time limit per test
2 seconds
memory limit per test
256 megabytes
A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, 2,3,1,5,4 is a permutation, but 1,2,2 and 1,3,4 are not permutations.
You are given a permutation p of length n. You can perform the following operation exactly once:
- Choose two integers l, r (1≤l≤r≤n).
- Reverse the segment l,r in the permutation p.
Your task is to output the lexicographically maximum permutation that can be obtained by performing this operation. A permutation a is lexicographically greater than a permutation b if for the first position i where they differ, it holds that ai>bi.
Input
Each test consists of several test cases. The first line contains a single integer t (1≤t≤104) --- the number of test cases. The description of the test cases follows.
The first line of each test case contains the number n (1≤n≤2⋅105).
The second line of each test case contains n distinct integers p1,p2,...,pn (1≤pi≤n).
It is guaranteed that the sum of the values of n across all test cases does not exceed 2⋅105.
Output
For each test case, output the lexicographically maximum permutation that can be obtained with one operation.
Example
Input
Copy
4
4
3 2 1 4
3
3 1 2
4
4 3 2 1
2
2 1
Output
Copy
4 1 2 3
3 2 1
4 3 2 1
2 1
Note
For the first test case, the best segment is 1,4. After reversing, a=4,1,2,3. For the second test case, the best segment is 2,3. After reversing, a=3,2,1.
解题说明：此题是一道数学题，采用贪心算法。给定一个排列 a，判断能否通过一次区间反转将其变为降序排列 [n, n-1, ..., 1]，并输出反转后的结果。采用贪心算法，从最大的值开始检查，找到第一个不在目标位置的值。
cpp
#include<iostream>
#include<algorithm>
using namespace std;
int n, a[200001], b[200001];
int main()
{
int ttt;
cin >> ttt;
while (ttt--)
{
cin >> n;
for (int i = 1; i <= n; i++)
{
cin >> a[i];
b[a[i]] = i;
}
for (int i = n; i >= 1; i--)
{
if (b[i] == n - i + 1)
{
continue;
}
reverse(a + n - i + 1, a + b[i] + 1);
break;
}
for (int i = 1; i <= n; i++)
{
cout << a[i] << " ";
}
cout << endl;
}
}