message stringlengths 2 433k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 113 108k | cluster float64 12 12 | __index_level_0__ int64 226 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Since you are the best Wraith King, Nizhniy Magazin «Mir» at the centre of Vinnytsia is offering you a discount.
You are given an array a of length n and an integer c.
The value of some array b of length k is the sum of its elements except for the <image> smallest. For example, the value of the array [3, 1, 6, 5, 2] with c = 2 is 3 + 6 + 5 = 14.
Among all possible partitions of a into contiguous subarrays output the smallest possible sum of the values of these subarrays.
Input
The first line contains integers n and c (1 ≤ n, c ≤ 100 000).
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of a.
Output
Output a single integer — the smallest possible sum of values of these subarrays of some partition of a.
Examples
Input
3 5
1 2 3
Output
6
Input
12 10
1 1 10 10 10 10 10 10 9 10 10 10
Output
92
Input
7 2
2 3 6 4 5 7 1
Output
17
Input
8 4
1 3 4 5 5 3 4 1
Output
23
Note
In the first example any partition yields 6 as the sum.
In the second example one of the optimal partitions is [1, 1], [10, 10, 10, 10, 10, 10, 9, 10, 10, 10] with the values 2 and 90 respectively.
In the third example one of the optimal partitions is [2, 3], [6, 4, 5, 7], [1] with the values 3, 13 and 1 respectively.
In the fourth example one of the optimal partitions is [1], [3, 4, 5, 5, 3, 4], [1] with the values 1, 21 and 1 respectively.
Submitted Solution:
```
def iter(ar, l, r, c, result, min_):
min_=min(ar[l:r])
if min_ == ar[l]:
sum_ = ar[l]
pos = -1
for i in range(l + 1, r):
if (r - i) < c:
break
elif (sum_ < ar[i]):
pos = i
sum_+=ar[i]
if (pos > 0 and (pos -l) >= min_):
# print(str(l)+" " + str(pos))
# print(str(pos)+" " + str(r))
iter(ar, l, pos, c, result, -1)
iter(ar, pos, r, c, result, 2)
return
if min_ == ar[r-1]:
sum_ = ar[r - 1]
pos = -1
for i in range(r - 2, l - 1, -1):
if (i - l + 1) < c:
break
# print(str(sum_)+" " +str(ar[i]))
if (sum_ < ar[i]):
pos = i + 1
sum_ += ar[i]
if (pos > 0):
# print(str(l) + " " + str(pos))
# print(str(pos) + " " + str(r))
iter(ar, l, pos, c, result, 2)
iter(ar, pos, r, c, result, -1)
return
delta = (r - l) // c
result[0] += sum(sorted(ar[l:r])[delta:])
pairs= [int(s) for s in input().split(" ")]
n= pairs[0]
c= pairs[1]
a= [int(s) for s in input().split(" ")]
#a.sort()
result=[0]
iter(a,0,n,c,result,-1)
print(result[0])
``` | instruction | 0 | 73,910 | 12 | 147,820 |
No | output | 1 | 73,910 | 12 | 147,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Since you are the best Wraith King, Nizhniy Magazin «Mir» at the centre of Vinnytsia is offering you a discount.
You are given an array a of length n and an integer c.
The value of some array b of length k is the sum of its elements except for the <image> smallest. For example, the value of the array [3, 1, 6, 5, 2] with c = 2 is 3 + 6 + 5 = 14.
Among all possible partitions of a into contiguous subarrays output the smallest possible sum of the values of these subarrays.
Input
The first line contains integers n and c (1 ≤ n, c ≤ 100 000).
The second line contains n integers ai (1 ≤ ai ≤ 109) — elements of a.
Output
Output a single integer — the smallest possible sum of values of these subarrays of some partition of a.
Examples
Input
3 5
1 2 3
Output
6
Input
12 10
1 1 10 10 10 10 10 10 9 10 10 10
Output
92
Input
7 2
2 3 6 4 5 7 1
Output
17
Input
8 4
1 3 4 5 5 3 4 1
Output
23
Note
In the first example any partition yields 6 as the sum.
In the second example one of the optimal partitions is [1, 1], [10, 10, 10, 10, 10, 10, 9, 10, 10, 10] with the values 2 and 90 respectively.
In the third example one of the optimal partitions is [2, 3], [6, 4, 5, 7], [1] with the values 3, 13 and 1 respectively.
In the fourth example one of the optimal partitions is [1], [3, 4, 5, 5, 3, 4], [1] with the values 1, 21 and 1 respectively.
Submitted Solution:
```
from functools import *
from time import time
from heapq import *
def measure_time(fun):
def wrapper(*args, **kwargs):
begin = time()
rv = fun(*args, **kwargs)
end = time()
print(end - begin)
return rv
return wrapper
def our_cache(func):
memo = [-1]*(max+1)
@wraps(func)
def wrapper(arg):
if memo[arg] == -1:
memo[arg] = func(arg)
return memo[arg]
return wrapper
def main():
n, c = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
if c >= n:
print(sum(a))
exit(0)
b = n * [0]
s = 0
h = []
for i in range(n):
s = s + a[i] if i <= c-1 else s + a[i] - a[i-c]
heappush(h, (a[i], i))
if i <= c-2:
b[i] = s
else:
while i-h[0][1] >= c:
heappop(h)
v1 = b[i-c] + s - h[0][0]
v2 = a[i] + b[i-1]
b[i] = min(v1, v2)
print(b[-1])
if __name__ == "__main__":
main()
``` | instruction | 0 | 73,911 | 12 | 147,822 |
No | output | 1 | 73,911 | 12 | 147,823 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ..., a_n. You can perform the following operation any number of times:
* Choose a pair of two neighboring equal elements a_i = a_{i + 1} (if there is at least one such pair).
* Replace them by one element with value a_i + 1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?
Input
The first line contains the single integer n (1 ≤ n ≤ 500) — the initial length of the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1000) — the initial array a.
Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.
Examples
Input
5
4 3 2 2 3
Output
2
Input
7
3 3 4 4 4 3 3
Output
2
Input
3
1 3 5
Output
3
Input
1
1000
Output
1
Note
In the first test, this is one of the optimal sequences of operations: 4 3 2 2 3 → 4 3 3 3 → 4 4 3 → 5 3.
In the second test, this is one of the optimal sequences of operations: 3 3 4 4 4 3 3 → 4 4 4 4 3 3 → 4 4 4 4 4 → 5 4 4 4 → 5 5 4 → 6 4.
In the third and fourth tests, you can't perform the operation at all. | instruction | 0 | 74,276 | 12 | 148,552 |
Tags: dp, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
n=int(input())
A=list(map(int,input().split()))
DP=[[-1]*(n+1) for i in range(n+1)]
for i in range(n):
DP[i][i]=A[i]
for mid in range(1,n):
for i in range(n):
j=i+mid
if j==n:
break
for k in range(i,j+1):
if DP[i][k]==DP[k+1][j] and DP[i][k]!=-1:
DP[i][j]=DP[i][k]+1
ANS=[2000]*(n+1)
ANS.append(0)
for i in range(n):
ANS[i]=min(ANS[i],ANS[i-1]+1)
for j in range(i,n):
if DP[i][j]!=-1:
ANS[j]=min(ANS[j],ANS[i-1]+1)
print(ANS[n-1])
``` | output | 1 | 74,276 | 12 | 148,553 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ..., a_n. You can perform the following operation any number of times:
* Choose a pair of two neighboring equal elements a_i = a_{i + 1} (if there is at least one such pair).
* Replace them by one element with value a_i + 1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?
Input
The first line contains the single integer n (1 ≤ n ≤ 500) — the initial length of the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1000) — the initial array a.
Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.
Examples
Input
5
4 3 2 2 3
Output
2
Input
7
3 3 4 4 4 3 3
Output
2
Input
3
1 3 5
Output
3
Input
1
1000
Output
1
Note
In the first test, this is one of the optimal sequences of operations: 4 3 2 2 3 → 4 3 3 3 → 4 4 3 → 5 3.
In the second test, this is one of the optimal sequences of operations: 3 3 4 4 4 3 3 → 4 4 4 4 3 3 → 4 4 4 4 4 → 5 4 4 4 → 5 5 4 → 6 4.
In the third and fourth tests, you can't perform the operation at all. | instruction | 0 | 74,277 | 12 | 148,554 |
Tags: dp, greedy
Correct Solution:
```
from sys import stdin, gettrace
if not gettrace():
def input():
return next(stdin)[:-1]
# def input():
# return stdin.buffer.readline()
INF = 10000
def main():
n = int(input())
aa = [int(a) for a in input().split()]
dp = [[0] * (n+1) for _ in range(n)]
def calc_dp(i, j):
if i + 1 == j:
dp[i][j] = aa[i]
if dp[i][j] != 0:
return dp[i][j]
dp[i][j] = -1
for k in range(i+1, j):
lf = calc_dp(i, k)
rg = calc_dp(k, j)
if lf > 0 and lf == rg:
dp[i][j] = lf + 1
break
return dp[i][j]
# for i in range(n):
# dp[i][i+1] = aa[i]
# for len in range(2, n):
# for i in range(n -len+1):
# j = i+len
# for k in range(i+1, j):
# seg = dp[i][k]
# if seg and seg == dp[k][j]:
# dp[i][j] = seg + 1
dp2 = list(range(0,n+1))
for i in range(n):
for j in range(i+1, n+1):
if calc_dp(i, j) > 0:
dp2[j] = min(dp2[j], dp2[i] + 1)
print(dp2[n])
if __name__ == "__main__":
main()
``` | output | 1 | 74,277 | 12 | 148,555 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ..., a_n. You can perform the following operation any number of times:
* Choose a pair of two neighboring equal elements a_i = a_{i + 1} (if there is at least one such pair).
* Replace them by one element with value a_i + 1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?
Input
The first line contains the single integer n (1 ≤ n ≤ 500) — the initial length of the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1000) — the initial array a.
Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.
Examples
Input
5
4 3 2 2 3
Output
2
Input
7
3 3 4 4 4 3 3
Output
2
Input
3
1 3 5
Output
3
Input
1
1000
Output
1
Note
In the first test, this is one of the optimal sequences of operations: 4 3 2 2 3 → 4 3 3 3 → 4 4 3 → 5 3.
In the second test, this is one of the optimal sequences of operations: 3 3 4 4 4 3 3 → 4 4 4 4 3 3 → 4 4 4 4 4 → 5 4 4 4 → 5 5 4 → 6 4.
In the third and fourth tests, you can't perform the operation at all. | instruction | 0 | 74,278 | 12 | 148,556 |
Tags: dp, greedy
Correct Solution:
```
import sys
from array import array # noqa: F401
from typing import List, Tuple, TypeVar, Generic, Sequence, Union # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
def main():
n = int(input())
a = list(map(int, input().split()))
dp = [array('h', [10000]) * (n + 1) for _ in range(n + 1)]
num = [array('h', [-1]) * (n + 1) for _ in range(n + 1)]
for i in range(n):
dp[i][i + 1] = 1
num[i][i + 1] = a[i]
for sublen in range(2, n + 1):
for l, r in zip(range(n), range(sublen, n + 1)):
for mid in range(l + 1, r):
if num[l][mid] == num[mid][r] != -1:
dp[l][r] = 1
num[l][r] = num[l][mid] + 1
break
dp[l][r] = min(dp[l][r], dp[l][mid] + dp[mid][r])
print(dp[0][-1])
if __name__ == '__main__':
main()
``` | output | 1 | 74,278 | 12 | 148,557 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ..., a_n. You can perform the following operation any number of times:
* Choose a pair of two neighboring equal elements a_i = a_{i + 1} (if there is at least one such pair).
* Replace them by one element with value a_i + 1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?
Input
The first line contains the single integer n (1 ≤ n ≤ 500) — the initial length of the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1000) — the initial array a.
Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.
Examples
Input
5
4 3 2 2 3
Output
2
Input
7
3 3 4 4 4 3 3
Output
2
Input
3
1 3 5
Output
3
Input
1
1000
Output
1
Note
In the first test, this is one of the optimal sequences of operations: 4 3 2 2 3 → 4 3 3 3 → 4 4 3 → 5 3.
In the second test, this is one of the optimal sequences of operations: 3 3 4 4 4 3 3 → 4 4 4 4 3 3 → 4 4 4 4 4 → 5 4 4 4 → 5 5 4 → 6 4.
In the third and fourth tests, you can't perform the operation at all. | instruction | 0 | 74,279 | 12 | 148,558 |
Tags: dp, greedy
Correct Solution:
```
n = int(input())
b = [int(_) for _ in input().split()]
e = [[-1] * (n+1) for _ in range(2002)]
d = [[] for _ in range(n)]
for i, v in enumerate(b):
e[v][i] = i
d[i].append(i)
for v in range(1, 2002):
for i in range(n):
j = e[v][i]
h = e[v][j+1] if j != -1 else -1
if j != -1 and h != -1:
e[v+1][i] = h
d[i].append(h)
a = [_ for _ in range(1, n+1)]
for s in range(n):
for e in d[s]:
a[e] = min(a[e], a[s-1]+1 if s > 0 else 1)
print(a[n-1])
``` | output | 1 | 74,279 | 12 | 148,559 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ..., a_n. You can perform the following operation any number of times:
* Choose a pair of two neighboring equal elements a_i = a_{i + 1} (if there is at least one such pair).
