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[][];
 int main()
 {
     int T;
     scanf("%d",&T);
     while(T--)
     {
         int i,j,len,max=,t;
         char str1[],str2[];
         memset(map,,sizeof(map));
         scanf("%s",str1);
         len = strlen(str1);
         for(i=;i<len;i++)
         str2[i]=str1[len--i];
         for(i=;i<len;i++)
         {
             for(j=;j<len;j++)
             {
                 if(str1[i]==str2[j])
                 map[i+][j+] = map[i][j]+;
                 if(map[i+][j+]>max)
                 {
                     max = map[i+][j+];
                     t = i+;
                 }
             }
         }
         for(i=t-max;i<t;i++)
         printf("%c",str1[i]);
         printf("\n");
     }
     return ;
 }

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