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[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.
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.
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
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：
对于一个1到n的排列，如果前K个数里没有X，那么剩余的数里一定含有X。
从它给的式子我们可以知道2P[i2]=P[i1]+P[i3],即P[i1]和P[i3]关于P[i2]对称。
首先将h数组标记为0，之后读入元素x，每读入一个元素，就将该元素对应的下标的h数组赋值为1。接着我们在h数组中以该元素的对称前方和后方找，即h[x-j]和h[x+j]，如果h[x+j]+h[x-j]==1，那么就找到了序列。
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int h[20010];//记录1到n是否出现了，出现了记为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]之中出现一个
{
flag=1;break;
}
}
}
if(flag) printf("Y\n");
else printf("N\n");
}
return 0;
}