离散化+排序

离散化统计人数就好，本来不难，但是测试点太丧心病狂了。。。CF还是大哥啊

#include <bits/stdc++.h>

#define INF 0x3f3f3f3f

using namespace std;

typedef long long ll;

inline int lowbit(int x){ return x & (-x); }

inline int read(){

    int X = 0, w = 0; char ch = 0;

    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }

    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();

    return w ? -X : X;

}

inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }

inline int lcm(int a, int b){ return a / gcd(a, b) * b; }

template<typename T>

inline T max(T x, T y, T z){ return max(max(x, y), z); }

template<typename T>

inline T min(T x, T y, T z){ return min(min(x, y), z); }

template<typename A, typename B, typename C>

inline A fpow(A x, B p, C yql){

    A ans = 1;

    for(; p; p >>= 1, x = 1LL * x * x % yql)if(p & 1)ans = 1LL * x * ans % yql;

    return ans;

}

const int N = 200005;

int a[N], b[N<<2], tot, num[N<<2];

struct Film{

    int x, y, cnt1, cnt2, id;

    bool operator < (const Film &rhs) const {

        if(cnt1 != rhs.cnt1) return cnt1 > rhs.cnt1;

        if(cnt2 != rhs.cnt2) return cnt2 > rhs.cnt2;

        return id < rhs.id;

    }

}film[N];

int main(){

    int n = read();

    for(int i = 0; i < n; i ++) a[i] = read(), b[++tot] = a[i];

    int m = read();

    for(int i = 0; i < m; i ++) film[i].x = read(), b[++tot] = film[i].x, film[i].id = i;

    for(int i = 0; i < m; i ++) film[i].y = read(), b[++tot] = film[i].y;

    sort(b + 1, b + tot + 1);

    tot = unique(b + 1, b + tot + 1) - b - 1;

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

        num[lower_bound(b + 1, b + tot + 1, a[i]) - b] ++;

    }

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

        film[i].cnt1 = num[lower_bound(b + 1, b + tot + 1, film[i].x) - b];

        film[i].cnt2 = num[lower_bound(b + 1, b + tot + 1, film[i].y) - b];

    }

    sort(film, film + m);

    printf("%d\n", film[0].id + 1);

    return 0;

}