codeforces 1335 E1 思维

题意：给一个长度为n（2000）的“字符串”，字符串只由26个组成。组成

 的最长长度为多少，x，y可以为0.

思路：数据范围很小，可以暴力枚举范围i，j，则三个范围分别为【1，i】【i+1，j-1】，【j，n】，然后枚举每一种字符，在使两端相同字符的时候，当前区间相同字符最多。

　　【x，y】的1的数量可以用前缀和维护。

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector> 
// #include <bits/stdc++.h>
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sp ' '
#define endl '
'
#define inf  0x3f3f3f3f;
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
#define P pair<int, int>
#define fi first
#define se second
#define pb(x) push_back(x)
#define ppb() pop_back()
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define sca3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define sca2(a,b) scanf("%d %d",&(a),&(b))
#define sca(a) scanf("%d",&(a));
#define sca3ll(a,b,c) scanf("%lld %lld %lld",&(a),&(b),&(c))
#define sca2ll(a,b) scanf("%lld %lld",&(a),&(b))
#define scall(a) scanf("%lld",&(a));


using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a, ll b, ll mod){ll sum = 1;while (b) {if (b & 1) {sum = (sum * a) % mod;b--;}b /= 2;a = a * a % mod;}return sum;}

const double Pi = acos(-1.0);
const double epsilon = Pi/180.0;
const int maxn = 2e3+10;
int sum[30][maxn];
int main()
{
    //freopen("input.txt", "r", stdin);
    int _;
    scanf("%d",&_);
    while(_--)
    {
        int n;
        cin>>n;
        rep(i,1,n){
            rep(j,1,26){
                sum[j][i] = 0;
            }
        }
        rep(i,1,n){
            rep(j,1,26){
                sum[j][i] = sum[j][i-1];
            }
            int x;
            cin>>x;
            sum[x][i]++;
        }
        int ans = 0;
        rep(i,1,26){
            ans = max(ans,sum[i][n]);
        }
        rep(i,1,n){
            rep(j,i+1,n){
                int lenlr = 0,lenmid = 0;
                rep(k,1,26){
                    int l = sum[k][i];
                    int r = sum[k][n]-sum[k][j-1];
                    lenlr = max(lenlr,min(l,r));
                }
                rep(k,1,26){
                    int mid = sum[k][j-1]-sum[k][i];
                    lenmid = max(lenmid,mid);
                }
                //cout<<lenlr<<sp<<lenmid<<endl;
                ans = max(lenlr*2+lenmid,ans);
            }
        }
        cout<<ans<<endl;
    }
}