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;
}
}
}