题目：
You are given an array a1,a2…an. Calculate the number of tuples (i,j,k,l) such that:

1≤i<j<k<l≤n;
ai=ak and aj=al;
Input
The first line contains a single integer t (1≤t≤100) — the number of test cases.

The first line of each test case contains a single integer n (4≤n≤3000) — the size of the array a.

The second line of each test case contains n integers a1,a2,…,an (1≤ai≤n) — the array a.

It’s guaranteed that the sum of n in one test doesn’t exceed 3000.

Output
For each test case, print the number of described tuples.

Example
inputCopy
2
5
2 2 2 2 2
6
1 3 3 1 2 3
outputCopy
5
2
Note
In the first test case, for any four indices i<j<k<l are valid, so the answer is the number of tuples.

In the second test case, there are 2 valid tuples:

(1,2,4,6): a1=a4 and a2=a6;
(1,3,4,6): a1=a4 and a3=a6.

题目大意： 找到满足条件的所有组合
1≤i<j<k<l≤n;
ai=ak and aj=al;

思路 ：枚举j和l，然后用num来统计 j到l 有多少种组合。
cnt数组 记录好 j下标前面 出现过多少次这个数。
因为数据比较小，直接o（n^2）.

#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstdlib>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <bitset>
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define FILL(a,b) (memset(a,b,sizeof(a)))
#define re register
#define lson rt<<1
#define rson rt<<1|1
#define lowbit(a) ((a)&-(a))
#define ios std::ios::sync_with_stdio(false);std::cin.tie(0);std::cout.tie(0);
#define fi first
#define rep(i,n) for(int i=0;(i)<(n);i++)
#define rep1(i,n) for(int i=1;(i)<=(n);i++)
#define se secondusing namespace std;
typedef long long  ll;
typedef unsigned long long  ull;
typedef pair<ll,ll> pii;
const ll mod=998244353;
const ll N =3e6+10;
const double eps = 1e-5;
const double pi=acos(-1);
ll gcd(ll a,ll b){return !b?a:gcd(b,a%b);}
int dx[4]= {-1,0,1,0}, dy[4] = {0,1,0,-1};
ll n,cnt[10005],a[10005],res,num;void solve()
{cin>>n;res=0;memset(cnt,0,sizeof(cnt));for(int i=1;i<=n;i++){cin>>a[i];}for(int j=1;j<n;j++){num=0;for(int l=j+1;l<=n;l++){if(a[j]==a[l]){res+=num;}num+=cnt[a[l]];}cnt[a[j]]++;}cout<<res<<endl;
}
int main()
{iosint T;cin>>T;//T=1;while(T--){solve();}return 0;
}