* Replace them by one element with value a_i + 1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?
Input
The first line contains the single integer n (1 ≤ n ≤ 500) — the initial length of the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1000) — the initial array a.
Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.
Examples
Input
5
4 3 2 2 3
Output
2
Input
7
3 3 4 4 4 3 3
Output
2
Input
3
1 3 5
Output
3
Input
1
1000
Output
1
Note
In the first test, this is one of the optimal sequences of operations: 4 3 2 2 3 → 4 3 3 3 → 4 4 3 → 5 3.
In the second test, this is one of the optimal sequences of operations: 3 3 4 4 4 3 3 → 4 4 4 4 3 3 → 4 4 4 4 4 → 5 4 4 4 → 5 5 4 → 6 4.
In the third and fourth tests, you can't perform the operation at all. | instruction | 0 | 74,280 | 12 | 148,560 |
Tags: dp, greedy
Correct Solution:
```
m = int(input())
a = list(map(int, input().split()))
dp = [[505]*m for _ in range(m)]
Max = [[0]*m for _ in range(m)]
for i in range(m):
dp[i][i] = 1
Max[i][i] = a[i]
for len in range(1, m+1):
for i in range(m-len+1):
j = i + len - 1
for k in range(i, j):
dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j])
if dp[i][k] == 1 and dp[k+1][j] == 1 and Max[i][k] == Max[k+1][j]:
dp[i][j] = 1
Max[i][j] = Max[i][k] + 1
print(dp[0][m-1])
``` | output | 1 | 74,280 | 12 | 148,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ..., a_n. You can perform the following operation any number of times:
* Choose a pair of two neighboring equal elements a_i = a_{i + 1} (if there is at least one such pair).
* Replace them by one element with value a_i + 1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?
Input
The first line contains the single integer n (1 ≤ n ≤ 500) — the initial length of the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1000) — the initial array a.
Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.
Examples
Input
5
4 3 2 2 3
Output
2
Input
7
3 3 4 4 4 3 3
Output
2
Input
3
1 3 5
Output
3
Input
1
1000
Output
1
Note
In the first test, this is one of the optimal sequences of operations: 4 3 2 2 3 → 4 3 3 3 → 4 4 3 → 5 3.
In the second test, this is one of the optimal sequences of operations: 3 3 4 4 4 3 3 → 4 4 4 4 3 3 → 4 4 4 4 4 → 5 4 4 4 → 5 5 4 → 6 4.
In the third and fourth tests, you can't perform the operation at all. | instruction | 0 | 74,281 | 12 | 148,562 |
Tags: dp, greedy
Correct Solution:
```
import sys
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def main():
inf=10**9
n=II()
aa=LI()
dp1=[[-1]*n for _ in range(n)]
dp2=[[inf]*n for _ in range(n)]
for i in range(n):
dp1[i][i]=aa[i]
dp2[i][i]=1
for w in range(2,n+1):
for l in range(n-w+1):
r=l+w-1
for m in range(l,r):
if dp1[l][m]!=-1 and dp1[l][m]==dp1[m+1][r]:
dp1[l][r]=dp1[l][m]+1
dp2[l][r]=1
for m in range(n):
for l in range(m+1):
for r in range(m+1,n):
dp2[l][r]=min(dp2[l][r],dp2[l][m]+dp2[m+1][r])
print(dp2[0][n-1])
main()
``` | output | 1 | 74,281 | 12 | 148,563 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ..., a_n. You can perform the following operation any number of times:
* Choose a pair of two neighboring equal elements a_i = a_{i + 1} (if there is at least one such pair).
* Replace them by one element with value a_i + 1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?
Input
The first line contains the single integer n (1 ≤ n ≤ 500) — the initial length of the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1000) — the initial array a.
Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.
Examples
Input
5
4 3 2 2 3
Output
2
Input
7
3 3 4 4 4 3 3
Output
2
Input
3
1 3 5
Output
3
Input
1
1000
Output
1
Note
In the first test, this is one of the optimal sequences of operations: 4 3 2 2 3 → 4 3 3 3 → 4 4 3 → 5 3.
In the second test, this is one of the optimal sequences of operations: 3 3 4 4 4 3 3 → 4 4 4 4 3 3 → 4 4 4 4 4 → 5 4 4 4 → 5 5 4 → 6 4.
In the third and fourth tests, you can't perform the operation at all. | instruction | 0 | 74,282 | 12 | 148,564 |
Tags: dp, greedy
Correct Solution:
```
import sys
readline = sys.stdin.buffer.readline
N = int(readline())
A = list(map(int, readline().split()))
dp = [[0]*N for _ in range(N)]
for j in range(N):
dp[j][0] = A[j]
for l in range(1, N):
for j in range(l, N):
for k in range(j-l, j):
if dp[k][k-j+l] == dp[j][j-k-1] > 0:
dp[j][l] = 1+dp[j][j-k-1]
break
dp = [None] + dp
Dp = [0]*(N+1)
for j in range(1, N+1):
res = N
for l in range(j):
if dp[j][l]:
res = min(res, 1+Dp[j-l-1])
Dp[j] = res
print(Dp[N])
``` | output | 1 | 74,282 | 12 | 148,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ..., a_n. You can perform the following operation any number of times:
* Choose a pair of two neighboring equal elements a_i = a_{i + 1} (if there is at least one such pair).
* Replace them by one element with value a_i + 1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?
Input
The first line contains the single integer n (1 ≤ n ≤ 500) — the initial length of the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1000) — the initial array a.
Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.
Examples
Input
5
4 3 2 2 3
Output
2
Input
7
3 3 4 4 4 3 3
Output
2
Input
3
1 3 5
Output
3
Input
1
1000
Output
1
Note
In the first test, this is one of the optimal sequences of operations: 4 3 2 2 3 → 4 3 3 3 → 4 4 3 → 5 3.
In the second test, this is one of the optimal sequences of operations: 3 3 4 4 4 3 3 → 4 4 4 4 3 3 → 4 4 4 4 4 → 5 4 4 4 → 5 5 4 → 6 4.
In the third and fourth tests, you can't perform the operation at all. | instruction | 0 | 74,283 | 12 | 148,566 |
Tags: dp, greedy
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
dp = [[505]*n for _ in range(n)]
Max = [[0]*n for _ in range(n)]
for i in range(n):
dp[i][i] = 1
Max[i][i] = a[i]
for len in range(1, n+1):
for i in range(n-len+1):
j = i + len - 1
for k in range(i, j):
dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j])
if dp[i][k] == 1 and dp[k+1][j] == 1 and Max[i][k] == Max[k+1][j]:
dp[i][j] = 1
Max[i][j] = Max[i][k] + 1
print(dp[0][n-1])
``` | output | 1 | 74,283 | 12 | 148,567 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_1, a_2, ..., a_n. You can perform the following operation any number of times:
* Choose a pair of two neighboring equal elements a_i = a_{i + 1} (if there is at least one such pair).
* Replace them by one element with value a_i + 1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?
Input
The first line contains the single integer n (1 ≤ n ≤ 500) — the initial length of the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1000) — the initial array a.
Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.
Examples
Input
5
4 3 2 2 3
Output
2
Input
7
3 3 4 4 4 3 3
Output
2
Input
3
1 3 5
Output
3
Input
1
1000
Output
1
Note
In the first test, this is one of the optimal sequences of operations: 4 3 2 2 3 → 4 3 3 3 → 4 4 3 → 5 3.
In the second test, this is one of the optimal sequences of operations: 3 3 4 4 4 3 3 → 4 4 4 4 3 3 → 4 4 4 4 4 → 5 4 4 4 → 5 5 4 → 6 4.
In the third and fourth tests, you can't perform the operation at all.
Submitted Solution:
```
N = int(input())
X = list(map(int, input().split()))
from collections import defaultdict
dp = defaultdict(lambda :-1)
for i in range(N):
dp[i+1001] = X[i]
for i in range(2, N+1):
for j in range(N-i+1):
for k in range(1, i):
u, v = dp[j+1001*k], dp[j+k+1001*(i-k)]
if u == -1 or v == -1 or u != v:
continue
dp[j+1001*i] = u+1;break
#print(dp)
dp2 = [0]*(N+1)
for i in range(N):
dp2[i+1] = dp2[i]+1
if dp[1001*(i+1)] != -1:
dp2[i+1] = 1
continue
for j in range(i+1):
if dp[j+(i+1-j)*1001] == -1:
continue
dp2[i+1] = min(dp2[i+1], dp2[j]+1)
print(dp2[-1])
``` | instruction | 0 | 74,284 | 12 | 148,568 |
Yes | output | 1 | 74,284 | 12 | 148,569 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_1, a_2, ..., a_n. You can perform the following operation any number of times:
* Choose a pair of two neighboring equal elements a_i = a_{i + 1} (if there is at least one such pair).
* Replace them by one element with value a_i + 1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?
Input
The first line contains the single integer n (1 ≤ n ≤ 500) — the initial length of the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1000) — the initial array a.
Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.
Examples
Input
5
4 3 2 2 3
Output
2
Input
7
3 3 4 4 4 3 3
Output
2
Input
3
1 3 5
Output
3
Input
1
1000
Output
1
Note
In the first test, this is one of the optimal sequences of operations: 4 3 2 2 3 → 4 3 3 3 → 4 4 3 → 5 3.
In the second test, this is one of the optimal sequences of operations: 3 3 4 4 4 3 3 → 4 4 4 4 3 3 → 4 4 4 4 4 → 5 4 4 4 → 5 5 4 → 6 4.
In the third and fourth tests, you can't perform the operation at all.
Submitted Solution:
```
import sys, math
import io, os
# data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from bisect import bisect_left as bl, bisect_right as br, insort
# from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
# from itertools import permutations,combinations
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var): sys.stdout.write(' '.join(map(str, var)) + '\n')
def out(var): sys.stdout.write(str(var) + '\n')
from decimal import Decimal
# from fractions import Fraction
# sys.setrecursionlimit(100000)
INF = float('inf')
mod = int(1e9) + 7
def cal(l,r):
if l==r:
dp1[l][r]=1
dp2[l][r]=a[l]
if dp1[l][r]:
return dp1[l][r]
for i in range(l,r):
if cal(l,i)==1 and cal(i+1,r)==1 and dp2[l][i]==dp2[i+1][r]:
dp1[l][r]=1
dp2[l][r]=dp2[l][i]+1
if not dp2[l][r]:
dp1[l][r]=2
return dp1[l][r]
def cal2(l,r):
if dp1[l][r]==1:
dp3[l][r]=1
return 1
elif dp3[l][r]:
return dp3[l][r]
ans=INF
for i in range(l,r):
ans=min(cal2(l,i)+cal2(i+1,r),ans)
dp3[l][r]=ans
return ans
n=int(data())
a=mdata()
ans=[n]
dp1=[[0]*n for i in range(n)]
dp2=[[0]*n for i in range(n)]
dp3=[[0]*n for i in range(n)]
cal(0,n-1)
cal2(0,n-1)
out(dp3[0][n-1])
``` | instruction | 0 | 74,285 | 12 | 148,570 |
Yes | output | 1 | 74,285 | 12 | 148,571 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_1, a_2, ..., a_n. You can perform the following operation any number of times:
* Choose a pair of two neighboring equal elements a_i = a_{i + 1} (if there is at least one such pair).
* Replace them by one element with value a_i + 1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?
Input
The first line contains the single integer n (1 ≤ n ≤ 500) — the initial length of the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1000) — the initial array a.
Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.
Examples
Input
5
4 3 2 2 3
Output
2
Input
7
3 3 4 4 4 3 3
Output
2
Input
3
1 3 5
Output
3
Input
1
1000
Output
1
Note
In the first test, this is one of the optimal sequences of operations: 4 3 2 2 3 → 4 3 3 3 → 4 4 3 → 5 3.
In the second test, this is one of the optimal sequences of operations: 3 3 4 4 4 3 3 → 4 4 4 4 3 3 → 4 4 4 4 4 → 5 4 4 4 → 5 5 4 → 6 4.
In the third and fourth tests, you can't perform the operation at all.
Submitted Solution:
```
# ------------------- fast io --------------------
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")
# ------------------- fast io --------------------
from math import gcd, ceil
def prod(a, mod=10**9+7):
ans = 1
for each in a:
ans = (ans * each) % mod
return ans
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
for _ in range(int(input()) if not True else 1):
n = int(input())
a = list(map(int, input().split()))
dp = [[False]*(n+2) for i in range(n+2)]
# dp[i][j] -> Can a[i-j] be reduced to a single element.
# If yes, then dp[i][j] contains value of that element. Else, false.
dp2 = [[600]*(n+2) for i in range(n+2)]
for i in range(n):
dp[i][i] = a[i]
dp2[i][i] = 1
for diff in range(1, n):
for i in range(n-diff):
# i -> i+diff
for j in range(i, i+diff):
if dp[i][j] == dp[j+1][i+diff] and dp[i][j]:
dp[i][i+diff] = dp[i][j] + 1
dp2[i][i+diff] = 1
dp2[i][i+diff] = min(dp2[i][i+diff], dp2[i][j]+dp2[j+1][i+diff])
if not dp2[i][i+diff]:
dp2[i][i+diff] = min(dp2[i+1][i+diff]+1, dp2[i][i+diff-1] + 1)
print(dp2[0][n-1])
``` | instruction | 0 | 74,286 | 12 | 148,572 |
Yes | output | 1 | 74,286 | 12 | 148,573 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_1, a_2, ..., a_n. You can perform the following operation any number of times:
* Choose a pair of two neighboring equal elements a_i = a_{i + 1} (if there is at least one such pair).
