（nyoj308）substring，nyoj308substring

（nyoj308）substring，nyoj308substring

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

这题很多人跪在题意上，表示我也贡献了一次wa，就是没读懂题意，这道题的意思是找最长子串反过来还是其子串，不是找最长的回文串，举个例子

abcdeba输出ab，因为ab是可以反过来的最长子串；

代码：

#include<iostream>
#include<cstdio>
#include<string>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<vector>
#include<cstring>
#include<stdlib.h>
#include<ctype.h>
using namespace std;


int main()
{
    string ch1;
    int t,maxlen=1,ans;
    cin>>t;
    while(t--)
    {
        cin>>ch1;
        string ch2=ch1;
        reverse(ch2.begin(),ch2.end());
        bool flag=0;
        for(int i=ch2.size();i>0;i--)
        {
             for(int j=0;j<=ch2.size()-i;j++)
            {
                string v=ch1.substr(j,i);
                if(ch2.find(v)!=string::npos)
                {
                    cout<<v<<endl;
                    flag=1;
                    break;
                }
            }
            if(flag==1)break;
        }


    }
}