国防科大A

Description

XueXX is a clever boy. And he always likes to do something with Palindrome String. What an interesting hobby!

A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. A P-P String is a string, which can be divided into three parts with the same length, and part one jointing(拼接) part two generates a palindrome,
and part two jointing part three generates a palindrome. That is to say, the string ”abccbaabc” is a P-P String because “abccba” is a palindrome and “cbaabc” is a palindrome.

XueXX’s friend Star can solve the problem whose string’s length is less than 100000. But XueXX cannot. Now give you a short string and can you help XueXX find the longest P-P String?

Input

The first line of input contains the number of test cases T(T<=10). The descriptions of the test cases follow: The first line of each test case contain a string which contains only lowercase letters. Note that the length of the string is less than 200.

Output

For each test case, output a single line containing the result standing the longest length.

Sample Input

3

aaabccbaabc

xxxxxxxxx

abcdefg

Sample Output

9

9

0

题目要求计算最大的回文重叠数，就是说给定的字符串，你去计算里面的回文数，然后两个如果重叠了，就算出第一个的开始位置和最后一个的位置，统计中间数

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int str[200];
int huiwen(int n)
{
int i;

int ans=0;
for(i=1;i<=(n+1)/3;i++)
{
if(jis(n,i))
ans=i;
}
return ans;
}
void jis(int n,int m)
{
int i=n,j=m;
for(j=1;j<=m;j++,i--)
{
if(str[i]==str[i-2*m]&&str[i]==str[i-2*m+2*j-1])
continue;
else
return 0;
}
return 1;
}
int main()
{
int t,max=0,m;
int i;

char a[200];
while(scanf("%d",&t)==1)
{
while(t--)
{
cin>>str;
m=strlen(str);
for(i=2;i<m;i++)
{
a[i]=huiwen(i);
}
for(i=2;i<m;i++)
{
if(max<a[i])
max=a[i];
}
cout<<max<<endl;
}
}
return 0;
}