POJ - 2406 Power Strings

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.

KMP的循环体问题

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>

using namespace std; 

char s[1000005];
int Next[1000005];

void getnext(char* s,int m)
{
    Next[0]=0;
    Next[1]=0;
    for(int i=1;i<m;i++)
    {
        int j=Next[i];
        while(j&&s[i]!=s[j])
            j=Next[j];
        if(s[i]==s[j])
            Next[i+1]=j+1;
            else
                Next[i+1]=0;
    }
}

int main()
{
    while(scanf("%s",s))
    {
        if(s[0]=='.'&&s[1]=='\0')
            break;
        memset(Next,0,sizeof(Next));
        int m=strlen(s);
        getnext(s,m);
        int x=1;
        if(m%(m-Next[m])==0)
            x=m/(m-Next[m]);
        printf("%d\n",x);
    }
    
    
    return 0;
}