电影

https://www.acwing.com/problem/content/105/

思路：这里可以采用离散化/map的方法来记录每种语言对应的人数，然后遍历每部电影进行比较，分别用两个变量记录当前某部电影喜欢和比较喜欢的人数进行更新

离散化：

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;

int n, m;
int a[N], b[N], c[N];
vector<int> alls, add;

int find(int x){
    int l = 0, r = alls.size() - 1;
    while(l < r){
        int mid = l + r >> 1;
        if(alls[mid] >= x) r = mid;
        else l = mid + 1;
    }
    if(alls[l] != x) return 0;
    return l + 1;
}

int main(){
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n;
    for(int i = 1; i <= n; i ++){
        int x;
        cin >> x;
        alls.push_back(x);
        add.push_back(x);
    }
    cin >> m;
    for(int i = 1; i <= m; i ++) cin >> b[i];
    for(int i = 1; i <= m; i ++) cin >> c[i];
    // 离散化
    sort(alls.begin(), alls.end());
    alls.erase(unique(alls.begin(), alls.end()), alls.end());
    for(auto x : add) a[find(x)] ++;

    int k1 = 0, k2 = 0, res = 1;
    for(int i = 1; i <= m; i ++){
        int x = a[find(b[i])], y = a[find(c[i])];
        if(x > k1) k1 = x, k2 = y, res = i;
        else if(x == k1 && y > k2) k1 = x, k2 = y, res = i;
    }
    cout << res << '\n';

    return 0;
}

map：

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;

int n, m;
int a[N], b[N], c[N];
map<int, int> mp;

int main(){
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n;
    for(int i = 1; i <= n; i ++){
        cin >> a[i];
        mp[a[i]] ++;
    }
    cin >> m;
    for(int i = 1; i <= m; i ++) cin >> b[i];
    for(int i = 1; i <= m; i ++) cin >> c[i];

    int k1 = 0, k2 = 0, res = 1;
    for(int i = 1; i <= m; i ++){
        if(mp[b[i]] > k1){
            k1 = mp[b[i]];
            k2 = mp[c[i]];
            res = i;
        }
        else if(mp[b[i]] == k1){
            if(mp[c[i]] > k2){
                k2 = mp[c[i]];
                res = i;
            }
        }
    }
    cout << res << '\n';

    return 0;
}