CF_1883_F. You Are So Beautiful

题目链接：Problem - 1883F - Codeforces

---

题目大意：

给定数组a，统计有多少连续子数组b，满足b作为子序列在a中只出现一次（即仅有b自身作为连续子数组时才能匹配）。

---

思路：

子数组一定是要连续的，但子序列不用连续.

观察图片规律，不难发现，要满足条件，该子数组左边是该数字第一次出现，右边是该数字最后一次出现

---

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<queue>
#include<deque>
#include<stack>
#include<set>
#include<map>
#include<unordered_set>
#include<unordered_map>
#include<bitset>
#include<tuple>
#include<array>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/numeric>
#define inf 72340172838076673
#define int long long
#define endl '\n'
#define F first
#define S second
#define mst(a,x) memset(a,x,sizeof (a))
#define gmap __gnu_pbds::gp_hash_table
#define power __gnu_cxx::power
using namespace std;
typedef pair<int, int> pii;

const int N = 200086, mod = 998244353;

int n, m;
int a[N];

void solve() {

    cin >> n;
    gmap<int, int> mp, cnt;
    vector<int> f(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        mp[a[i]]++;
    }
    
    int res = 0;
    for (int i = 1; i <= n; i++) {
        f[i] = f[i - 1];
        if (++cnt[a[i]] == 1) f[i]++;
        if (cnt[a[i]] == mp[a[i]]) {
            res += f[i];
        }
        
    }
    
    cout << res << endl;

}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    
    int T = 1;
    cin >> T;
    while (T--) solve();
    
    return 0;
}