思路：1≤i,j,k≤N Ai<Bj<Ck可以枚举Bj,然后在Ai中找到最大的满足<Bj的，Ck中找最小满足的（找的时候用二分找）

#include <iostream>
#include <algorithm>

using namespace std;

long long ll = 1;

int n;
const int N = 1e5 + 10;
int a[N],b[N],c[N];
long long res;

int main(){


    cin>>n;
    for(int i = 0;i<n;i++)
        scanf("%d",&a[i]);

    for(int i = 0;i<n;i++)
        scanf("%d",&b[i]);

    for(int i = 0;i<n;i++)
        scanf("%d",&c[i]);

    sort(a,a+n);
    sort(b,b+n);
    sort(c,c+n);

    for(int i = 0;i<n;i++){

        int l = -1,r = n-1;
        while(l < r){
            int mid = l + r + 1 >> 1;
            if(a[mid] < b[i])
                l = mid ;
            else
                r = mid - 1;

        }

        int t = l + 1;

        l = 0,r = n;
        while(l < r){
            int mid = l + r >> 1;
            if(c[mid] > b[i])
                r = mid;
            else
                l = mid +1;
        }


        res += ll*t * (n-l );

    }

    cout<<res<<endl;
    return 0;
}