NYOJ，

NYOJ，

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

#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

int main()
{
	string s1,s2;
	int n;
	cin >> n;
	while(n--)
	{
		cin >> s1;
		s2 = s1;
		reverse(s2.begin(),s2.end());  // 反转字符串 
		bool flag = false;
		for(int i = s1.size(); i > 0; i--)   // 可以从最长字符串遍历 
		{
			for(int j = 0; j <= s1.size()-i; j++)
			{
				string t = s1.substr(j,i);
				string::size_type pos = s2.find(t);
				if(pos != string::npos)
				{
					cout << t << endl;
					flag = true;
					break;
				}
			}
			if(flag) break;
		}
	}
	return 0;
}