* Replace them by one element with value a_i + 1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?
Input
The first line contains the single integer n (1 ≤ n ≤ 500) — the initial length of the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1000) — the initial array a.
Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.
Examples
Input
5
4 3 2 2 3
Output
2
Input
7
3 3 4 4 4 3 3
Output
2
Input
3
1 3 5
Output
3
Input
1
1000
Output
1
Note
In the first test, this is one of the optimal sequences of operations: 4 3 2 2 3 → 4 3 3 3 → 4 4 3 → 5 3.
In the second test, this is one of the optimal sequences of operations: 3 3 4 4 4 3 3 → 4 4 4 4 3 3 → 4 4 4 4 4 → 5 4 4 4 → 5 5 4 → 6 4.
In the third and fourth tests, you can't perform the operation at all.
Submitted Solution:
```
rr = lambda: input().rstrip()
rri = lambda: int(rr())
rrm = lambda: list(map(int, rr().split()))
from functools import lru_cache;memo=lru_cache(None)
from sys import setrecursionlimit as srl;srl(10**5)
def solve(N, A):
@memo
def dp(i, j, left=0):
if i == j:
if left == 0:
return 1
if A[i] == left:
return 1
return 2
if i > j:
return 0 if left == 0 else 1
# Try to greedy run up until left
ans = 1 + dp(i+1, j, A[i])
if left >= 1:
stack = []
for k in range(i, j+1):
stack.append(A[k])
while len(stack) >= 2 and stack[-1] == stack[-2]:
stack.pop()
stack[-1] += 1
if len(stack) == 1 and left == stack[-1]:
cand = dp(k+1, j, left+1)
if cand < ans:
ans = cand
return ans
return dp(1, N-1, A[0])
#print(solve(2,[2,2]))
#print(solve(3,[3,2,2]))
#print(solve(4,[3,2,2,3]))
#print(solve(5,[4,3,2,2,3]))
#print(solve(4,[4,3,2,2]))
#import random;random.seed(0);ri=random.randint
#print(solve(500, [ri(1,5) for _ in range(500)]))
print(solve(rri(), rrm()))
``` | instruction | 0 | 74,287 | 12 | 148,574 |
Yes | output | 1 | 74,287 | 12 | 148,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_1, a_2, ..., a_n. You can perform the following operation any number of times:
* Choose a pair of two neighboring equal elements a_i = a_{i + 1} (if there is at least one such pair).
* Replace them by one element with value a_i + 1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?
Input
The first line contains the single integer n (1 ≤ n ≤ 500) — the initial length of the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1000) — the initial array a.
Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.
Examples
Input
5
4 3 2 2 3
Output
2
Input
7
3 3 4 4 4 3 3
Output
2
Input
3
1 3 5
Output
3
Input
1
1000
Output
1
Note
In the first test, this is one of the optimal sequences of operations: 4 3 2 2 3 → 4 3 3 3 → 4 4 3 → 5 3.
In the second test, this is one of the optimal sequences of operations: 3 3 4 4 4 3 3 → 4 4 4 4 3 3 → 4 4 4 4 4 → 5 4 4 4 → 5 5 4 → 6 4.
In the third and fourth tests, you can't perform the operation at all.
Submitted Solution:
```
def solve(a):
n = len(a)
b = 0
for i in range(n-1):
if(a[i] == a[i+1]):
a[i+1] += 1
a.pop(i)
b = 1
break
if(b == 0):
print(len(a))
return
else:
solve(a)
N = int(input())
a = [int(x) for x in input().split()]
solve(a)
``` | instruction | 0 | 74,288 | 12 | 148,576 |
No | output | 1 | 74,288 | 12 | 148,577 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_1, a_2, ..., a_n. You can perform the following operation any number of times:
* Choose a pair of two neighboring equal elements a_i = a_{i + 1} (if there is at least one such pair).
* Replace them by one element with value a_i + 1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?
Input
The first line contains the single integer n (1 ≤ n ≤ 500) — the initial length of the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1000) — the initial array a.
Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.
Examples
Input
5
4 3 2 2 3
Output
2
Input
7
3 3 4 4 4 3 3
Output
2
Input
3
1 3 5
Output
3
Input
1
1000
Output
1
Note
In the first test, this is one of the optimal sequences of operations: 4 3 2 2 3 → 4 3 3 3 → 4 4 3 → 5 3.
In the second test, this is one of the optimal sequences of operations: 3 3 4 4 4 3 3 → 4 4 4 4 3 3 → 4 4 4 4 4 → 5 4 4 4 → 5 5 4 → 6 4.
In the third and fourth tests, you can't perform the operation at all.
Submitted Solution:
```
import sys
input=sys.stdin.readline
n=int(input())
a=list(map(int,input().split()))
for rep in range(n*2):
MIN=1001
idx=-1
for i in range(len(a)-1):
if a[i]==a[i+1] and MIN>a[i]:
MIN=a[i]
idx=i
if idx==-1:
break
else:
b=[]
for i in range(idx):
b.append(a[i])
b.append(a[idx]+1)
for i in range(idx+2,len(a)):
b.append(a[i])
a=[b[i] for i in range(len(b))]
print(len(a))
``` | instruction | 0 | 74,289 | 12 | 148,578 |
No | output | 1 | 74,289 | 12 | 148,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_1, a_2, ..., a_n. You can perform the following operation any number of times:
* Choose a pair of two neighboring equal elements a_i = a_{i + 1} (if there is at least one such pair).
* Replace them by one element with value a_i + 1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?
Input
The first line contains the single integer n (1 ≤ n ≤ 500) — the initial length of the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1000) — the initial array a.
Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.
Examples
Input
5
4 3 2 2 3
Output
2
Input
7
3 3 4 4 4 3 3
Output
2
Input
3
1 3 5
Output
3
Input
1
1000
Output
1
Note
In the first test, this is one of the optimal sequences of operations: 4 3 2 2 3 → 4 3 3 3 → 4 4 3 → 5 3.
In the second test, this is one of the optimal sequences of operations: 3 3 4 4 4 3 3 → 4 4 4 4 3 3 → 4 4 4 4 4 → 5 4 4 4 → 5 5 4 → 6 4.
In the third and fourth tests, you can't perform the operation at all.
Submitted Solution:
```
N = int(input())
X = list(map(int, input().split()))
from collections import defaultdict
dp = defaultdict(lambda :-1)
for i in range(N):
dp[i+1001] = X[i]
for i in range(2, N+1):
for j in range(N-i+1):
for k in range(1, i):
u, v = dp[j+1001*k], dp[j+k+1001*(i-k)]
if u == -1 or v == -1 or u != v:
continue
dp[j+1001*i] = u+1;break
#print(dp)
dp2 = [0]*(N+1)
for i in range(N):
dp2[i+1] = dp[i]+1
if dp[1001*(i+1)] != -1:
dp2[i+1] = 1
continue
for j in range(i+1):
if dp[j+(i+1-j)*1001] == -1:
continue
dp2[i+1] = min(dp2[i+1], dp2[j]+1)
print(dp2[-1])
``` | instruction | 0 | 74,290 | 12 | 148,580 |
No | output | 1 | 74,290 | 12 | 148,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a_1, a_2, ..., a_n. You can perform the following operation any number of times:
* Choose a pair of two neighboring equal elements a_i = a_{i + 1} (if there is at least one such pair).
* Replace them by one element with value a_i + 1.
After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array a you can get?
Input
The first line contains the single integer n (1 ≤ n ≤ 500) — the initial length of the array a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 1000) — the initial array a.
Output
Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.
Examples
Input
5
4 3 2 2 3
Output
2
Input
7
3 3 4 4 4 3 3
Output
2
Input
3
1 3 5
Output
3
Input
1
1000
Output
1
Note
In the first test, this is one of the optimal sequences of operations: 4 3 2 2 3 → 4 3 3 3 → 4 4 3 → 5 3.
In the second test, this is one of the optimal sequences of operations: 3 3 4 4 4 3 3 → 4 4 4 4 3 3 → 4 4 4 4 4 → 5 4 4 4 → 5 5 4 → 6 4.
In the third and fourth tests, you can't perform the operation at all.
Submitted Solution:
```
def substitute(arr, i):
arr.remove(arr[i])
arr.remove(arr[i + 1])
arr.insert(i, arr[i] + 1)
return arr
def check(LIMIT, arr):
for i in range(LIMIT):
if i == 0 and arr[i] == arr[i + 1]:
LIMIT -= 1
arr = substitute(arr, i)
elif i == LIMIT - 1 and arr[i] == arr[i - 1]:
LIMIT -= 1
arr = substitute(arr,i - 1)
elif i != 0 and i != LIMIT - 1 and arr[i] == arr[i+1] and arr[i - 1] == arr[i] + 1:
LIMIT -= 1
arr = substitute(arr, i)
elif i != LIMIT - 1 and i != LIMIT - 2 and arr[i] == arr[i + 1] and arr[i + 2] == arr[i] + 1:
LIMIT -= 1
arr = substitute(arr, i)
else:
continue
print(arr)
``` | instruction | 0 | 74,291 | 12 | 148,582 |
No | output | 1 | 74,291 | 12 | 148,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n, which initially is a permutation of numbers from 1 to n. In one operation, you can choose an index i (1 ≤ i < n) such that a_i < a_{i + 1}, and remove either a_i or a_{i + 1} from the array (after the removal, the remaining parts are concatenated).
For example, if you have the array [1, 3, 2], you can choose i = 1 (since a_1 = 1 < a_2 = 3), then either remove a_1 which gives the new array [3, 2], or remove a_2 which gives the new array [1, 2].
Is it possible to make the length of this array equal to 1 with these operations?
Input
The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n, a_i are pairwise distinct) — elements of the array.
It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
Output
For each test case, output on a single line the word "YES" if it is possible to reduce the array to a single element using the aforementioned operation, or "NO" if it is impossible to do so.
Example
Input
4
3
1 2 3
4
3 1 2 4
3
2 3 1
6
2 4 6 1 3 5
Output
YES
YES
NO
YES
Note
For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):
[1, 2, 3] → [1, 2] → [1]
[3, 1, 2, 4] → [3, 1, 4] → [3, 4] → [4]
[2, 4, 6, 1, 3, 5] → [4, 6, 1, 3, 5] → [4, 1, 3, 5] → [4, 1, 5] → [4, 5] → [4] | instruction | 0 | 74,325 | 12 | 148,650 |
Tags: constructive algorithms, data structures, greedy
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
a=list((map(int,input().split())))
if a[0]<a[-1]:
print("YES")
else:
print("NO")
``` | output | 1 | 74,325 | 12 | 148,651 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n, which initially is a permutation of numbers from 1 to n. In one operation, you can choose an index i (1 ≤ i < n) such that a_i < a_{i + 1}, and remove either a_i or a_{i + 1} from the array (after the removal, the remaining parts are concatenated).
For example, if you have the array [1, 3, 2], you can choose i = 1 (since a_1 = 1 < a_2 = 3), then either remove a_1 which gives the new array [3, 2], or remove a_2 which gives the new array [1, 2].
Is it possible to make the length of this array equal to 1 with these operations?
Input
The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n, a_i are pairwise distinct) — elements of the array.
It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
Output
For each test case, output on a single line the word "YES" if it is possible to reduce the array to a single element using the aforementioned operation, or "NO" if it is impossible to do so.
Example
Input
4
3
1 2 3
4
3 1 2 4
3
2 3 1
6
2 4 6 1 3 5
Output
YES
YES
NO
YES
Note
For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):
[1, 2, 3] → [1, 2] → [1]
[3, 1, 2, 4] → [3, 1, 4] → [3, 4] → [4]
[2, 4, 6, 1, 3, 5] → [4, 6, 1, 3, 5] → [4, 1, 3, 5] → [4, 1, 5] → [4, 5] → [4] | instruction | 0 | 74,326 | 12 | 148,652 |
Tags: constructive algorithms, data structures, greedy
Correct Solution:
```
# from functools import lru_cache
from sys import stdin, stdout
import sys
from math import *
# from collections import deque
# sys.setrecursionlimit(3*int(1e5))
# input = stdin.buffer.readline
# print = stdout.write
# @lru_cache()
for __ in range(int(input())):
n=int(input())
ar=list(map(int,input().split()))
if(ar[-1]==1 or ar[0]==n):
print("NO")
elif(ar[-1]==n or ar[0]==1):
print("YES")
else:
if(ar[0]<ar[-1]):
print("YES")
else:
print("NO")
``` | output | 1 | 74,326 | 12 | 148,653 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n, which initially is a permutation of numbers from 1 to n. In one operation, you can choose an index i (1 ≤ i < n) such that a_i < a_{i + 1}, and remove either a_i or a_{i + 1} from the array (after the removal, the remaining parts are concatenated).
For example, if you have the array [1, 3, 2], you can choose i = 1 (since a_1 = 1 < a_2 = 3), then either remove a_1 which gives the new array [3, 2], or remove a_2 which gives the new array [1, 2].
Is it possible to make the length of this array equal to 1 with these operations?
Input
The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n, a_i are pairwise distinct) — elements of the array.
