nyoj308，

nyoj308，

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

上传者

张云聪

string的函数的使用

代码如下：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<stack>
using namespace std;

int main(){
    int n,len;
    string s,s1;
    scanf("%d",&n);
    while(n--){
        cin>>s;
        len=s.length();
        string ans;
        for(int i=0;i<len;i++){
            for(int j=1;j<=len;j++){
                s1=s.substr(i,j);
                string s2(s1.rbegin(),s1.rend());
                if(ans.length()<s1.length()&&s.find(s1)!=string::npos&&s.find(s2)!=string::npos){
                    ans=s1;
                }
            }
        }
        cout<<ans<<endl;
    }
}