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;}