POJ 2406 Power Strings (求字符串循环节出现的次数)

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 44217		Accepted: 18449

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

/*
 * POJ 2406 Power Strings
 * 求字符串循环节出现的次数
 * 
 * 结论：字符串S有最小循环节(p<len)的充要条件是(len-next[len])|len,next[len]>0
 * 其中p是循环节长度，len是字符串S的长度
 */

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;

const int MAXN = 1000000+10;

void Get_Next(char* P,int* next)
{
    int i=0,k=-1;
    int plen=strlen(P);
    next[0]=-1;
    while(i<plen)
    {
        if(k==-1||P[k]==P[i])
        {
            i++,k++;
            next[i]=k;
        }
        else k=next[k];
    }
}
char str[MAXN];
int nxt[MAXN];

int main()
{
    while(scanf("%s",str)==1)
    {
        if(str[0]=='.') break;
        Get_Next(str,nxt);
        int n=strlen(str);
        int k=n-nxt[n];
        if(n%k==0&&n/k>1) printf("%d
",n/k);
        else printf("1
");
    }
    return 0;
}