nyoj 308 Substring

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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

2015,4,17
最长公共子序列

#include<stdio.h>
#include<string.h>
char s[55],c[55];
int a[55][55];
int main(){
	int t,i,j,k,len,m;
	scanf("%d",&t);
	while(t--){
		m=0;
		memset(a,0,sizeof(a));
		scanf("%s",s);
		len=strlen(s);
		for(i=0,j=len-1;i<len;i++,j--)//翻转字符串 
			c[i]=s[j];
		for(i=1;i<=len;i++)
			for(j=1;j<
			=len;j++)
				if(s[i-1]==c[j-1]){
					a[i][j]=a[i-1][j-1]+1;//a[i][j]代表如果s[i-1]=c[j-1]的话，那么a[i][j]就等于a[i-1][j-1]+1； 
					if(m<a[i][j]){//因为如果前两个不匹配，那么a[i-1][j-1]就是0，那么a[i][j]就是1；如果前边两个匹配 
						m=a[i][j];//那么a[i][j]就等于a[i-1][j-1]再加 1，就是2，a[i][j]就是指到ij的时候有几个字符匹配 
						k=i;//然后记录位置，输出就可以了，，这个过程如果还不懂就模拟一下，我也是看大神代码模拟一遍才理解 
					}
				}
		for(i=k-m;i<k;i++)
			printf("%c",s[i]);
		printf("\n");
	}
	return 0;
}

nyoj 308 Substring

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原文地址：http://blog.csdn.net/ling_du/article/details/45098067