hdu 3833 YY's new problem

YY's new problem

Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 5443    Accepted Submission(s): 1530

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

 

Source

2011 Multi-University Training Contest 1 - Host by HNU

 

思路：

對於一個1到n的排列，如果前K個數裏沒有X，那麼剩餘的數裏一定含有X。

從它給的式子我們可以知道2P[i2]=P[i1]+P[i3],即P[i1]和P[i3]關於P[i2]對稱。

首先將h數組標記爲0，之後讀入元素x，每讀入一個元素，就將該元素對應的下標的h數組賦值爲1。接着我們在h數組中以該元素的對稱前方和後方找，即h[x-j]和h[x+j]，如果h[x+j]+h[x-j]==1，那麼就找到了序列。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int h[20010];//記錄1到n是否出現了，出現了記爲1。
int main()
{
    int t,n,i,j,x;
    scanf("%d",&t);
    while(t--)
    {
        int flag=0;
        scanf("%d",&n);
        memset(h,0,sizeof(h));
        for(i=1;i<=n;i++)
        {
            scanf("%d",&x);
            h[x]=1;
            if(flag) continue;
            for(j=1;x-j>0&&x+j<=n;j++)
            {
                if(h[x-j]+h[x+j]==1)//p[i1],p[i3]之中出現一個
                {
                    flag=1;break;
                }
            }
        }
        if(flag) printf("Y\n");
        else printf("N\n");
    }
    return 0;
}