1 /*Substring
 2 时间限制：1000 ms  |  内存限制：65535 KB 
 3 难度：1
 4 描述 
 5 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, 
 6 return the string that occurs earliest in input. 
 7 
 8 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
 9  one character substring, so we implement the tie-breaker rule of taking the earliest one first.
10 
11 输入
12 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').
13 输出
14 Output for each test case the longest substring of input such that the reversal of the substring is also a substring of input
15 样例输入
16 3                   
17 ABCABA
18 XYZ
19 XCVCX
20 样例输出
21 ABA
22 X
23 XCVCX
24 来源
25 第四届河南省程序设计大赛
26 上传者
27 张云聪
28 
29 */
30 #include<stdio.h>
31 #include<string.h>
32 int main()
33 {
34     int n, i, j, k, m, real, l;
35     char a[52], b[52];
36     scanf("%d", &n);
37     while(n--)
38     {
39         real = 0;
40         scanf("%s", a);
41         m = strlen(a);
42         for(i = m; i>=1; i--){   // 字符串长度 
43         for(j = 0; j <= m-i; j++)// 字符串起始位置 
44         {
45             for(k = 0; k < i; k++)
46             b[k] = a[i+j-k-1];
47             b[i] = '\0';             
48             if( strstr( a , b ) )    //求出substring 
49             {
50                 for(l = i-1; l>=0; l--)
51                 printf("%c", b[l]);      // 输出 
52                 printf("\n");
53                 real = 1;
54                 break;
55             }
56         }
57         if(real)   break;
58         }
59     }
60     return 0;
61 }