POJ2406 Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 42637		Accepted: 17787

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.

Source

Waterloo local 2002.07.01

KMP求出next数组，设字符串长度为len，若len/(len-next[len])为整数，那么这个数就是答案。

类似的题见http://www.cnblogs.com/SilverNebula/p/5550595.html

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int next[1000010];
char s[1000010];
int len,ans;
void getnext(){
    int i,j;
    next[0]=0;
    next[1]=0;
    for(i=1,j=0;i<len;i++){
        while(s[j]!=s[i] && j)j=next[j];
        if(s[j]==s[i])j++;
        next[i+1]=j;
    }
    return;
}
int main(){
    while(scanf("%s",s)!=EOF){
        if(s[0]=='.')break;
        len=strlen(s);
//        memset(next,0,sizeof next);
        getnext();
        ans=0;
        if(len%(len-next[len])==0)
            printf("%d
",len/(len-next[len]));
        else printf("1
");
    }
    return 0;
}