nyoj_308_Substring_201405091611

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Substring

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

 1 #include <stdio.h>
 2 #include <string.h>
 3 int map[60][60];
 4 int main()
 5 {
 6     int T;
 7     scanf("%d",&T);
 8     while(T--)
 9     {
10         int i,j,len,max=0,t;
11         char str1[60],str2[60];
12         memset(map,0,sizeof(map));
13         scanf("%s",str1);
14         len = strlen(str1);
15         for(i=0;i<len;i++)
16         str2[i]=str1[len-1-i];
17         for(i=0;i<len;i++)
18         {
19             for(j=0;j<len;j++)
20             {
21                 if(str1[i]==str2[j])
22                 map[i+1][j+1] = map[i][j]+1;
23                 if(map[i+1][j+1]>max)
24                 {
25                     max = map[i+1][j+1];
26                     t = i+1;
27                 }
28             }
29         }
30         for(i=t-max;i<t;i++)
31         printf("%c",str1[i]);
32         printf("\n");
33     }
34     return 0;
35 }

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nyoj_308_Substring_201405091611

标签：style   blog   class   code   java   color   

原文地址：http://www.cnblogs.com/xl1027515989/p/3719326.html