HDOJ 题目3833 YY's new problem（hash）

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本题要求编写程序判断一个从1到N的排列中是否存在三个元素构成等差数列。输入包括多个测试用例，每个用例包含一个整数N及一个由1到N组成的排列。输出对于每个测试用例，如果存在这样的三个元素，则输出'Y'，否则输出'N'。

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YY's new problem

Time Limit: 12000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4968 Accepted Submission(s): 1388

Problem Description

Given a permutation P of 1 to N, YY wants to know whether there exists such three elements P[i ₁], P[i ₂], P[i ₃] that
P[i ₁]-P[i ₂]=P[i ₂]-P[i ₃], 1<=i ₁<i ₂<i ₃<=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 i ₁, i ₂, i ₃ can be found, else 'N'.

Sample Input

Sample Output

N
Y

Source

2011 Multi-University Training Contest 1 - Host by HNU

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ac代码

#include<stdio.h>
#include<string.h>
int hash[10010];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		int flag=0;
		memset(hash,0,sizeof(hash));
		scanf("%d",&n);
		for(int i=1;i<=n;i++)
		{
			int x;
			scanf("%d",&x);
			hash[x]=1;
			if(flag)
				continue;
			for(int j=1;j<x&&j+x<=n;j++)
			{
				if(hash[x-j]+hash[x+j]==1)
				{
					flag=1;
					break;
				}
			}
		}
		if(flag)
			printf("Y\n");
		else
			printf("N\n");
	}
}