参考文献 榜一大哥的题解

//训练离散化算法，当然这道题用map和构造结构体排序更方便
#include<bits/stdc++.h>
using namespace std;
const int N = 2e5+50;
int n,m,a[N],b[N],c[N];
int ans[3*N];
int ans1=-1,ans2=-1;
int id;
vector<int> unq;
//离散化   稀疏编号转换成稠密编号
int find(int x) {
    return lower_bound(unq.begin(),unq.end(),x) - unq.begin();
} 
int main() {
    cin >> n;
    for(int i = 1; i <= n; i++)
    {    
        cin >> a[i];
        unq.push_back(a[i]);
    }
    cin >> m;
    for(int i = 1; i <= m; i++) {
        cin >> b[i];
        unq.push_back(b[i]);
    }
    for(int i = 1; i <= m; i++) {
        cin >> c[i];
        unq.push_back(c[i]);
    }
    //排序后进行去重操作
    sort(unq.begin(),unq.end());
    //unique是将重复的元素丢到unq.end()去，.erase两个参数分别是删除区间的起点和终点+1，即end()
    unq.erase(unique(unq.begin(),unq.end()),unq.end());
    for(int i = 1; i <= n; i++)
    {    
        ans[find(a[i])]++;
    }
    for(int i = 1; i <= m; i++)
    {    
        int t1 = ans[find(b[i])], t2 = ans[find(c[i])];
        if(t1 > ans1 || t1==ans1 && t2> ans2) {
            ans1 = t1;
            ans2 = t2;
            id = i;
        }
    }
    cout << id;
}