It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
Output
For each test case, output on a single line the word "YES" if it is possible to reduce the array to a single element using the aforementioned operation, or "NO" if it is impossible to do so.
Example
Input
4
3
1 2 3
4
3 1 2 4
3
2 3 1
6
2 4 6 1 3 5
Output
YES
YES
NO
YES
Note
For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):
[1, 2, 3] → [1, 2] → [1]
[3, 1, 2, 4] → [3, 1, 4] → [3, 4] → [4]
[2, 4, 6, 1, 3, 5] → [4, 6, 1, 3, 5] → [4, 1, 3, 5] → [4, 1, 5] → [4, 5] → [4] | instruction | 0 | 74,327 | 12 | 148,654 |
Tags: constructive algorithms, data structures, greedy
Correct Solution:
```
# t=int(input)
for _ in range(int(input())):
n=int(input())
arr=[int(c) for c in input().split()]
if arr[-1] > arr[0]:
print("YES")
else:
print("NO")
``` | output | 1 | 74,327 | 12 | 148,655 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n, which initially is a permutation of numbers from 1 to n. In one operation, you can choose an index i (1 ≤ i < n) such that a_i < a_{i + 1}, and remove either a_i or a_{i + 1} from the array (after the removal, the remaining parts are concatenated).
For example, if you have the array [1, 3, 2], you can choose i = 1 (since a_1 = 1 < a_2 = 3), then either remove a_1 which gives the new array [3, 2], or remove a_2 which gives the new array [1, 2].
Is it possible to make the length of this array equal to 1 with these operations?
Input
The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n, a_i are pairwise distinct) — elements of the array.
It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
Output
For each test case, output on a single line the word "YES" if it is possible to reduce the array to a single element using the aforementioned operation, or "NO" if it is impossible to do so.
Example
Input
4
3
1 2 3
4
3 1 2 4
3
2 3 1
6
2 4 6 1 3 5
Output
YES
YES
NO
YES
Note
For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):
[1, 2, 3] → [1, 2] → [1]
[3, 1, 2, 4] → [3, 1, 4] → [3, 4] → [4]
[2, 4, 6, 1, 3, 5] → [4, 6, 1, 3, 5] → [4, 1, 3, 5] → [4, 1, 5] → [4, 5] → [4] | instruction | 0 | 74,328 | 12 | 148,656 |
Tags: constructive algorithms, data structures, greedy
Correct Solution:
```
I=lambda :list(map(int,input().split(' ')))
def main(lst):
stack=[]
for i in range(len(lst)):
if len(stack)==0 :
stack.append(lst[i])
elif len(stack)>0 and stack[-1]<lst[i]:
while len(stack)>0 and stack[-1]<lst[i]:
m=stack.pop()
stack.append(m)
else:
stack.append(lst[i])
if len(stack)<=1:
print('YES')
else:
print('NO')
for _ in range(int(input())):
n=int(input())
lst=I()
main(lst)
``` | output | 1 | 74,328 | 12 | 148,657 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n, which initially is a permutation of numbers from 1 to n. In one operation, you can choose an index i (1 ≤ i < n) such that a_i < a_{i + 1}, and remove either a_i or a_{i + 1} from the array (after the removal, the remaining parts are concatenated).
For example, if you have the array [1, 3, 2], you can choose i = 1 (since a_1 = 1 < a_2 = 3), then either remove a_1 which gives the new array [3, 2], or remove a_2 which gives the new array [1, 2].
Is it possible to make the length of this array equal to 1 with these operations?
Input
The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n, a_i are pairwise distinct) — elements of the array.
It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
Output
For each test case, output on a single line the word "YES" if it is possible to reduce the array to a single element using the aforementioned operation, or "NO" if it is impossible to do so.
Example
Input
4
3
1 2 3
4
3 1 2 4
3
2 3 1
6
2 4 6 1 3 5
Output
YES
YES
NO
YES
Note
For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):
[1, 2, 3] → [1, 2] → [1]
[3, 1, 2, 4] → [3, 1, 4] → [3, 4] → [4]
[2, 4, 6, 1, 3, 5] → [4, 6, 1, 3, 5] → [4, 1, 3, 5] → [4, 1, 5] → [4, 5] → [4] | instruction | 0 | 74,329 | 12 | 148,658 |
Tags: constructive algorithms, data structures, greedy
Correct Solution:
```
import sys
from collections import defaultdict
# input=sys.stdin.readline
for i in range(int(input())):
n=int(input())
lis=list(map(int,input().split()))
if lis[0]>lis[-1]:
print("NO")
else:
print("YES")
``` | output | 1 | 74,329 | 12 | 148,659 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n, which initially is a permutation of numbers from 1 to n. In one operation, you can choose an index i (1 ≤ i < n) such that a_i < a_{i + 1}, and remove either a_i or a_{i + 1} from the array (after the removal, the remaining parts are concatenated).
For example, if you have the array [1, 3, 2], you can choose i = 1 (since a_1 = 1 < a_2 = 3), then either remove a_1 which gives the new array [3, 2], or remove a_2 which gives the new array [1, 2].
Is it possible to make the length of this array equal to 1 with these operations?
Input
The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n, a_i are pairwise distinct) — elements of the array.
It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
Output
For each test case, output on a single line the word "YES" if it is possible to reduce the array to a single element using the aforementioned operation, or "NO" if it is impossible to do so.
Example
Input
4
3
1 2 3
4
3 1 2 4
3
2 3 1
6
2 4 6 1 3 5
Output
YES
YES
NO
YES
Note
For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):
[1, 2, 3] → [1, 2] → [1]
[3, 1, 2, 4] → [3, 1, 4] → [3, 4] → [4]
[2, 4, 6, 1, 3, 5] → [4, 6, 1, 3, 5] → [4, 1, 3, 5] → [4, 1, 5] → [4, 5] → [4] | instruction | 0 | 74,330 | 12 | 148,660 |
Tags: constructive algorithms, data structures, greedy
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
l = input().split()
for i in range(n):
l[i] = int(l[i])
if l[0] == n or l[n-1] == 1:
print('NO')
else:
if l[0] > l[n-1]:
print('No')
else:
print('Yes')
``` | output | 1 | 74,330 | 12 | 148,661 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n, which initially is a permutation of numbers from 1 to n. In one operation, you can choose an index i (1 ≤ i < n) such that a_i < a_{i + 1}, and remove either a_i or a_{i + 1} from the array (after the removal, the remaining parts are concatenated).
For example, if you have the array [1, 3, 2], you can choose i = 1 (since a_1 = 1 < a_2 = 3), then either remove a_1 which gives the new array [3, 2], or remove a_2 which gives the new array [1, 2].
Is it possible to make the length of this array equal to 1 with these operations?
Input
The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n, a_i are pairwise distinct) — elements of the array.
It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
Output
For each test case, output on a single line the word "YES" if it is possible to reduce the array to a single element using the aforementioned operation, or "NO" if it is impossible to do so.
Example
Input
4
3
1 2 3
4
3 1 2 4
3
2 3 1
6
2 4 6 1 3 5
Output
YES
YES
NO
YES
Note
For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):
[1, 2, 3] → [1, 2] → [1]
[3, 1, 2, 4] → [3, 1, 4] → [3, 4] → [4]
[2, 4, 6, 1, 3, 5] → [4, 6, 1, 3, 5] → [4, 1, 3, 5] → [4, 1, 5] → [4, 5] → [4] | instruction | 0 | 74,331 | 12 | 148,662 |
Tags: constructive algorithms, data structures, greedy
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
if a[n-1]>a[0]:print('YES')
else:print('NO')
``` | output | 1 | 74,331 | 12 | 148,663 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n, which initially is a permutation of numbers from 1 to n. In one operation, you can choose an index i (1 ≤ i < n) such that a_i < a_{i + 1}, and remove either a_i or a_{i + 1} from the array (after the removal, the remaining parts are concatenated).
For example, if you have the array [1, 3, 2], you can choose i = 1 (since a_1 = 1 < a_2 = 3), then either remove a_1 which gives the new array [3, 2], or remove a_2 which gives the new array [1, 2].
Is it possible to make the length of this array equal to 1 with these operations?
Input
The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n, a_i are pairwise distinct) — elements of the array.
It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
Output
For each test case, output on a single line the word "YES" if it is possible to reduce the array to a single element using the aforementioned operation, or "NO" if it is impossible to do so.
Example
Input
4
3
1 2 3
4
3 1 2 4
3
2 3 1
6
2 4 6 1 3 5
Output
YES
YES
NO
YES
Note
For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):
[1, 2, 3] → [1, 2] → [1]
[3, 1, 2, 4] → [3, 1, 4] → [3, 4] → [4]
[2, 4, 6, 1, 3, 5] → [4, 6, 1, 3, 5] → [4, 1, 3, 5] → [4, 1, 5] → [4, 5] → [4] | instruction | 0 | 74,332 | 12 | 148,664 |
Tags: constructive algorithms, data structures, greedy
Correct Solution:
```
"""
____ _ _____
/ ___|___ __| | ___| ___|__ _ __ ___ ___ ___
| | / _ \ / _` |/ _ \ |_ / _ \| '__/ __/ _ \/ __|
| |__| (_) | (_| | __/ _| (_) | | | (_| __/\__ \
\____\___/ \__,_|\___|_| \___/|_| \___\___||___/
"""
"""
░░██▄░░░░░░░░░░░▄██
░▄▀░█▄░░░░░░░░▄█░░█░
░█░▄░█▄░░░░░░▄█░▄░█░
░█░██████████████▄█░
░█████▀▀████▀▀█████░
▄█▀█▀░░░████░░░▀▀███
██░░▀████▀▀████▀░░██
██░░░░█▀░░░░▀█░░░░██
███▄░░░░░░░░░░░░▄███
░▀███▄░░████░░▄███▀░
░░░▀██▄░▀██▀░▄██▀░░░
░░░░░░▀██████▀░░░░░░
░░░░░░░░░░░░░░░░░░░░
"""
import sys
import math
import collections
import operator as op
from collections import deque
from math import gcd, inf, sqrt, pi, cos, sin
from bisect import bisect_right, bisect_left
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
from functools import reduce
from sys import stdin, stdout, setrecursionlimit
setrecursionlimit(2**20)
def factorial(n):
if n == 0:
return 1
return n * factorial(n - 1)
def ncr(n, r):
r = min(r, n - r)
numer = reduce(op.mul, range(n, n - r, -1), 1)
denom = reduce(op.mul, range(1, r + 1), 1)
return numer // denom # or / in Python 2
def prime_factors(n):
i = 2
factors = []
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return len(set(factors))
def isPowerOfTwo(x):
return (x and (not(x & (x - 1))))
def factors(n):
return list(set(reduce(list.__add__, ([i, n // i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
MOD = 1000000007
T = 1
T = int(stdin.readline())
for _ in range(T):
# n, m = list(map(int, stdin.readline().split()))
# s1 = list(stdin.readline().strip('\n'))
n = int(stdin.readline())
a = list(map(int, stdin.readline().rstrip().split()))
# s = str(stdin.readline().strip('\n'))
# a = list(stdin.readline().strip('\n'))
#n = int(stdin.readline())
if(a[0] == 1 or a[-1] == n):
print('YES')
else:
if(a[0] < a[n - 1]):
print('YES')
else:
print('NO')
``` | output | 1 | 74,332 | 12 | 148,665 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n, which initially is a permutation of numbers from 1 to n. In one operation, you can choose an index i (1 ≤ i < n) such that a_i < a_{i + 1}, and remove either a_i or a_{i + 1} from the array (after the removal, the remaining parts are concatenated).
For example, if you have the array [1, 3, 2], you can choose i = 1 (since a_1 = 1 < a_2 = 3), then either remove a_1 which gives the new array [3, 2], or remove a_2 which gives the new array [1, 2].
Is it possible to make the length of this array equal to 1 with these operations?
Input
The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n, a_i are pairwise distinct) — elements of the array.
It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
Output
For each test case, output on a single line the word "YES" if it is possible to reduce the array to a single element using the aforementioned operation, or "NO" if it is impossible to do so.
Example
Input
4
3
1 2 3
4
3 1 2 4
3
2 3 1
6
2 4 6 1 3 5
Output
YES
YES
NO
YES
Note
For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):
[1, 2, 3] → [1, 2] → [1]
[3, 1, 2, 4] → [3, 1, 4] → [3, 4] → [4]
[2, 4, 6, 1, 3, 5] → [4, 6, 1, 3, 5] → [4, 1, 3, 5] → [4, 1, 5] → [4, 5] → [4]
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
arr = list(map(int,input().split()))
if arr[0]<arr[n-1]:
print("YES")
else:
print("NO")
``` | instruction | 0 | 74,333 | 12 | 148,666 |
Yes | output | 1 | 74,333 | 12 | 148,667 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n, which initially is a permutation of numbers from 1 to n. In one operation, you can choose an index i (1 ≤ i < n) such that a_i < a_{i + 1}, and remove either a_i or a_{i + 1} from the array (after the removal, the remaining parts are concatenated).
For example, if you have the array [1, 3, 2], you can choose i = 1 (since a_1 = 1 < a_2 = 3), then either remove a_1 which gives the new array [3, 2], or remove a_2 which gives the new array [1, 2].
Is it possible to make the length of this array equal to 1 with these operations?
Input
The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n, a_i are pairwise distinct) — elements of the array.
