算法1

(离散化) $O(nlogn)$

这题还是很容易想的，就是有个细节要注意，电影的使用的语言可能没有科学家会。

C++ 代码

#include <iostream>
#include <algorithm>
using namespace std;
const int N = 200005;
int b[N][2];
int a[N], cp[N], times[N], n, m, tmp;
int main() {
    cin.tie(0), ios::sync_with_stdio(false);
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i], cp[i] = a[i];
    sort(a+1, a+1+n);
    int len = unique(a+1, a+1+n) - a - 1;
    //算出每种语言懂的人的数量
    for (int i = 1; i <= n; i++) times[lower_bound(a+1, a+1+len, cp[i]) - a]++;
    cin >> m;
    for (int i = 1; i <= m; i++) {
        cin >> tmp;
        int p = lower_bound(a+1, a+1+len, tmp) - a;
        if (a[p] != tmp) p = 0;//这里是易错点，可能找不到这门语言
        b[i][0] = times[p];
    }
    for (int i = 1; i <= m; i++) {
        cin >> tmp;
        int p = lower_bound(a+1, a+1+len, tmp)-a;
        if (a[p] != tmp) p = 0;
        b[i][1] = times[p];
    }
    int j = 1;
    for (int i = 2; i <= m; i++) {
        if (b[i][0] > b[j][0]) j = i;
        else if (b[i][0] == b[j][0] && b[i][1] > b[j][1]) j = i;
    }
    cout << j;
    return 0;
}