King of Karaoke

King of Karaoke
Time Limit: 1 Second Memory Limit: 65536 KB
It's Karaoke time! DreamGrid is performing the song Powder Snow in the game King of Karaoke. The song performed by DreamGrid can be considered as an integer sequence \(D_1, D_2, \dots, D_n\), and the standard version of the song can be considered as another integer sequence \(S_1, S_2, \dots, S_n\). The score is the number of integers \(i\) satisfying \(1 \le i \le n\) and \(S_i = D_i\).

As a good tuner, DreamGrid can choose an integer \(K\) (can be positive, 0, or negative) as his tune and add \(K\) to every element in \(D\). Can you help him maximize his score by choosing a proper tune?

Input
There are multiple test cases. The first line of the input contains an integer \(T\) (about 100), indicating the number of test cases. For each test case:

The first line contains one integer \(n\) (\(1 \le n \le 10^5\)), indicating the length of the sequences \(D\) and \(S\).

The second line contains \(n\) integers \(D_1, D_2, \dots, D_n\) (\(-10^5 \le D_i \le 10^5\)), indicating the song performed by DreamGrid.

The third line contains \(n\) integers \(S_1, S_2, \dots, S_n\) (\(-10^5 \le S_i \le 10^5\)), indicating the standard version of the song.

It's guaranteed that at most 5 test cases have \(n > 100\).

Output
For each test case output one line containing one integer, indicating the maximum possible score.

Sample Input
2
4
1 2 3 4
2 3 4 6
5
-5 -4 -3 -2 -1
5 4 3 2 1
Sample Output
3
1
Hint
For the first sample test case, DreamGrid can choose \(K = 1\) and changes \(D\) to \(\{2,3,4,5\}\).

For the second sample test case, no matter which \(K\) DreamGrid chooses, he can only get at most 1 match.

【题意】：通过在第一个序列中的每个元素增加k/减去k/不变来最大化分数，分数为上下序列对应相等的个数。

#include<bits/stdc++.h>

using namespace std;
int a[1000005];
int b[1000005];
vector<int> str;
map<int,int> mp;
int main()
{
    int t,n,num;
    cin>>t;
    while(t--){
        int f=1;
        cin>>n;
        mp.clear();
        for(int i=0;i<n;i++){
            cin>>a[i];
        }
        for(int i=0;i<n;i++){
            cin>>b[i];
        }
        for(int i=0;i<n;i++){
            if(a[i]!=b[i])  f=0;

        }
        if(f) cout<<n<<endl;
        if(!f){
        for(int i=0;i<n;i++){
            mp[b[i]-a[i]]++; //map里面int可以放负数作为value-key
        }
        num=0;
        int Max=0;
        for(map<int,int>::iterator it = mp.begin();it!=mp.end();it++)
            if(it->second > num)
            {
                num = it->second;
            }
            cout<<num<<endl;
       }
    }

}
/*
sort(v.begin(),v.end(),less<int>());
        num=v[0];
        int count=1,min_num=v[0],min_count=1;
        for(i=1;i<N-1;i++)
        {
            if(v[i]>v[i-1])
            {
                count=1;
                if(count>min_count)
                   min_num=v[i];
            }
            else if(v[i]==v[i-1])
            {
                count++;
                if(count>min_count)
                {
                    min_count=count;
                    min_num=v[i];
                }
            }
        }

        if(v[N-1]==min_num)
            min_count++;

        cout<<min_num<<min_count<<endl;

        /*
        sort(v.begin(),v.end());
        int Max=0,num=1,value=v[0],pre=v[0];
        int cur;
        //-1 -1 -1 2
        for(int i=1;i<v.size();i++){
            cur=v[i];
            if(cur==pre) num++;
            else{
                if(num>Max){
                    Max=num;
                    value=pre;
                }
                num=1;
                pre=cur;
            }
        }
        cout<<Max<<endl;
        */


        /*
        int Max=1;
        for(int i=0;i<n;i++){
            num = count(v.begin(),v.end(),a[i]);
            Max=max(num,Max);
        }
        */