第四届河南省程序设计大赛 Substring nyoj 308

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

比较坑，第一次以为是求自身最长回文串，感觉省赛签到题好水，很快写完一提交，wa，%>_<%，又把题目读了好几遍（其实根本没读懂），还是不知道哪儿错了，（样例坑爹，老是误导人），后来无奈百度了下，看别人博客把题意看懂了，原来是求正序与倒序的最长公共子序列，如果样例是这样子（453254），结果应该是45，就不是4了，自己想了想，四个for循环，AC~

01. #include<stdio.h>

02. #include<string.h>

03. int main()

04. {

05. int N;

06. scanf("%d",&N);

07. while(N--)

08. {

09. char a[100],b[100];

10. scanf("%s",a);

11. int i,j,k,v,max=0,qi,zhong,n=strlen(a);

12. for(i=0; i<n; i++)

13. b[n-i-1]=a[i];

14. for(i=0; i<n-1; i++)

15. for(j=i+1; j<n; j++)

16. {

17. for(k=0; k<n-(j-i); k++)

18. {

19. int f=0,d=i;

20. for(v=0; v<j-i+1; v++)

21. if(a[d+v]!=b[k+v])

22. {

23. f=1;

24. break;

25. }

26. if(!f&&j-i+1>max)

27. qi=i,zhong=j,max=j-i+1;

28. }

29. }

30. if(max)

31. for(i=qi; i<=zhong; i++)

32. printf("%c",a[i]);

33. else

34. printf("%c",a[0]);

35. printf("\n");

36. }

37. }