nyoj_308_Substring_201405091611

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

来源

第四届河南省程序设计大赛

上传者

张云聪

#include <stdio.h>
#include <string.h>
int map[60][60];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int i,j,len,max=0,t;
        char str1[60],str2[60];
        memset(map,0,sizeof(map));
        scanf("%s",str1);
        len = strlen(str1);
        for(i=0;i<len;i++)
        str2[i]=str1[len-1-i];
        for(i=0;i<len;i++)
        {
            for(j=0;j<len;j++)
            {
                if(str1[i]==str2[j])
                map[i+1][j+1] = map[i][j]+1;
                if(map[i+1][j+1]>max)
                {
                    max = map[i+1][j+1];
                    t = i+1;
                }
            }
        }
        for(i=t-max;i<t;i++)
        printf("%c",str1[i]);
        printf("
");
    }
    return 0;
}

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