Moscow is hosting a major international conference, which is attended by n scientists from different countries. Each of the scientists knows exactly one language. For convenience, we enumerate all languages of the world with integers from 1 to 109.
In the evening after the conference, all n scientists decided to go to the cinema. There are m movies in the cinema they came to. Each of the movies is characterized by two distinct numbers — the index of audio language and the index of subtitles language. The scientist, who came to the movie, will be very pleased if he knows the audio language of the movie, will be almost satisfied if he knows the language of subtitles and will be not satisfied if he does not know neither one nor the other (note that the audio language and the subtitles language for each movie are always different).
Scientists decided to go together to the same movie. You have to help them choose the movie, such that the number of very pleased scientists is maximum possible. If there are several such movies, select among them one that will maximize the number of almost satisfied scientists.
The first line of the input contains a positive integer n (1 ≤ n ≤ 200 000) — the number of scientists.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the index of a language, which the i-th scientist knows.
The third line contains a positive integer m (1 ≤ m ≤ 200 000) — the number of movies in the cinema.
The fourth line contains m positive integers b1, b2, ..., bm (1 ≤ bj ≤ 109), where bj is the index of the audio language of the j-th movie.
The fifth line contains m positive integers c1, c2, ..., cm (1 ≤ cj ≤ 109), where cj is the index of subtitles language of the j-th movie.
It is guaranteed that audio languages and subtitles language are different for each movie, that is bj ≠ cj.
Print the single integer — the index of a movie to which scientists should go. After viewing this movie the number of very pleased scientists should be maximum possible. If in the cinema there are several such movies, you need to choose among them one, after viewing which there will be the maximum possible number of almost satisfied scientists.
If there are several possible answers print any of them.
3 2 3 2 2 3 2 2 3
2
6 6 3 1 1 3 7 5 1 2 3 4 5 2 3 4 5 1
1
In the first sample, scientists must go to the movie with the index 2, as in such case the 1-th and the 3-rd scientists will be very pleased and the 2-nd scientist will be almost satisfied.
In the second test case scientists can go either to the movie with the index 1 or the index 3. After viewing any of these movies exactlytwo scientists will be very pleased and all the others will be not satisfied.
題意：有很多科學家，每個科學家懂得一種語言，他們一起去看電影，電影聲音和字幕是兩種不同語言，只要能看懂字幕或者聽懂語言，科學家就不會不高興，問看哪個電影可以儘量滿足更多科學家。對於多種情況的，任意輸出一種即可。輸入時聲音一行，字幕一行，豎着兩個爲一組。
思路：用map處理一下各種語言的人數，判斷一下同一種聲音的哪種語言最多，然後在去判斷在最多聲音人數的情況下哪種字幕人最多。可能存在聲音或者字幕科學家們都不懂的情況。
#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#define ll __int64
using namespace std;
int n,m,t;
int x[200009],y[200009],pos[200009];
int a;
map<int,int>mp;
int main()
{
while(~scanf("%d",&n))
{
for(int i=0;i<n;i++)
{
scanf("%d",&a);
mp[a]++;
}
int mmax=0;
int tmp=0;
scanf("%d",&m);
for(int i=0;i<m;i++)
{
scanf("%d",&x[i]);
if(mp[x[i]] && mp[x[i]]>mmax )
{
mmax=mp[x[i]];
tmp=x[i];
}
}
//cout<<"tmp="<<tmp<<endl;
// cout<<"mmax="<<mmax<<endl;
int j=0;
for(int i=0;i<m;i++) //記錄一下懂得聲音最多的情況下科學家的標號
{
if(mp[x[i]] && mmax==mp[x[i]])
pos[j++]=i;
}
for(int i=0;i<m;i++)
{
scanf("%d",&y[i]);
}
int ans=0;
mmax=0;
if(j==0)
{
for(int i=0;i<m;i++)
{
if(mp[y[i]] && mp[y[i]]>mmax)
{
mmax=mp[y[i]];
ans=i;
}
}
printf("%d\n",ans+1);
continue;
}
ans=pos[0];
mmax=mp[y[ans]];
for(int i=0;i<j;i++)
{
if(mp[y[pos[i]]] && mp[y[pos[i]]]>mmax)
{
mmax=mp[y[pos[i]]];
ans=pos[i]; //注意這裏輸出位置
}
}
printf("%d\n",ans+1);
}
return 0;
}