_2, \ dots, a_N $ separated by blanks. Each $ a_i $ fills $ M \ leq a_i \ leq T $ and is also $ a_i <a_ {i + 1} $ ($ 1 \ leq i <N $).
Output
Output an integer representing the maximum amount of time Taro can study on one line.
Sample Input 1
1 1 5
3
Output for the Sample Input 1
3
Sample Input 2
2 1 10
2 7
Output for the Sample Input 2
6
Sample Input 3
2 4 10
6 8
Output for the Sample Input 3
2
Example
Input
1 1 5
3
Output
3 | instruction | 0 | 30,010 | 8 | 60,020 |
"Correct Solution:
```
#21:25
n,m,t = map(int,input().split())
arrive_time = list(map(int,input().split()))
#
list_time = [0]*(t+m+1)
for i in arrive_time:
# out of bounsの場合どうしよう
list_time[i-m] += 1
list_time[i+m] -= 1
for i in range(1,t):
list_time[i] += list_time[i-1]
print(sum(x==0 for x in list_time[:t]))
``` | output | 1 | 30,010 | 8 | 60,021 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
B-Mansion and courier
Problem Statement
Taro lives alone in a mansion. Taro, who loves studying, intends to study in his study in the house today. Taro can't concentrate outside the study, so he always studies in the study.
However, on this day, $ N $ of courier service to Taro arrived. $ i $ ($ 1 \ leq i \ leq N $) The arrival time of the third courier is $ a_i $. It is painful to have the delivery person wait at the front door, so Taro decided to be at the front door by the time the courier arrives. Due to the large size of the mansion, it takes $ M $ one way to move between the study and the entrance.
On the other hand, Taro wants to study for as long as possible. Find the maximum amount of time Taro can study in the study from time $ 0 $ to time $ T $.
Taro is in the study at time $ 0 $, and the courier does not arrive earlier than the time $ M $, and the courier does not arrive later than the time $ T $. Also, the time it takes for Taro to receive the courier can be ignored.
Input
Each dataset consists of two lines. The first line consists of three integers $ N, M, T $ separated by blanks. These integers satisfy $ 1 \ leq N \ leq 100 $, $ 1 \ leq M \ leq 10 {,} 000 $, $ 1 \ leq T \ leq 10 {,} 000 $. The second line consists of $ N $ integers $ a_1, a_2, \ dots, a_N $ separated by blanks. Each $ a_i $ fills $ M \ leq a_i \ leq T $ and is also $ a_i <a_ {i + 1} $ ($ 1 \ leq i <N $).
Output
Output an integer representing the maximum amount of time Taro can study on one line.
Sample Input 1
1 1 5
3
Output for the Sample Input 1
3
Sample Input 2
2 1 10
2 7
Output for the Sample Input 2
6
Sample Input 3
2 4 10
6 8
Output for the Sample Input 3
2
Example
Input
1 1 5
3
Output
3
Submitted Solution:
```
n, m, t = map(int, input().split())
a = list(map(int, input().split()))
ts = [0 for i in range(t)]
for i in a:
ts[max(i - m, 0)] = 2 * m
temp = 0
count = 0
for i in range(1, t):
ts[i] = max(ts[i - 1] - 1, ts[i])
print(ts.count(0))
``` | instruction | 0 | 30,011 | 8 | 60,022 |
Yes | output | 1 | 30,011 | 8 | 60,023 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
B-Mansion and courier
Problem Statement
Taro lives alone in a mansion. Taro, who loves studying, intends to study in his study in the house today. Taro can't concentrate outside the study, so he always studies in the study.
However, on this day, $ N $ of courier service to Taro arrived. $ i $ ($ 1 \ leq i \ leq N $) The arrival time of the third courier is $ a_i $. It is painful to have the delivery person wait at the front door, so Taro decided to be at the front door by the time the courier arrives. Due to the large size of the mansion, it takes $ M $ one way to move between the study and the entrance.
On the other hand, Taro wants to study for as long as possible. Find the maximum amount of time Taro can study in the study from time $ 0 $ to time $ T $.
Taro is in the study at time $ 0 $, and the courier does not arrive earlier than the time $ M $, and the courier does not arrive later than the time $ T $. Also, the time it takes for Taro to receive the courier can be ignored.
Input
Each dataset consists of two lines. The first line consists of three integers $ N, M, T $ separated by blanks. These integers satisfy $ 1 \ leq N \ leq 100 $, $ 1 \ leq M \ leq 10 {,} 000 $, $ 1 \ leq T \ leq 10 {,} 000 $. The second line consists of $ N $ integers $ a_1, a_2, \ dots, a_N $ separated by blanks. Each $ a_i $ fills $ M \ leq a_i \ leq T $ and is also $ a_i <a_ {i + 1} $ ($ 1 \ leq i <N $).
Output
Output an integer representing the maximum amount of time Taro can study on one line.
Sample Input 1
1 1 5
3
Output for the Sample Input 1
3
Sample Input 2
2 1 10
2 7
Output for the Sample Input 2
6
Sample Input 3
2 4 10
6 8
Output for the Sample Input 3
2
Example
Input
1 1 5
3
Output
3
Submitted Solution:
```
N,M,T = map(int,input().split())
src = list(map(int,input().split()))
back_time = ans = 0
for t in src:
if t - M > back_time:
ans += (t - M - back_time)
back_time = t + M
ans += max(0, T - back_time)
print(ans)
``` | instruction | 0 | 30,012 | 8 | 60,024 |
Yes | output | 1 | 30,012 | 8 | 60,025 |
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