poj 2406

http://poj.org/problem?id=2406

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions:66981		Accepted: 27644

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<cstdio>
#include<cstring>
const int maxn=1e7+3;
int len;
int next[maxn];
char str[maxn];
void getnext(){
    int k=-1;
    int j=0;
    next[0]=-1;
    while(j<len){
        if(k==-1||str[j]==str[k]){
            k++;
            j++;
            next[j]=k;
        }else{
            k=next[k];
        }
    }
    return;
}
int main(){
    while(scanf("%s",str)&&str[0]!='.'){
        len=strlen(str);
        getnext();
         int tmp=len-next[len];
        if(len%tmp!=0){
            printf("1
");
        }else{
            printf("%d
",len/tmp);
        }
    }
    return 0;
}