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 ;

 }

//最长公共子串

//http://i.cnblogs.com/EditPosts.aspx?postid=3582994

巴特西

nyoj_308_Substring_201405091611

Substring

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