枚举b[i]数组，在a[]数组中找有多少个数比b[i]小，在c[]数组中找有多少个数比b[i]大
然后两个数相乘
这个寻找的过程可以用两种方法实现
二分和前缀和

二分法

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int a[4][N];
int n;
int main()
{
    cin >> n;
    for (int i = 1; i <= 3; i++) {
        for (int j = 0; j < n; j++) {
            cin >> a[i][j];
        }
    }
    sort(a[1], a[1] + n);
    sort(a[3], a[3] + n);

    // for(int i=1;i<=n;i++)
    //     cout<<a[1][i]<<' ';
    // cout<<endl;
    //     for(int i=1;i<=n;i++)
    //     cout<<a[2][i]<<' ';
    // cout<<endl;
    long long ans = 0;
    for (int i = 0; i < n; i++) {
        int num1, num2;
        int l = 0, r = n - 1;
        while (l < r) {
            int mid = l + r + 1 >> 1;
            if (a[1][mid] < a[2][i]) {
                l = mid;
            }
            else r = mid - 1;
        }
        num1 = l;
        if (a[1][l] >= a[2][i])
            l = -1;
        num1 = l;
        l = 0, r = n - 1;
        while (l < r) {
            int mid = l + r >> 1;
            if (a[3][mid] > a[2][i]) {
                r = mid;
            }
            else l = mid + 1;
        }
        num2 = l;
        if (a[3][num2] <= a[2][i]) {
            num2 = n;
        }
        //cout<<num1<<' '<<num2<<endl;
        ans += (long long)(num1 + 1) * (n - num2);
        //cout<<ans<<endl;

    }
    cout << ans << endl;
    //system("PAUSE");
    return 0;
}

前缀和法

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int a[N], b[N], c[N];
int as[N], cs[N], cnt[N], s[N];
int n;
int main() {
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        a[i]++;
    }
    for (int i = 0; i < n; i++) {
        cin >> b[i];
        b[i]++;
    }
    for (int i = 0; i < n; i++) {
        cin >> c[i];
        c[i] ++;
    }
    for (int i = 0; i < n; i++) {
        cnt[a[i]]++;
    }
    for (int i = 1; i < N; i++) {
        s[i] = s[i - 1] + cnt[i];
    }
    for (int i = 0; i < n; i++) {
        as[i] = s[b[i]-1];
    }
    memset(cnt, 0, sizeof(cnt));
    memset(s, 0, sizeof(s));
    for (int i = 0; i < n; i++) {
        cnt[c[i]]++;
    }
    for (int i = 1; i < N; i++) {
        s[i] = s[i - 1] + cnt[i];
    }
    for (int i = 0; i < n; i++) {
        cs[i] = s[N - 1] - s[b[i]];
    }
    long long ans = 0;
    for (int i = 0; i < n; i++) {
        ans += (long long)as[i] * cs[i];
    }
    cout << ans << endl;
    return 0;
}