（nyoj308）substring

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

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;        }    }}