poj2406--Power Strings（KMP求最小循环节）

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 33178		Accepted: 13792

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

用next数组求出整个数组的最大前缀。假设整个串是用循环节组成的，那么 n - next[n] 也就是最小循环节，验证最小循环节会被n整出。

#include <cstdio>
#include <cstring>
#include <algorithm>
int next[1100000] ;
char str[1100000] ;
void getnext(int l)
{
    int j = 0 , k = -1 ;
    next[0] = -1 ;
    while(j < l)
    {
        if( k == -1 || str[j] == str[k] )
        {
            j++ ;
            k++ ;
            next[j] = k ;
        }
        else
            k = next[k] ;
    }
}
int main()
{
    int l , m ;
    while(scanf("%s", str)!=EOF)
    {
        if( str[0] == '.' ) break;
        l = strlen(str);
        getnext(l) ;
        m = next[l];
        if( l % (l-m) != 0 )
            printf("1
");
        else
        {
            m = l / ( l-m );
            printf("%d
", m);
        }
        memset(str,0,sizeof(str));
    }
    return 0;
}