NYOJ 308 Substring 字符串处理问题

      这道题我也是花了好长时间才写出来，本来想出来方法了，带式写着太麻烦，所以就一直没写，一直想用string类写的，但是由于对string类中的函数不清楚，所以基本上把每个函数都试了一遍，，，，好在写出来了。这道题题意容易搞错，去年省赛时就有许多队伍理解错题意了，此题不是让求最长回文子串的，而是让求一个最长的子串，该子串的逆串也在原串中。举个例子来说，abcdba,若是求最长回文子串，应为a，对此题来说，应为ab，因为ab的逆串ba也在原串中。题目：

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                   ABCABAXYZXCVCX

样例输出

ABAXXCVCX

ac代码：

#include <iostream>#include <cstdio>#include <string>#include <algorithm>#include <string.h>using namespace std;int main(){//freopen("1.txt","r",stdin);int numcase;scanf("%d",&numcase);string s1,s2,s3;while(numcase--){  cin>>s1;  s2=s1;  int max=0;  reverse(s2.begin(),s2.end());  int len=s1.size();  for(int i=0;i<len;++i){  for(int j=1;j<=len-i;++j){  string::size_type pos=s2.find(s1.substr(i,j));  if(pos!=string::npos){  if(max<j){    max=j;s3=s1.substr(i,j);  }  }  }  }  cout<<s3<<endl;}return 0;}