The 15th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple - B King of Karaoke

King of Karaoke

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Time Limit: 1 Second       Memory Limit: 65536 KB

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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 , and the standard version of the song can be considered as another integer sequence . The score is the number of integers  satisfying  and .

As a good tuner, DreamGrid can choose an integer  (can be positive, 0, or negative) as his tune and add  to every element in . 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  (about 100), indicating the number of test cases. For each test case:

The first line contains one integer  (), indicating the length of the sequences  and .

The second line contains  integers  (), indicating the song performed by DreamGrid.

The third line contains  integers  (), indicating the standard version of the song.

It's guaranteed that at most 5 test cases have .

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  and changes  to .

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

原题地址：http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5753

题意：给你两个数组A，B,问你A数组所有元素同时加一个数或减一个数或不变之后与B数组相同的最大个数；

思路：因为数组A加的数是相同的所以A数组变化之后差值个数还是一样，所以直接计算A数组和B数组之间的差值最多的个数就行，因为可能有负数，所以用map又轻松又简单，map真是个好东西

代码：

#include<bits/stdc++.h>
using namespace std;

int a[5000000];
int b[5000000];
int main()
{
    std::ios::sync_with_stdio(false);
    int t;
    cin>>t;
    while(t--){
        map<int,int>mp;
        int n;
        cin>>n;
        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++){
                mp[a[i]-b[i]]++;
        }
        int maxn=0;
        map<int,int>::iterator it;
        for(it=mp.begin();it!=mp.end();it++){
                if(it->second>maxn){
                        maxn=it->second;
                }
        }
        cout<<maxn<<endl;
    }
    return 0;
}