sort+二分，没啥技巧的做法

//O(nlog(n)log(n))
#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
typedef long long LL;
int a[N], b[N], c[N];
int main()
{
    long long n, ans = 0;
    cin >> n;
    for(int i = 0;i < n;i++) cin >> a[i];
    for(int i = 0;i < n;i++) cin >> b[i];
    for(int i = 0;i < n;i++) cin >> c[i];
    sort(a, a + n), sort(b, b + n), sort(c, c + n);
    for(int i = 0;i < n;i++) {
        int pos1 = lower_bound(a, a + n, b[i]) - a;
        if(pos1 == 0) continue;
        int pos2 = upper_bound(c, c + n, b[i]) - c;
        if(pos2 == n) continue;
        ans += 1ll * pos1 * (n - pos2);
    }
    cout << ans << endl;
    return 0;
}