hdu3833（暴力）

YY's new problem

Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 5384    Accepted Submission(s): 1508

Problem Description

Given a permutation P of 1 to N, 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

2
3
1 3 2
4
3 2 4 1

 

Sample Output

N
Y

暴力要有方法。。。。。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
using namespace std;
typedef long long ll;
/*const int M=200000;
int ip;
int head[M],cd[M],rd[M];
struct data
{
    int v,next;
}tu[M];
void init(int n)
{
    memset(head,-1,sizeof(head));
    memset(rd,0,sizeof(rd));
    memset(cd,0,sizeof(cd));
    ip=0;
}
void add(int u,int v)
{
    tu[ip].v=v,tu[ip].next=head[u],head[u]=ip++;
}
*/
int zb[10010],a[10010];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        for(int i=0; i<n; i++)
            {scanf("%d",&a[i]);zb[a[i]]=i;}
        int flag=0;
        for(int i=0; i<n; i++)
        {
            for(int j=i+1; j<n; j++)
            {
                if((a[i]+a[j])%2==0)
                {
                    if(zb[(a[i]+a[j])/2]>i&&zb[(a[i]+a[j])/2]<j)
                    {
                        flag=1;
                        break;
                    }
                }
            }
            if(flag)break;
        }
        if(flag)printf("Y\n");
        else printf("N\n");
    }
    return 0;
}