poj2406 Power Strings(最大重复子串)

K - Power Strings

Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;
const int maxm=1000500;
char p[maxm];
int m,next[maxm];
void getnext()
{
    int i=0,j=next[0]=-1;
    while(i<m)
    {
        if(j==-1||p[i]==p[j])
        {
            ++i;++j;
            next[i]=p[i]!=p[j]?j:next[j];
        }
        else j=next[j];
    }
}

int main()
{
    while(~scanf("%s",p))
    {
        if(!strcmp(p,"."))break;
        m=strlen(p);
        getnext();
        int cc=1;
        if(m%(m-next[m])==0)
            cc=m/(m-next[m]);
        cout<<cc<<endl;
    }
    return 0;
}