It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
Output
For each test case, output on a single line the word "YES" if it is possible to reduce the array to a single element using the aforementioned operation, or "NO" if it is impossible to do so.
Example
Input
4
3
1 2 3
4
3 1 2 4
3
2 3 1
6
2 4 6 1 3 5
Output
YES
YES
NO
YES
Note
For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):
[1, 2, 3] → [1, 2] → [1]
[3, 1, 2, 4] → [3, 1, 4] → [3, 4] → [4]
[2, 4, 6, 1, 3, 5] → [4, 6, 1, 3, 5] → [4, 1, 3, 5] → [4, 1, 5] → [4, 5] → [4]
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
print("YES" if a[0]<a[-1] else "NO")
``` | instruction | 0 | 74,334 | 12 | 148,668 |
Yes | output | 1 | 74,334 | 12 | 148,669 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n, which initially is a permutation of numbers from 1 to n. In one operation, you can choose an index i (1 ≤ i < n) such that a_i < a_{i + 1}, and remove either a_i or a_{i + 1} from the array (after the removal, the remaining parts are concatenated).
For example, if you have the array [1, 3, 2], you can choose i = 1 (since a_1 = 1 < a_2 = 3), then either remove a_1 which gives the new array [3, 2], or remove a_2 which gives the new array [1, 2].
Is it possible to make the length of this array equal to 1 with these operations?
Input
The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n, a_i are pairwise distinct) — elements of the array.
It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
Output
For each test case, output on a single line the word "YES" if it is possible to reduce the array to a single element using the aforementioned operation, or "NO" if it is impossible to do so.
Example
Input
4
3
1 2 3
4
3 1 2 4
3
2 3 1
6
2 4 6 1 3 5
Output
YES
YES
NO
YES
Note
For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):
[1, 2, 3] → [1, 2] → [1]
[3, 1, 2, 4] → [3, 1, 4] → [3, 4] → [4]
[2, 4, 6, 1, 3, 5] → [4, 6, 1, 3, 5] → [4, 1, 3, 5] → [4, 1, 5] → [4, 5] → [4]
Submitted Solution:
```
import sys
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))
n,=I()
for _ in range(n):
m,=I()
arr=I()
if arr[0]<arr[-1]:
print('YES')
else:
print('NO')
``` | instruction | 0 | 74,335 | 12 | 148,670 |
Yes | output | 1 | 74,335 | 12 | 148,671 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n, which initially is a permutation of numbers from 1 to n. In one operation, you can choose an index i (1 ≤ i < n) such that a_i < a_{i + 1}, and remove either a_i or a_{i + 1} from the array (after the removal, the remaining parts are concatenated).
For example, if you have the array [1, 3, 2], you can choose i = 1 (since a_1 = 1 < a_2 = 3), then either remove a_1 which gives the new array [3, 2], or remove a_2 which gives the new array [1, 2].
Is it possible to make the length of this array equal to 1 with these operations?
Input
The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n, a_i are pairwise distinct) — elements of the array.
It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
Output
For each test case, output on a single line the word "YES" if it is possible to reduce the array to a single element using the aforementioned operation, or "NO" if it is impossible to do so.
Example
Input
4
3
1 2 3
4
3 1 2 4
3
2 3 1
6
2 4 6 1 3 5
Output
YES
YES
NO
YES
Note
For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):
[1, 2, 3] → [1, 2] → [1]
[3, 1, 2, 4] → [3, 1, 4] → [3, 4] → [4]
[2, 4, 6, 1, 3, 5] → [4, 6, 1, 3, 5] → [4, 1, 3, 5] → [4, 1, 5] → [4, 5] → [4]
Submitted Solution:
```
import sys
# from collections import deque
# from collections import Counter
# from math import sqrt
# from math import log
# from math import ceil
# from bisect import bisect_left, bisect_right
# alpha=['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
# mod=10**9+7
# mod=998244353
# def BinarySearch(a,x):
# i=bisect_left(a,x)
# if(i!=len(a) and a[i]==x):
# return i
# else:
# return -1
# def sieve(n):
# prime=[True for i in range(n+1)]
# p=2
# while(p*p<=n):
# if (prime[p]==True):
# for i in range(p*p,n+1,p):
# prime[i]=False
# p+=1
# prime[0]=False
# prime[1]=False
# s=set()
# for i in range(len(prime)):
# if(prime[i]):
# s.add(i)
# return s
# def gcd(a, b):
# if(a==0):
# return b
# return gcd(b%a,a)
fast_reader=sys.stdin.readline
fast_writer=sys.stdout.write
def input():
return fast_reader().strip()
def print(*argv):
fast_writer(' '.join((str(i)) for i in argv))
fast_writer('\n')
#____________________________________________________________________________________________________________________________________
for _ in range(int(input())):
n=int(input())
l=list(map(int, input().split()))
if(l[0]>l[-1]):
print('NO')
else:
print('YES')
``` | instruction | 0 | 74,336 | 12 | 148,672 |
Yes | output | 1 | 74,336 | 12 | 148,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n, which initially is a permutation of numbers from 1 to n. In one operation, you can choose an index i (1 ≤ i < n) such that a_i < a_{i + 1}, and remove either a_i or a_{i + 1} from the array (after the removal, the remaining parts are concatenated).
For example, if you have the array [1, 3, 2], you can choose i = 1 (since a_1 = 1 < a_2 = 3), then either remove a_1 which gives the new array [3, 2], or remove a_2 which gives the new array [1, 2].
Is it possible to make the length of this array equal to 1 with these operations?
Input
The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n, a_i are pairwise distinct) — elements of the array.
It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
Output
For each test case, output on a single line the word "YES" if it is possible to reduce the array to a single element using the aforementioned operation, or "NO" if it is impossible to do so.
Example
Input
4
3
1 2 3
4
3 1 2 4
3
2 3 1
6
2 4 6 1 3 5
Output
YES
YES
NO
YES
Note
For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):
[1, 2, 3] → [1, 2] → [1]
[3, 1, 2, 4] → [3, 1, 4] → [3, 4] → [4]
[2, 4, 6, 1, 3, 5] → [4, 6, 1, 3, 5] → [4, 1, 3, 5] → [4, 1, 5] → [4, 5] → [4]
Submitted Solution:
```
from sys import stdin,stdout
def INPUT():return list(int(i) for i in stdin.readline().split())
import math
def inp():return stdin.readline()
def out(x):return stdout.write(x)
import math as M
MOD=10**9+7
from random import randint as R
import sys
#####################################
for _ in range(int(input())):
n=int(input())
A=INPUT()
#max for no of elements at even positions
i=A.index(max(A))
if i==n-1:
print("YES")
elif i==0:
print("NO")
else:
if A[0]>max(A[i+1:]):
print("NO")
else:
print("YES")
#print()
``` | instruction | 0 | 74,337 | 12 | 148,674 |
No | output | 1 | 74,337 | 12 | 148,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n, which initially is a permutation of numbers from 1 to n. In one operation, you can choose an index i (1 ≤ i < n) such that a_i < a_{i + 1}, and remove either a_i or a_{i + 1} from the array (after the removal, the remaining parts are concatenated).
For example, if you have the array [1, 3, 2], you can choose i = 1 (since a_1 = 1 < a_2 = 3), then either remove a_1 which gives the new array [3, 2], or remove a_2 which gives the new array [1, 2].
Is it possible to make the length of this array equal to 1 with these operations?
Input
The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n, a_i are pairwise distinct) — elements of the array.
It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
Output
For each test case, output on a single line the word "YES" if it is possible to reduce the array to a single element using the aforementioned operation, or "NO" if it is impossible to do so.
Example
Input
4
3
1 2 3
4
3 1 2 4
3
2 3 1
6
2 4 6 1 3 5
Output
YES
YES
NO
YES
Note
For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):
[1, 2, 3] → [1, 2] → [1]
[3, 1, 2, 4] → [3, 1, 4] → [3, 4] → [4]
[2, 4, 6, 1, 3, 5] → [4, 6, 1, 3, 5] → [4, 1, 3, 5] → [4, 1, 5] → [4, 5] → [4]
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
ar = list(map(int,input().strip().split()))[:n]
if ar == sorted(ar):
print('YES')
else:
x = []
i = 0
while i < (n):
t = i
while t < (n-1) and ar[t] < ar[t+1]:
t += 1
x.append(ar[i : (t+1)])
i = (t+1)
#print(x)
h = 0
flag = 0
while h < (len(x)-1):
if max(x[h+1]) <= min(x[h]):
flag = 1
print('NO')
break
h += 1
if flag == 0:
print('YES')
``` | instruction | 0 | 74,338 | 12 | 148,676 |
No | output | 1 | 74,338 | 12 | 148,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n, which initially is a permutation of numbers from 1 to n. In one operation, you can choose an index i (1 ≤ i < n) such that a_i < a_{i + 1}, and remove either a_i or a_{i + 1} from the array (after the removal, the remaining parts are concatenated).
For example, if you have the array [1, 3, 2], you can choose i = 1 (since a_1 = 1 < a_2 = 3), then either remove a_1 which gives the new array [3, 2], or remove a_2 which gives the new array [1, 2].
Is it possible to make the length of this array equal to 1 with these operations?
Input
The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n, a_i are pairwise distinct) — elements of the array.
It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
Output
For each test case, output on a single line the word "YES" if it is possible to reduce the array to a single element using the aforementioned operation, or "NO" if it is impossible to do so.
Example
Input
4
3
1 2 3
4
3 1 2 4
3
2 3 1
6
2 4 6 1 3 5
Output
YES
YES
NO
YES
Note
For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):
[1, 2, 3] → [1, 2] → [1]
[3, 1, 2, 4] → [3, 1, 4] → [3, 4] → [4]
[2, 4, 6, 1, 3, 5] → [4, 6, 1, 3, 5] → [4, 1, 3, 5] → [4, 1, 5] → [4, 5] → [4]
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
if (a[0]==1):
print('YES')
continue
else:
f=0
d = a.index(1)
d2 = a.index(2)
if d2==0:
for i in range(d):
if a[i]!=(i*2)+2:
f=1
break
x=3
for i in range(d+1,n):
if a[i]!=x:
f=1
break
else:
x+=2
elif d2==d+1:
x=2
for i in range(d+1,n):
if a[i]!=x:
f=1
break
else:
x+=2
for i in range(d):
if a[i]!=((i+1)*2)+1:
f=1
break
else:
f=1
if f==1:
print('NO')
else:
print('YES')
``` | instruction | 0 | 74,339 | 12 | 148,678 |
No | output | 1 | 74,339 | 12 | 148,679 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n, which initially is a permutation of numbers from 1 to n. In one operation, you can choose an index i (1 ≤ i < n) such that a_i < a_{i + 1}, and remove either a_i or a_{i + 1} from the array (after the removal, the remaining parts are concatenated).
For example, if you have the array [1, 3, 2], you can choose i = 1 (since a_1 = 1 < a_2 = 3), then either remove a_1 which gives the new array [3, 2], or remove a_2 which gives the new array [1, 2].
Is it possible to make the length of this array equal to 1 with these operations?
Input
The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
The first line of each test case contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n, a_i are pairwise distinct) — elements of the array.
It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
Output
For each test case, output on a single line the word "YES" if it is possible to reduce the array to a single element using the aforementioned operation, or "NO" if it is impossible to do so.
Example
Input
4
3
1 2 3
4
3 1 2 4
3
2 3 1
6
2 4 6 1 3 5
Output
YES
YES
NO
YES
Note
For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):
[1, 2, 3] → [1, 2] → [1]
[3, 1, 2, 4] → [3, 1, 4] → [3, 4] → [4]
[2, 4, 6, 1, 3, 5] → [4, 6, 1, 3, 5] → [4, 1, 3, 5] → [4, 1, 5] → [4, 5] → [4]
Submitted Solution:
```
#!/usr/bin/env python3
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
a = [int(item) for item in input().split()]
one_id = a.index(1)
n_id = a.index(n)
if one_id < n_id:
print("YES")
continue
left = min(a[:n_id+1])
right = max(a[one_id:])
mid = a[n_id+1:one_id]
comp = [left] + mid + [right]
ok = True
for c1, c2 in zip(comp, comp[1:]):
if c1 > c2:
ok = False
break
if ok:
print("YES")
else:
print("NO")
``` | instruction | 0 | 74,340 | 12 | 148,680 |
No | output | 1 | 74,340 | 12 | 148,681 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A matrix of size n × m is called nice, if all rows and columns of the matrix are palindromes. A sequence of integers (a_1, a_2, ... , a_k) is a palindrome, if for any integer i (1 ≤ i ≤ k) the equality a_i = a_{k - i + 1} holds.
Sasha owns a matrix a of size n × m. In one operation he can increase or decrease any number in the matrix by one. Sasha wants to make the matrix nice. He is interested what is the minimum number of operations he needs.
Help him!
Input
The first line contains a single integer t — the number of test cases (1 ≤ t ≤ 10). The t tests follow.
The first line of each test contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the matrix.
Each of the next n lines contains m integers a_{i, j} (0 ≤ a_{i, j} ≤ 10^9) — the elements of the matrix.
Output
For each test output the smallest number of operations required to make the matrix nice.
