题目链接：https://www.luogu.com.cn/problem/P8667

二分

用lower_bound和upper_bound

#include <iostream>
#include <vector>
#include <algorithm>

#define ll long long


const ll N = 1e5;
ll op1[N],op2[N],op3[N];

void solve(){
    ll n;
    std::cin >> n;
    for(ll i = 1; i <= n; i ++){
        std::cin >> op1[i];
    }
    for(ll i = 1; i <= n; i ++){
        std::cin >> op2[i];
    }
    for(ll i = 1; i <= n; i ++){
        std::cin >> op3[i];
    }
    std::sort(op1 + 1,op1 + n + 1);
    std::sort(op2 + 1,op2 + n + 1);
    std::sort(op3 + 1,op3 + n + 1);
    ll res = 0;
    for(ll i = 1; i <= n; i ++){
        ll re = std::lower_bound(op1 + 1,op1 + n + 1,op2[i]) - op1 - 1;
//      std::cout << re << " ";
//      ll *jj = std::upper_bound(op2 + 1,op2 + n + 1,op1[i]);
        ll rk = std::upper_bound(op3 + 1,op3 + n + 1,op2[i]) - op3;
        rk = n - rk + 1;
//      std::cout << rk << "\n";
        res += (re * rk);       
    }
    std::cout << res << "\n";
}

int main(){
    ll t = 1;
    while(t --)
    solve();
    return 0;
}