NYOJ 308

 

Substring

时间限制： 1000 ms | 内存限制： 65535 KB

难度： 1

 

描述

You are given a string input. You are to find the longest substring of input such that the reversal of the substring is also a substring of input. In case of a tie, return the string that occurs earliest in input.

Note well: The substring and its reversal may overlap partially or completely. The entire original string is itself a valid substring . The best we can do is find a one character substring, so we implement the tie-breaker rule of taking the earliest one first.

 

输入

The first line of input gives a single integer, 1 ≤ N ≤ 10, the number of test cases. Then follow, for each test case, a line containing between 1 and 50 characters, inclusive. Each character of input will be an uppercase letter ('A'-'Z').

输出

Output for each test case the longest substring of input such that the reversal of the substring is also a substring of input

样例输入

3                   
ABCABA
XYZ
XCVCX

样例输出

ABA
X
XCVCX

//利用最长连续公共子序列长度算法 
#include<stdio.h>
#include<string.h>
int main()
{
	int T,i,j;int len,max,k;
	char a[51],b[51],c[51][51];
	scanf("%d%*c",&T);
	while(T--)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		memset(c,0,sizeof(c));
		scanf("%s",a);
		len=strlen(a);
		for(i=len-1;i>=0;i--)
			b[len-1-i]=a[i];
		max=0;
		for(i=1;i<=len;i++)
			for(j=1;j<=len;j++)
				if(a[i-1]==b[j-1])
				{
					c[i][j]=c[i-1][j-1]+1;
					if(max<c[i][j])
					{
						max=c[i][j];
						k=i;
					}
				}
		/*max=c[len][len];,这条不行*/ 
		for(i=k-max+1;i<=k;i++)
			printf("%c",a[i-1]);
		printf("\n");
	}
	return 0;
}