Example
Input
2
4 2
4 2
2 4
4 2
2 4
3 4
1 2 3 4
5 6 7 8
9 10 11 18
Output
8
42
Note
In the first test case we can, for example, obtain the following nice matrix in 8 operations:
2 2
4 4
4 4
2 2
In the second test case we can, for example, obtain the following nice matrix in 42 operations:
5 6 6 5
6 6 6 6
5 6 6 5
| instruction | 0 | 74,353 | 12 | 148,706 |
Tags: greedy, implementation, math
Correct Solution:
```
for _ in range(int(input())):
n,m = map(int,input().split())
arr = [list(map(int,input().split())) for i in range(n)]
ans = 0
for i in range(n):
for j in range(m):
t = [arr[i][j],arr[i][m-1-j],arr[n-1-i][j]]
t.sort()
# print(t)
med = t[1]
ans += abs(arr[i][j]-med) + abs(arr[i][m-1-j]-med) + abs(arr[n-1-i][j]-med)
arr[i][j] = med
arr[i][m-1-j] = med
arr[n-1-i][j] = med
print(ans)
``` | output | 1 | 74,353 | 12 | 148,707 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A matrix of size n × m is called nice, if all rows and columns of the matrix are palindromes. A sequence of integers (a_1, a_2, ... , a_k) is a palindrome, if for any integer i (1 ≤ i ≤ k) the equality a_i = a_{k - i + 1} holds.
Sasha owns a matrix a of size n × m. In one operation he can increase or decrease any number in the matrix by one. Sasha wants to make the matrix nice. He is interested what is the minimum number of operations he needs.
Help him!
Input
The first line contains a single integer t — the number of test cases (1 ≤ t ≤ 10). The t tests follow.
The first line of each test contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the matrix.
Each of the next n lines contains m integers a_{i, j} (0 ≤ a_{i, j} ≤ 10^9) — the elements of the matrix.
Output
For each test output the smallest number of operations required to make the matrix nice.
Example
Input
2
4 2
4 2
2 4
4 2
2 4
3 4
1 2 3 4
5 6 7 8
9 10 11 18
Output
8
42
Note
In the first test case we can, for example, obtain the following nice matrix in 8 operations:
2 2
4 4
4 4
2 2
In the second test case we can, for example, obtain the following nice matrix in 42 operations:
5 6 6 5
6 6 6 6
5 6 6 5
| instruction | 0 | 74,354 | 12 | 148,708 |
Tags: greedy, implementation, math
Correct Solution:
```
for _ in range(int(input())):
n,m = map(int,input().split())
arr = [list(map(int,input().split())) for i in range(n)]
# rows = []
# cols = []
# for i in range(m):
# t =[]
# for j in range(n):
# t.append(arr[j][i])
# t.sort()
# cols.append(t[n//2])
# for i in range(n):
# t = arr[i][:]
# t.sort()
# rows.append()
ans = 0
# print(arr[-2])
mid = n//2
for c in range(n//2+n%2):
i = 0
j = m-1
while i<=j:
t = []
if i!=j:
if c!=mid or c==mid and n%2==0:
t += [arr[c][i],arr[c][j],arr[-c-1][i],arr[-c-1][j]]
else:
t += [arr[c][i],arr[c][j]]
else:
if c!=mid or c==mid and n%2==0:
t += [arr[c][i],arr[-c-1][i]]
else:
t += [arr[c][i]]
med = 0
t.sort()
if len(t)==1:
med = t[0]
else:
x = len(t)//2
med = int((t[x]+t[x-1])/2)
for k in t:
ans += abs(med-k)
i+=1
j-=1
print(ans)
``` | output | 1 | 74,354 | 12 | 148,709 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A matrix of size n × m is called nice, if all rows and columns of the matrix are palindromes. A sequence of integers (a_1, a_2, ... , a_k) is a palindrome, if for any integer i (1 ≤ i ≤ k) the equality a_i = a_{k - i + 1} holds.
Sasha owns a matrix a of size n × m. In one operation he can increase or decrease any number in the matrix by one. Sasha wants to make the matrix nice. He is interested what is the minimum number of operations he needs.
Help him!
Input
The first line contains a single integer t — the number of test cases (1 ≤ t ≤ 10). The t tests follow.
The first line of each test contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the matrix.
Each of the next n lines contains m integers a_{i, j} (0 ≤ a_{i, j} ≤ 10^9) — the elements of the matrix.
Output
For each test output the smallest number of operations required to make the matrix nice.
Example
Input
2
4 2
4 2
2 4
4 2
2 4
3 4
1 2 3 4
5 6 7 8
9 10 11 18
Output
8
42
Note
In the first test case we can, for example, obtain the following nice matrix in 8 operations:
2 2
4 4
4 4
2 2
In the second test case we can, for example, obtain the following nice matrix in 42 operations:
5 6 6 5
6 6 6 6
5 6 6 5
| instruction | 0 | 74,355 | 12 | 148,710 |
Tags: greedy, implementation, math
Correct Solution:
```
from sys import stdin
for _ in range(int(stdin.readline())):
x, y = map(int, stdin.readline().split())
ans = 0
array = []
for i in range(x):
array.append(list(map(int, stdin.readline().split())))
for i in range((x + 1) // 2):
for j in range((y + 1) // 2):
a = array[i][j]
b = array[i][y - j - 1]
c = array[x - i - 1][j]
d = array[x - i - 1][y - j - 1]
sorts = sorted([a, b, c, d])
value = (sorts[1] + sorts[2]) // 2
for aeg in [a, b, c, d]:
ans += abs(aeg - value)
if i == x - i - 1:
ans -= abs(value - c)
ans -= abs(value - d)
if j == y - j - 1:
ans -= abs(value - b)
ans -= abs(value - d)
print(ans)
``` | output | 1 | 74,355 | 12 | 148,711 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A matrix of size n × m is called nice, if all rows and columns of the matrix are palindromes. A sequence of integers (a_1, a_2, ... , a_k) is a palindrome, if for any integer i (1 ≤ i ≤ k) the equality a_i = a_{k - i + 1} holds.
Sasha owns a matrix a of size n × m. In one operation he can increase or decrease any number in the matrix by one. Sasha wants to make the matrix nice. He is interested what is the minimum number of operations he needs.
Help him!
Input
The first line contains a single integer t — the number of test cases (1 ≤ t ≤ 10). The t tests follow.
The first line of each test contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the matrix.
Each of the next n lines contains m integers a_{i, j} (0 ≤ a_{i, j} ≤ 10^9) — the elements of the matrix.
Output
For each test output the smallest number of operations required to make the matrix nice.
Example
Input
2
4 2
4 2
2 4
4 2
2 4
3 4
1 2 3 4
5 6 7 8
9 10 11 18
Output
8
42
Note
In the first test case we can, for example, obtain the following nice matrix in 8 operations:
2 2
4 4
4 4
2 2
In the second test case we can, for example, obtain the following nice matrix in 42 operations:
5 6 6 5
6 6 6 6
5 6 6 5
| instruction | 0 | 74,356 | 12 | 148,712 |
Tags: greedy, implementation, math
Correct Solution:
```
import math
import sys
class Read:
@staticmethod
def string():
return input()
@staticmethod
def int():
return int(input())
@staticmethod
def list(sep=' '):
return input().split(sep)
@staticmethod
def list_int(sep=' '):
return list(map(int, input().split(sep)))
def solve():
n, m = Read.list_int()
a = []
for i in range(n):
a.append(Read.list_int())
res = 0
for i in range(math.floor(n / 2)):
for j in range(math.floor(m / 2)):
l = list(sorted([a[i][j], a[i][m - j - 1], a[n - i - 1][j], a[n - i - 1][m - j - 1]]))
sr = l[1]
res += abs(a[i][j] - sr)
res += abs(a[i][m - j - 1] - sr)
res += abs(a[n - i - 1][j] - sr)
res += abs(a[n - i - 1][m - j - 1] - sr)
if n % 2 == 1:
idx = math.floor(n / 2)
for i in range(0, math.floor(m / 2)):
res += abs(a[idx][m - i - 1] - a[idx][i])
if m % 2 == 1:
idx = math.floor(m / 2)
for i in range(0, math.floor(n / 2)):
res += abs(a[n - i - 1][idx] - a[i][idx])
print(res)
# query_count = 1
query_count = Read.int()
while query_count:
query_count -= 1
solve()
``` | output | 1 | 74,356 | 12 | 148,713 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A matrix of size n × m is called nice, if all rows and columns of the matrix are palindromes. A sequence of integers (a_1, a_2, ... , a_k) is a palindrome, if for any integer i (1 ≤ i ≤ k) the equality a_i = a_{k - i + 1} holds.
Sasha owns a matrix a of size n × m. In one operation he can increase or decrease any number in the matrix by one. Sasha wants to make the matrix nice. He is interested what is the minimum number of operations he needs.
Help him!
Input
The first line contains a single integer t — the number of test cases (1 ≤ t ≤ 10). The t tests follow.
The first line of each test contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the matrix.
Each of the next n lines contains m integers a_{i, j} (0 ≤ a_{i, j} ≤ 10^9) — the elements of the matrix.
Output
For each test output the smallest number of operations required to make the matrix nice.
Example
Input
2
4 2
4 2
2 4
4 2
2 4
3 4
1 2 3 4
5 6 7 8
9 10 11 18
Output
8
42
Note
In the first test case we can, for example, obtain the following nice matrix in 8 operations:
2 2
4 4
4 4
2 2
In the second test case we can, for example, obtain the following nice matrix in 42 operations:
5 6 6 5
6 6 6 6
5 6 6 5
| instruction | 0 | 74,357 | 12 | 148,714 |
Tags: greedy, implementation, math
Correct Solution:
```
testcases = int(input())
for testcae in range(testcases):
temparr = input()
temparr = temparr.split()
row = int(temparr[0])
col = int(temparr[1])
ans = 0
if row == 1 and col == 1:
vh = input()
print(0)
continue
if row == 1 :
temparr = input()
temparr = temparr.split()
dp = []
for i in temparr:
dp.append(int(i))
left = 0
right = col - 1
while left < right:
leftitem = dp[left]
rightitem = dp[right]
if leftitem == rightitem:
left += 1
right -= 1
continue
else:
ans += abs(leftitem - rightitem)
left += 1
right -= 1
print(ans)
continue
temparr = [0 for _ in range(col)]
dp = [temparr[::] for _ in range(row)]
for i in range(row):
temparr = input()
temparr = temparr.split()
for j in range(col):
dp[i][j] = int(temparr[j])
toprow = 0
bottomrow = row - 1
left = 0
right = col - 1
while True:
left = 0
right = col - 1
if toprow > bottomrow:
break
if toprow == bottomrow:
left = 0
right = col - 1
while left < right:
leftitem = dp[toprow][left]
rightitem = dp[toprow][right]
if leftitem == rightitem:
left += 1
right -= 1
continue
else:
ans += abs(leftitem - rightitem)
left += 1
right -= 1
break
while left <= right:
if left == right:
ans += abs(dp[toprow][left] - dp[bottomrow][right])
break
avg = (dp[toprow][left] + dp[toprow][right] + dp[bottomrow][left] + dp[bottomrow][right]) // 4
curbest = 0
curbest1 = 0
curbest2 = 0
curbest3 = 0
curbest4 = 0
newarr = []
newarr.append(dp[toprow][left])
newarr.append(dp[toprow][right])
newarr.append(dp[bottomrow][right])
newarr.append(dp[bottomrow][left])
newarr = sorted(newarr)
firstdiff = newarr[1] - newarr[0]
secondiff = newarr[3] - newarr[2]
ans += firstdiff
ans += secondiff
differencemid = newarr[2] - newarr[1]
ans += differencemid * 2
# curbest += abs(avg - dp[toprow][left])
# curbest += abs(avg - dp[toprow][right])
# curbest += abs(avg - dp[bottomrow][left])
# curbest += abs(avg - dp[bottomrow][right])
# avg1 = avg + 1
# curbest1 += abs(avg1 - dp[toprow][left])
# curbest1 += abs(avg1 - dp[toprow][right])
# curbest1 += abs(avg1 - dp[bottomrow][left])
# curbest1 += abs(avg1 - dp[bottomrow][right])
# avg2 = avg - 1
# curbest2 += abs(avg2 - dp[toprow][left])
# curbest2 += abs(avg2 - dp[toprow][right])
# curbest2 += abs(avg2 - dp[bottomrow][left])
# curbest2 += abs(avg2 - dp[bottomrow][right])
# avg3 = avg + 2
# curbest3 += abs(avg3 - dp[toprow][left])
# curbest3 += abs(avg3 - dp[toprow][right])
# curbest3 += abs(avg3 - dp[bottomrow][left])
# curbest3 += abs(avg3 - dp[bottomrow][right])
# avg4 = avg - 2
# curbest4 += abs(avg4 - dp[toprow][left])
# curbest4 += abs(avg4 - dp[toprow][right])
# curbest4 += abs(avg4 - dp[bottomrow][left])
# curbest4 += abs(avg4 - dp[bottomrow][right])
#ans += min(curbest4,curbest3,curbest2,curbest1)
left += 1
right -= 1
toprow += 1
bottomrow -= 1
print(ans)
``` | output | 1 | 74,357 | 12 | 148,715 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A matrix of size n × m is called nice, if all rows and columns of the matrix are palindromes. A sequence of integers (a_1, a_2, ... , a_k) is a palindrome, if for any integer i (1 ≤ i ≤ k) the equality a_i = a_{k - i + 1} holds.
