国防科大校赛决赛-final(镜像赛) Problem A: XueXX and P-P String

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

#include <iostream>
#include <string.h>
using namespace std;
bool judge(int n, int m);
char str[200];
int Palindrome(int n)
{
    int i;
    int current = 0;
    for(i = 1; i <= (n+1)/3; i++)
    {
        if(judge(n, i))
            current = i;
    }
    return current;
}
bool judge(int n, int m)
{
    int i = n;
    int 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()
{
    char arr[200];
    int max;
    int m;
    int t;
    cin >>t;
    while(t--)
    {
        max = 0;
        for(int i = 0; i < 200; i++)
            arr[i] = 0;
        cin >> str;
        m = strlen(str);
        for(int i= 2; i < m; i++)
        {
            arr[i] = Palindrome(i);
        }
        for(int i = 2; i< m; i++)
        {
            if(max < arr[i])
                max = arr[i];
        }
        cout << max * 3 << endl;
    }
    return 0;
}