UVA 455 periodic Strings

Original question:
A character string is said to having period K if it can be formed by concatenating one or more repetitions of another string of length K. For example, the string "ABCABCABCABC" had period 3, since it was formed by 4 repetitions of the string "abc". It also has periods 6 (both repetitions of "ABCABC") and (one
Repetition of "abcabcabcabc"). Write a program to read a character string and determine its smallest period.
Input
The first line oif the input file would contain a single integer N indicating what many test case which your program would TES T followed by a blank line. Each of the test case would contain a single character string of Non-blank characters. Consecutive input would separated by a blank line.
Output
An integer denoting the smallest period of the input string for each input. The consecutive output is separated by a blank line.
Sample Input
1
Hohoho

Sample Output
2

English
Give you a string that lets you find the smallest cycle

#include <bits/stdc++.h>
using namespace std;
const int maxnum=1000000;
int Next[maxnum];
Char P[maxnum];
int Plen;
void GetNext (char* p,int plen)
{
    Next[0] =-1;
    int k =-1;
    int j = 0;
    while (J < Plen)
    {
        if (k = =-1 | | p[j] = = P[k])
        {
            ++k;
            ++j;
            NEXT[J] = k;
        }
        else
        {
            k = next[k];
}}} int main ()
{
    Ios::sync_with_stdio (false);
    int t;
    cin>>t;
    while (t--)
    {
        cin>>p;
        int ans;
        Plen=strlen (p);
        GetNext (P,plen);
        ans=0;
        if (plen% (Plen-next[plen));
            cout<<plen<<endl;
        else
        {
            ans= (Plen-next[plen]);//plen-next[plen] is to find the length of a looping section
            cout<<ans<<endl;
        }
        if (t)
            cout<<endl;
    }
    return 0;
}

idea:
found a simple string on the purple book to practice practiced hand, and found that KMP can solve ~ using the KMP algorithm prefix array to find the smallest loop section
The total length of the string is Plen
period is plen-next[ Lpen]
First judge is not a periodic string plen% (Plen-next[plen]), if divisible by the period string, otherwise is a non-periodic string, output the original length can be