hdu 3833 YY's new problem hash

Problem Description

Given a permutation P of 1 to N  (表明出现的数字从1到N，只是顺序打乱), YY wants to know whether there exists such three elements P[i₁], P[i₂], P[i₃] that   
P[i₁]-P[i₂]=P[i₂]-P[i₃], 1<=i₁<i₂<i₃<=N.

 

Input

The first line is T(T<=60), representing the total test cases.
Each test case comes two lines, the former one is N, 3<=N<=10000, the latter is a permutation of 1 to N.

 

Output

For each test case, just output 'Y' if such i₁, i₂, i₃ can be found, else 'N'

 

Sample Input

231 3 243 2 4 1

 

Sample Output

NY

由上面的红字可知，对于一个数p[i2] ,如果p[i1]在前面出现一次

 hash[a]=1;

则p[i3]在后面肯定会出现，所以我们只需要找到p[i1]（即hash[p[i1]]=1）,并且要求p[i3]在前面没有出现(hash[p[i1]]=0) 或者反过来 于是

if(hash[a-j]+hash[a+j]==1) //表示找到了

#include <iostream>
#include <cstring>
#include <queue>
#include <vector>
#define N 23541
using namespace std;

int main()
{
   // priority_queue<int,vector<int>,greater<int> > q;
    int num,m,a,hash[10000];
    cin>>num;
    while(num--)
    {
        cin>>m;
        memset(hash,0,sizeof(hash));
        int flag=0;

        for(int i=0;i<m;i++)
        {
            cin>>a;
           hash[a]=1;
           if(!flag){
           for(int j=1;a-j>0&&j+a<m;j++)
           {
               if(hash[a-j]+hash[a+j]==1)
               {
                   flag=1;
                   break;
               }
           }
           }
        }
        if(flag) cout<<"Y"<<endl;
        else cout<<"N"<<endl;


    }

    return 0;
}