This is the easy version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.
You are given a string s, consisting of lowercase English letters. Find the longest string, t, which satisfies the following conditions:
The length of t does not exceed the length of s.
t is a palindrome.
There exists two strings a and b (possibly empty), such that t=a+b ( “+” represents concatenation), and a is prefix of s while b is suffix of s.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤t≤1000), the number of test cases. The next t lines each describe a test case.
Each test case is a non-empty string s, consisting of lowercase English letters.
It is guaranteed that the sum of lengths of strings over all test cases does not exceed 5000.
Output
For each test case, print the longest string which satisfies the conditions described above. If there exists multiple possible solutions, print any of them.
Example
Input
5
a
abcdfdcecba
abbaxyzyx
codeforces
acbba
Output
a
abcdfdcba
xyzyx
c
abba
Note
In the first test, the string s=“a” satisfies all conditions.
In the second test, the string “abcdfdcba” satisfies all conditions, because:
Its length is 9, which does not exceed the length of the string s, which equals 11.
It is a palindrome.
“abcdfdcba” = “abcdfdc” + “ba”, and “abcdfdc” is a prefix of s while “ba” is a suffix of s.
It can be proven that there does not exist a longer string which satisfies the conditions.
In the fourth test, the string “c” is correct, because “c” = “c” + “” and a or b can be empty. The other possible solution for this test is “s”.
思路：简单版本还是很好想的。首先，我们把前缀后缀最大匹配的找出来，然后中间的，再去寻找最长回文前缀和最长回文后缀，比较长短，取最长。
代码如下：
#include<bits/stdc++.h>
#define ll long long
using namespace std;
string s,t1,t2,s1,s2,ss,_max;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
cin>>s;
int l=0,r=s.length()-1,L,R;
while(l<r&&s[l]==s[r]) l++,r--;
L=l,R=r;
if(l>r) swap(l,r),l++;
else if(l<r) r++;
t1=s.substr(0,l);
t2=s.substr(r);
if(L<R)
{
ss=s.substr(L,R-L+1);
l=0,r=ss.length()-1;
while(l<=r)
{
if(ss[l]==ss[r])
{
s1=ss.substr(l,r-l+1);
s2=s1;
reverse(s1.begin(),s1.end());
if(s1==s2) break;
}
r--;
}
_max=t1+s1+t2;
l=0,r=ss.length()-1;
while(l<=r)
{
if(ss[l]==ss[r])
{
s1=ss.substr(l,r-l+1);
s2=s1;
reverse(s1.begin(),s1.end());
if(s1==s2) break;
}
l++;
}
string f=t1+s1+t2;
if(f.length()>_max.length()) _max=f;
}
else _max=t1+t2;
cout<<_max<<endl;
}
return 0;
}
努力加油a啊，(o)/~