Sasha owns a matrix a of size n × m. In one operation he can increase or decrease any number in the matrix by one. Sasha wants to make the matrix nice. He is interested what is the minimum number of operations he needs.
Help him!
Input
The first line contains a single integer t — the number of test cases (1 ≤ t ≤ 10). The t tests follow.
The first line of each test contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the matrix.
Each of the next n lines contains m integers a_{i, j} (0 ≤ a_{i, j} ≤ 10^9) — the elements of the matrix.
Output
For each test output the smallest number of operations required to make the matrix nice.
Example
Input
2
4 2
4 2
2 4
4 2
2 4
3 4
1 2 3 4
5 6 7 8
9 10 11 18
Output
8
42
Note
In the first test case we can, for example, obtain the following nice matrix in 8 operations:
2 2
4 4
4 4
2 2
In the second test case we can, for example, obtain the following nice matrix in 42 operations:
5 6 6 5
6 6 6 6
5 6 6 5
| instruction | 0 | 74,358 | 12 | 148,716 |
Tags: greedy, implementation, math
Correct Solution:
```
#!/usr/bin/env python3
import sys
def main():
t = int(input())
for _ in range(t):
n, m = map(int, sys.stdin.readline().split())
mat = [ list(map(int, sys.stdin.readline().split())) for _ in range(n)]
n_is_odd = n%2 != 0
m_is_odd = m%2 != 0
ops = 0
for i in range(n // 2):
for j in range(m // 2):
elems = [mat[i][j], mat[n -1 - i][j], mat[i][m - 1 - j], mat[n - 1 - i][m - 1 - j]]
med = median(elems)
ops += sum(map(lambda e: abs(med- e), elems))
if n_is_odd:
i = n // 2
for j in range(m//2):
elems = [mat[i][j], mat[i][m - 1 - j]]
med = sum(elems) // 2
ops += sum(map(lambda e: abs(med-e), elems))
if m_is_odd:
j = m // 2
for i in range(n//2):
elems = [mat[i][j], mat[n - 1 - i][j]]
med = sum(elems) // 2
ops += sum(map(lambda e: abs(med-e), elems))
print(ops)
def median(ary4):
s = sorted(ary4)
return (s[1] + s[2]) // 2
if __name__ == '__main__':
main()
``` | output | 1 | 74,358 | 12 | 148,717 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A matrix of size n × m is called nice, if all rows and columns of the matrix are palindromes. A sequence of integers (a_1, a_2, ... , a_k) is a palindrome, if for any integer i (1 ≤ i ≤ k) the equality a_i = a_{k - i + 1} holds.
Sasha owns a matrix a of size n × m. In one operation he can increase or decrease any number in the matrix by one. Sasha wants to make the matrix nice. He is interested what is the minimum number of operations he needs.
Help him!
Input
The first line contains a single integer t — the number of test cases (1 ≤ t ≤ 10). The t tests follow.
The first line of each test contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the matrix.
Each of the next n lines contains m integers a_{i, j} (0 ≤ a_{i, j} ≤ 10^9) — the elements of the matrix.
Output
For each test output the smallest number of operations required to make the matrix nice.
Example
Input
2
4 2
4 2
2 4
4 2
2 4
3 4
1 2 3 4
5 6 7 8
9 10 11 18
Output
8
42
Note
In the first test case we can, for example, obtain the following nice matrix in 8 operations:
2 2
4 4
4 4
2 2
In the second test case we can, for example, obtain the following nice matrix in 42 operations:
5 6 6 5
6 6 6 6
5 6 6 5
| instruction | 0 | 74,359 | 12 | 148,718 |
Tags: greedy, implementation, math
Correct Solution:
```
from math import *
from bisect import *
from collections import *
from random import *
from decimal import *
import sys
input=sys.stdin.readline
def inp():
return int(input())
def st():
return input().rstrip('\n')
def lis():
return list(map(int,input().split()))
def ma():
return map(int,input().split())
t=inp()
while(t):
t-=1
n,m=ma()
r=[]
for i in range(n):
r.append(lis())
res=0
for i in range((n+1)//2):
for j in range((m+1)//2):
l=[]
l.append(r[i][j])
if(j!=(m-j-1)):
l.append(r[i][-1-j])
if(i!=n-i-1):
l.append(r[-i-1][j])
if(i!=(n-i-1) and j!=(m-j-1)):
l.append(r[-1-i][-j-1])
re=1000000000000000000000000
for k1 in l:
lo=0
for k in l:
lo+=abs(k1-k)
re=min(re,lo)
res+=re
print(res)
``` | output | 1 | 74,359 | 12 | 148,719 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A matrix of size n × m is called nice, if all rows and columns of the matrix are palindromes. A sequence of integers (a_1, a_2, ... , a_k) is a palindrome, if for any integer i (1 ≤ i ≤ k) the equality a_i = a_{k - i + 1} holds.
Sasha owns a matrix a of size n × m. In one operation he can increase or decrease any number in the matrix by one. Sasha wants to make the matrix nice. He is interested what is the minimum number of operations he needs.
Help him!
Input
The first line contains a single integer t — the number of test cases (1 ≤ t ≤ 10). The t tests follow.
The first line of each test contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the matrix.
Each of the next n lines contains m integers a_{i, j} (0 ≤ a_{i, j} ≤ 10^9) — the elements of the matrix.
Output
For each test output the smallest number of operations required to make the matrix nice.
Example
Input
2
4 2
4 2
2 4
4 2
2 4
3 4
1 2 3 4
5 6 7 8
9 10 11 18
Output
8
42
Note
In the first test case we can, for example, obtain the following nice matrix in 8 operations:
2 2
4 4
4 4
2 2
In the second test case we can, for example, obtain the following nice matrix in 42 operations:
5 6 6 5
6 6 6 6
5 6 6 5
| instruction | 0 | 74,360 | 12 | 148,720 |
Tags: greedy, implementation, math
Correct Solution:
```
import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools,itertools,operator,bisect,fractions,statistics
from collections import deque,defaultdict,OrderedDict,Counter
from fractions import Fraction
from decimal import Decimal
from sys import stdout
def main():
starttime=datetime.datetime.now()
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
for _ in range(ri()):
n,m=ria()
ar=[]
for i in range(n):
ar.append(ria())
z=0
ans=0
index1=0
index2=0
while index2<=m-1-index2:
index1=0
while index1<=n-1-index1:
a=ar[index1][index2]
b=ar[index1][m-1-index2]
c=ar[n-1-index1][index2]
d=ar[n-1-index1][m-1-index2]
if index1==n-1-index1 and index2==m-1-index2:
ans+=0
elif index1==n-1-index1 and index2!=m-1-index2:
z=statistics.median([a,b])
ans+=abs(a-z)+abs(b-z)
elif index1!=n-1-index1 and index2==m-1-index2:
z=statistics.median([a,c])
ans+=abs(a-z)+abs(c-z)
else:
z=statistics.median([a,b,c,d])
ans+=abs(a-z)+abs(c-z)+abs(d-z)+abs(b-z)
index1+=1
index2+=1
print(int(ans))
#<--Solving Area Ends
endtime=datetime.datetime.now()
time=(endtime-starttime).total_seconds()*1000
if(os.path.exists('input.txt')):
print("Time:",time,"ms")
class FastReader(io.IOBase):
newlines = 0
def __init__(self, fd, chunk_size=1024 * 8):
self._fd = fd
self._chunk_size = chunk_size
self.buffer = io.BytesIO()
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
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, size=-1):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
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()
class FastWriter(io.IOBase):
def __init__(self, fd):
self._fd = fd
self.buffer = io.BytesIO()
self.write = self.buffer.write
def flush(self):
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class FastStdin(io.IOBase):
def __init__(self, fd=0):
self.buffer = FastReader(fd)
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
class FastStdout(io.IOBase):
def __init__(self, fd=1):
self.buffer = FastWriter(fd)
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.flush = self.buffer.flush
if __name__ == '__main__':
sys.stdin = FastStdin()
sys.stdout = FastStdout()
main()
``` | output | 1 | 74,360 | 12 | 148,721 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A matrix of size n × m is called nice, if all rows and columns of the matrix are palindromes. A sequence of integers (a_1, a_2, ... , a_k) is a palindrome, if for any integer i (1 ≤ i ≤ k) the equality a_i = a_{k - i + 1} holds.
Sasha owns a matrix a of size n × m. In one operation he can increase or decrease any number in the matrix by one. Sasha wants to make the matrix nice. He is interested what is the minimum number of operations he needs.
Help him!
Input
The first line contains a single integer t — the number of test cases (1 ≤ t ≤ 10). The t tests follow.
The first line of each test contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the matrix.
Each of the next n lines contains m integers a_{i, j} (0 ≤ a_{i, j} ≤ 10^9) — the elements of the matrix.
Output
For each test output the smallest number of operations required to make the matrix nice.
Example
Input
2
4 2
4 2
2 4
4 2
2 4
3 4
1 2 3 4
5 6 7 8
9 10 11 18
Output
8
42
Note
In the first test case we can, for example, obtain the following nice matrix in 8 operations:
2 2
4 4
4 4
2 2
In the second test case we can, for example, obtain the following nice matrix in 42 operations:
5 6 6 5
6 6 6 6
5 6 6 5
Submitted Solution:
```
from collections import defaultdict, Counter
from sys import stdin
from math import ceil, floor
input = stdin.readline
for _ in range(int(input())):
n, m = map(int, input().split())
mat = []
for i in range(n):
mat.append(list(map(int, input().split())))
ans = 0
for i in range((n) // 2):
for j in range((m) // 2):
c = [mat[i][j], mat[n - i - 1][j], mat[i][m - j - 1], mat[n - i - 1][m - j - 1]]
f = sorted(c)[2]
for k in range(4):
ans += abs(f - c[k])
if n % 2 == 1:
for i in range(m // 2):
ans+= abs(mat[(n) // 2][i] - mat[(n) // 2][m - i - 1])
if m % 2 == 1:
for i in range(n // 2):
ans+=abs( mat[i][(m) // 2] - mat[n - i - 1][(m) // 2])
print(ans)
``` | instruction | 0 | 74,361 | 12 | 148,722 |
Yes | output | 1 | 74,361 | 12 | 148,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A matrix of size n × m is called nice, if all rows and columns of the matrix are palindromes. A sequence of integers (a_1, a_2, ... , a_k) is a palindrome, if for any integer i (1 ≤ i ≤ k) the equality a_i = a_{k - i + 1} holds.
Sasha owns a matrix a of size n × m. In one operation he can increase or decrease any number in the matrix by one. Sasha wants to make the matrix nice. He is interested what is the minimum number of operations he needs.
Help him!
Input
The first line contains a single integer t — the number of test cases (1 ≤ t ≤ 10). The t tests follow.
The first line of each test contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the matrix.
Each of the next n lines contains m integers a_{i, j} (0 ≤ a_{i, j} ≤ 10^9) — the elements of the matrix.
Output
For each test output the smallest number of operations required to make the matrix nice.
Example
Input
2
4 2
4 2
2 4
4 2
2 4
3 4
1 2 3 4
5 6 7 8
9 10 11 18
Output
8
42
Note
In the first test case we can, for example, obtain the following nice matrix in 8 operations:
2 2
4 4
4 4
2 2
In the second test case we can, for example, obtain the following nice matrix in 42 operations:
5 6 6 5
6 6 6 6
5 6 6 5
Submitted Solution:
```
for _ in range(int(input())):
n,m=map(int,input().split())
mat=[]
for i in range(n):
mat.append(list(map(int,input().split())))
out=0
for i in range(n):
for j in range(m):
x=mat[i][j]
y=mat[i][m-1-j]
z=mat[n-1-i][j]
a=abs(x-y)+abs(z-x)
b=abs(x-y)+abs(z-y)
c=abs(y-z)+abs(z-x)
if a<=b and a<=c:
out+=a
mat[i][j]=x
mat[i][m-1-j]=x
mat[n-1-i][j]=x
elif b<=a and b<=c:
out+=b
mat[i][j]=y
mat[i][m-1-j]=y
mat[n-1-i][j]=y
else:
out+=c
mat[i][j]=z
mat[i][m-1-j]=z
mat[n-1-i][j]=z
print(out)
``` | instruction | 0 | 74,362 | 12 | 148,724 |
Yes | output | 1 | 74,362 | 12 | 148,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A matrix of size n × m is called nice, if all rows and columns of the matrix are palindromes. A sequence of integers (a_1, a_2, ... , a_k) is a palindrome, if for any integer i (1 ≤ i ≤ k) the equality a_i = a_{k - i + 1} holds.
Sasha owns a matrix a of size n × m. In one operation he can increase or decrease any number in the matrix by one. Sasha wants to make the matrix nice. He is interested what is the minimum number of operations he needs.
Help him!
Input
The first line contains a single integer t — the number of test cases (1 ≤ t ≤ 10). The t tests follow.
The first line of each test contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the matrix.
Each of the next n lines contains m integers a_{i, j} (0 ≤ a_{i, j} ≤ 10^9) — the elements of the matrix.
Output
For each test output the smallest number of operations required to make the matrix nice.
