poj-2406 Power Strings-蒲公英云

poj-2406 Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 65854		Accepted: 27197

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.

思路：next[i]表示以第i个字符结束的最大后缀和最大前缀相等。n-next[n]则为最小的循环节。若n%(n-next[n])==0,那么n/(n-next[n])则为循环节出现的最大次数。

#include <iostream>
#include<string.h>
using namespace std;
const int maxn = 1e6+10;
char a[maxn];
int ne[maxn];
int main()
{
    while(scanf("%s",a+1)&&a[1]!='.')
    {
        memset(ne,0,sizeof(ne));
        int len = strlen(a+1);
        for(int i=2,j=0;i<=len;i++)
        {
            while(j&&a[i]!=a[j+1]) j=ne[j];
            if(a[i]==a[j+1]) j++;
            ne[i]=j;
        }
        if(len%(len-ne[len])==0) printf("%d\n",len/(len-ne[len]));
        else
            printf("1\n");
    }
    return 0;
}