POJ——T 2406 Power Strings

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

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 50627		Accepted: 21118

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

 

求最短循环节出现次数

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

using namespace std;

const int N(2333333);
int ans,len,p[N];
char s[N];

inline void Get_next()
{
    for(int i=2,j=0;i<=len;i++)
    {
        for(;s[i]!=s[j+1]&&j>0;) j=p[j];
        if(s[i]==s[j+1]) j++;
        p[i]=j;
    }
}

int main()
{
    for(scanf("%s",s+1);s[1]!='.';scanf("%s",s+1))
    {
        memset(p,0,sizeof(p));
        len=strlen(s+1);
        Get_next();
        int tmp=len-p[len];
        if(len%tmp) puts("1");
        else printf("%d
",len/tmp);
    }
    return 0;
}