B - Yet Another Palindrome Problem

标签：tle   read   sum   sub   order   algorithm   ril   ace   etc   

You are given an array aa consisting of nn integers.

Your task is to determine if aa has some subsequence of length at least 33 that is a palindrome.

Recall that an array bb is called a subsequence of the array aa if bb can be obtained by removing some (possibly, zero) elements from aa (not necessarily consecutive) without changing the order of remaining elements. For example, [2][2], [1,2,1,3][1,2,1,3] and [2,3][2,3] are subsequences of [1,2,1,3][1,2,1,3], but [1,1,2][1,1,2]and [4][4] are not.

Also, recall that a palindrome is an array that reads the same backward as forward. In other words, the array aa of length nn is the palindrome if ai=an−i−1ai=an−i−1 for all ii from 11 to nn. For example, arrays [1234][1234], [1,2,1][1,2,1], [1,3,2,2,3,1][1,3,2,2,3,1] and [10,100,10][10,100,10] are palindromes, but arrays [1,2][1,2] and [1,2,3,1][1,2,3,1] are not.

You have to answer tt independent test cases.

Input

The first line of the input contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases.

Next 2t2t lines describe test cases. The first line of the test case contains one integer nn (3≤n≤50003≤n≤5000) — the length of aa. The second line of the test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤n1≤ai≤n), where aiai is the ii-th element of aa.

It is guaranteed that the sum of nn over all test cases does not exceed 50005000 (∑n≤5000∑n≤5000).

Output

For each test case, print the answer — "YES" (without quotes) if aa has some subsequence of length at least 33 that is a palindrome and "NO" otherwise.

Example

Input

5
3
1 2 1
5
1 2 2 3 2
3
1 1 2
4
1 2 2 1
10
1 1 2 2 3 3 4 4 5 5

Output

YES
YES
NO
YES
NO

简单的回文字符串，反序求最长公共子序列，如果能大于三就是yes，感觉可以在dp内判断是否大于三，可以尝试一下

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;



int main()
{
    int t;
    scanf("%d", &t);
    while (t--) {
        int n;
        scanf("%d", &n);
        int a[5001] = { 0 };
        int b[5001] = { 0 };

        for (int i = 0; i < n; i++) {
            scanf("%d", &a[i]);
            b[n - i - 1 ] = a[i];
        }


        int dp[5001][5001] = { 0 };
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (a[i] == b[j]) {
                    dp[i + 1][j + 1] = dp[i][j] + 1;
                }
                else {
                    dp[i + 1][j + 1] = max(dp[i][j + 1], dp[i + 1][j]);
                }
            }
        }

        if (dp[n][n] >= 3) {
            printf("YES\n");
        }
        else {
            printf("NO\n");
        }
    }




    return 0;
}

B - Yet Another Palindrome Problem

标签：tle   read   sum   sub   order   algorithm   ril   ace   etc   

原文地址：https://www.cnblogs.com/Vetsama/p/12495006.html