Example
Input
2
4 2
4 2
2 4
4 2
2 4
3 4
1 2 3 4
5 6 7 8
9 10 11 18
Output
8
42
Note
In the first test case we can, for example, obtain the following nice matrix in 8 operations:
2 2
4 4
4 4
2 2
In the second test case we can, for example, obtain the following nice matrix in 42 operations:
5 6 6 5
6 6 6 6
5 6 6 5
Submitted Solution:
```
R=lambda:map(int,input().split())
t,=R()
for _ in[0]*t:
n,m=R();a=[[*R()]for _ in[0]*n];r=0
for i in range(n):
b=*zip(a[i],a[n-i-1]),
for j in range(m):c=sorted(b[j]+b[m-j-1]);r+=c[2]+c[3]-c[0]-c[1]
print(r//4)
``` | instruction | 0 | 74,363 | 12 | 148,726 |
Yes | output | 1 | 74,363 | 12 | 148,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A matrix of size n × m is called nice, if all rows and columns of the matrix are palindromes. A sequence of integers (a_1, a_2, ... , a_k) is a palindrome, if for any integer i (1 ≤ i ≤ k) the equality a_i = a_{k - i + 1} holds.
Sasha owns a matrix a of size n × m. In one operation he can increase or decrease any number in the matrix by one. Sasha wants to make the matrix nice. He is interested what is the minimum number of operations he needs.
Help him!
Input
The first line contains a single integer t — the number of test cases (1 ≤ t ≤ 10). The t tests follow.
The first line of each test contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the matrix.
Each of the next n lines contains m integers a_{i, j} (0 ≤ a_{i, j} ≤ 10^9) — the elements of the matrix.
Output
For each test output the smallest number of operations required to make the matrix nice.
Example
Input
2
4 2
4 2
2 4
4 2
2 4
3 4
1 2 3 4
5 6 7 8
9 10 11 18
Output
8
42
Note
In the first test case we can, for example, obtain the following nice matrix in 8 operations:
2 2
4 4
4 4
2 2
In the second test case we can, for example, obtain the following nice matrix in 42 operations:
5 6 6 5
6 6 6 6
5 6 6 5
Submitted Solution:
```
for _ in range(int(input())):
n,m=map(int,input().split())
mat=[]
for _ in range(n):
mat.append(list(map(int,input().split())))
ans=0
for i in range(n//2):
for j in range(m//2):
i2,j2=i,m-j-1
i3,j3=n-i-1,j
i4,j4=n-i-1,m-j-1
diff=[mat[i][j],mat[i2][j2],mat[i3][j3],mat[i4][j4]]
diff.sort()
d1=diff[1]-diff[0]
d2=diff[2]-diff[1]
d3=diff[3]-diff[2]
change=min(3*d1+2*d2+d3,min(d1+2*d2+d3,d1+2*d2+3*d3))
ans+=change
if n&1:
i=n//2
j1=0
j2=m-1
while j1<j2:
ans+=abs(mat[i][j1]-mat[i][j2])
j1+=1
j2-=1
if m&1:
i1=0
i2=n-1
j=m//2
while i1<i2:
ans+=abs(mat[i1][j]-mat[i2][j])
i1+=1
i2-=1
print(ans)
``` | instruction | 0 | 74,364 | 12 | 148,728 |
Yes | output | 1 | 74,364 | 12 | 148,729 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A matrix of size n × m is called nice, if all rows and columns of the matrix are palindromes. A sequence of integers (a_1, a_2, ... , a_k) is a palindrome, if for any integer i (1 ≤ i ≤ k) the equality a_i = a_{k - i + 1} holds.
Sasha owns a matrix a of size n × m. In one operation he can increase or decrease any number in the matrix by one. Sasha wants to make the matrix nice. He is interested what is the minimum number of operations he needs.
Help him!
Input
The first line contains a single integer t — the number of test cases (1 ≤ t ≤ 10). The t tests follow.
The first line of each test contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the matrix.
Each of the next n lines contains m integers a_{i, j} (0 ≤ a_{i, j} ≤ 10^9) — the elements of the matrix.
Output
For each test output the smallest number of operations required to make the matrix nice.
Example
Input
2
4 2
4 2
2 4
4 2
2 4
3 4
1 2 3 4
5 6 7 8
9 10 11 18
Output
8
42
Note
In the first test case we can, for example, obtain the following nice matrix in 8 operations:
2 2
4 4
4 4
2 2
In the second test case we can, for example, obtain the following nice matrix in 42 operations:
5 6 6 5
6 6 6 6
5 6 6 5
Submitted Solution:
```
# If you do what you've always done, you'll get what you've always gotten. Tony Robbins
# by : Blue Edge - Create some chaos
def f(a,t):
ans = 0
for x in a:
ans+=abs(x-t)
return ans
for _ in range(int(input())):
n,m=map(int,input().split())
a=[list(map(int,input().split())) for __ in range(n)]
# i,j
# i,m-1-j
# n-1-i,j
# n-1-i,m-1-j
b = []
for i in range(n):
b+=[1]*m,
ans = 0
for i in range(n):
for j in range(m):
if b[i][j]:
s = set()
for x in sorted([i,n-1-i]):
for y in sorted([j,m-1-j]):
b[x][y]=0
s.add(str(x)+","+str(y))
d = []
for c in s:
c=list(map(int,c.split(",")))
d+=a[c[0]][c[1]],
mn = sum(d)//len(d)
pv = []
for ik in range(-10,10):
pv+=f(d,mn+ik),
ans+=min(pv)
print(ans)
``` | instruction | 0 | 74,365 | 12 | 148,730 |
No | output | 1 | 74,365 | 12 | 148,731 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A matrix of size n × m is called nice, if all rows and columns of the matrix are palindromes. A sequence of integers (a_1, a_2, ... , a_k) is a palindrome, if for any integer i (1 ≤ i ≤ k) the equality a_i = a_{k - i + 1} holds.
Sasha owns a matrix a of size n × m. In one operation he can increase or decrease any number in the matrix by one. Sasha wants to make the matrix nice. He is interested what is the minimum number of operations he needs.
Help him!
Input
The first line contains a single integer t — the number of test cases (1 ≤ t ≤ 10). The t tests follow.
The first line of each test contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the matrix.
Each of the next n lines contains m integers a_{i, j} (0 ≤ a_{i, j} ≤ 10^9) — the elements of the matrix.
Output
For each test output the smallest number of operations required to make the matrix nice.
Example
Input
2
4 2
4 2
2 4
4 2
2 4
3 4
1 2 3 4
5 6 7 8
9 10 11 18
Output
8
42
Note
In the first test case we can, for example, obtain the following nice matrix in 8 operations:
2 2
4 4
4 4
2 2
In the second test case we can, for example, obtain the following nice matrix in 42 operations:
5 6 6 5
6 6 6 6
5 6 6 5
Submitted Solution:
```
import sys
input=sys.stdin.readline
for _ in range(int(input())):
n,m=map(int,input().split())
ar=[]
for i in range(n):
ar.append(list(map(int,input().split())))
ans=0
for i in range(n//2):
for j in range(m//2):
x1=ar[i][j]
x2=ar[i][-(j+1)]
x3=ar[-(i+1)][j]
x4=ar[-(i+1)][-(j+1)]
li=[x1,x2,x3,x4]
li.sort()
for ii in range(4):
ans+=abs(li[ii]-li[1])
if(n%2!=0 and m%2==0):
li=[]
for i in range(m):
li.append(ar[n//2][i])
li.sort()
for i in range(m):
ans+=abs(li[i]-li[m//2])
elif(n%2==0 and m%2!=0):
li=[]
for i in range(n):
li.append(ar[i][m//2])
for i in range(n):
ans+=abs(li[i]-li[n//2])
elif(n%2!=0 and m%2!=0):
li=[]
for i in range(n):
li.append(ar[i][m//2])
for j in range(m):
if(j!=m//2):
li.append(ar[n//2][j])
li.sort()
le=len(li)
for i in range(le):
ans+=abs(li[i]-li[le//2])
print(ans)
``` | instruction | 0 | 74,366 | 12 | 148,732 |
No | output | 1 | 74,366 | 12 | 148,733 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A matrix of size n × m is called nice, if all rows and columns of the matrix are palindromes. A sequence of integers (a_1, a_2, ... , a_k) is a palindrome, if for any integer i (1 ≤ i ≤ k) the equality a_i = a_{k - i + 1} holds.
Sasha owns a matrix a of size n × m. In one operation he can increase or decrease any number in the matrix by one. Sasha wants to make the matrix nice. He is interested what is the minimum number of operations he needs.
Help him!
Input
The first line contains a single integer t — the number of test cases (1 ≤ t ≤ 10). The t tests follow.
The first line of each test contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the matrix.
Each of the next n lines contains m integers a_{i, j} (0 ≤ a_{i, j} ≤ 10^9) — the elements of the matrix.
Output
For each test output the smallest number of operations required to make the matrix nice.
Example
Input
2
4 2
4 2
2 4
4 2
2 4
3 4
1 2 3 4
5 6 7 8
9 10 11 18
Output
8
42
Note
In the first test case we can, for example, obtain the following nice matrix in 8 operations:
2 2
4 4
4 4
2 2
In the second test case we can, for example, obtain the following nice matrix in 42 operations:
5 6 6 5
6 6 6 6
5 6 6 5
Submitted Solution:
```
t = int(input())
def najdi(a,b,c,d):
priemer = (a+b+c+d)//4
return (min(abs(a - priemer) + abs(b-priemer) +abs(c-priemer) + abs(d-priemer), abs(a-(priemer+1)) + abs(b-(priemer+1)) + abs(c-(priemer+1))+abs(d-(priemer+1))))
def najdi2(a,b):
priemer = (a+b)//2
return (min(abs(a - priemer) + abs(b-priemer), abs(a-(priemer+1)) + abs(b-(priemer+1))))
for _ in range(t):
[n, m] = [int(x) for x in input().split()]
a = []
for i in range(n):
a.append([int(x) for x in input().split()])
out = 0
if n % 2 == 0 and m % 2 == 0:
for i in range(n//2):
for j in range(m//2):
out = out + najdi(a[i][j], a[n-i-1][j], a[i][m-j-1], a[n-i-1][m-j-1])
elif n%2 == 1 and m%2 == 0:
for i in range(n//2):
for j in range(m//2):
out = out + najdi(a[i][j], a[n-i-1][j], a[i][m-j-1], a[n-i-1][m-j-1])
i = n//2
for j in range(m//2):
out = out + najdi2(a[i][j], a[i][m-j-1])
elif n%2 == 0 and m%2 == 1:
for i in range(n//2):
for j in range(m//2):
out = out + najdi(a[i][j], a[n-i-1][j], a[i][m-j-1], a[n-i-1][m-j-1])
j = m//2
out = out + najdi2(a[i][j], a[n-i-1][j])
else:
for i in range(n//2):
for j in range(m//2):
out = out + najdi(a[i][j], a[n-i-1][j], a[i][m-j-1], a[n-i-1][m-j-1])
j = m//2
out = out + najdi2(a[i][j], a[n-i-1][j])
i = n//2
for j in range(m//2):
out = out + najdi2(a[i][j], a[i][m-j-1])
print(out)
``` | instruction | 0 | 74,367 | 12 | 148,734 |
No | output | 1 | 74,367 | 12 | 148,735 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A matrix of size n × m is called nice, if all rows and columns of the matrix are palindromes. A sequence of integers (a_1, a_2, ... , a_k) is a palindrome, if for any integer i (1 ≤ i ≤ k) the equality a_i = a_{k - i + 1} holds.
Sasha owns a matrix a of size n × m. In one operation he can increase or decrease any number in the matrix by one. Sasha wants to make the matrix nice. He is interested what is the minimum number of operations he needs.
Help him!
Input
The first line contains a single integer t — the number of test cases (1 ≤ t ≤ 10). The t tests follow.
The first line of each test contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the matrix.
Each of the next n lines contains m integers a_{i, j} (0 ≤ a_{i, j} ≤ 10^9) — the elements of the matrix.
Output
For each test output the smallest number of operations required to make the matrix nice.
Example
Input
2
4 2
4 2
2 4
4 2
2 4
3 4
1 2 3 4
5 6 7 8
9 10 11 18
Output
8
42
Note
In the first test case we can, for example, obtain the following nice matrix in 8 operations:
2 2
4 4
4 4
2 2
In the second test case we can, for example, obtain the following nice matrix in 42 operations:
5 6 6 5
6 6 6 6
5 6 6 5
Submitted Solution:
```
from math import *
from bisect import *
from collections import *
from random import *
from decimal import *
import sys
input=sys.stdin.readline
def inp():
return int(input())
def st():
return input().rstrip('\n')
def lis():
return list(map(int,input().split()))
def ma():
return map(int,input().split())
t=inp()
while(t):
t-=1
n,m=ma()
r=[]
for i in range(n):
r.append(lis())
res=0
for i in range((n+1)//2):
for j in range((m+1)//2):
l=[]
l.append(r[i][j])
if(j!=(m-j-1)):
l.append(r[i][-1-j])
if(i!=n-i-1):
l.append(r[-i-1][j])
if(i!=(n-i-1) and j!=(m-j-1)):
l.append(r[-1-i][-j-1])
req=(min(l)+max(l))//2
#print(l,req)
for k in l:
res+=abs(req-k)
print(res)
``` | instruction | 0 | 74,368 | 12 | 148,736 |
No | output | 1 | 74,368 | 12 | 148,737 |
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