SUBSTRING
Problem Description
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.
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.
Input
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').
Output
Output for each test case the longest substring of input such that the reversal of the substring is also a substring of input
Sample Input
3 ABCABA XYZ XCVCX
Sample Output
ABA X XCVCX
注意:题目意思不是求最长连续可逆字符串
#include<iostream>
#include<string>
#include<algorithm>
#include<string.h>
using namespace std;
int main()
{
int n,i,j,flag;
string s1,ss;
cin>>n;
string s;
while(n--)
{
cin>>s;
flag=0;
for(i=s.length();i>0;i--)
{
for(j=0;j+i<=s.length();j++)
{
ss=s.substr(j,i);
s1=ss;
reverse(s1.begin(),s1.end());
if(s.find(s1)!=-1) {cout<<ss<<endl;flag=1;break;}
}
if(flag==1) break;
}
}
return 0;
}
为了明天所以选择坚定的执着今天。