复杂度O(nt)
code:
#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
typedef pair<int, int>pii;
int t;
int a[N];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while(t--){
        map<pii, int>p1, p2, p3;
        map<vector<int>, int>cnt;
        int n;
        cin >> n;
        int ans = 0;
        for(int i = 1;i <= n;i++)cin >> a[i];
        for(int i = 1;i <= n - 2;i++){
            ans += p1[{a[i], a[i + 1]}]++;
            ans += p2[{a[i], a[i + 2]}]++;
            ans += p3[{a[i + 1], a[i + 2]}]++;
            vector<int> v = {a[i], a[i + 1], a[i + 2]};
            // cout << v[1] << endl;
            ans -= 3 * cnt[v];
            cnt[v]++; 
        }
        cout << ans << endl;
    }

    return 0;
}