POJ 2406 Power Strings 求连续重复字串（kmp）

Description

Input

Output

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

Hint

Source

//5248K	110MS
#include<stdio.h>
#include<string.h>
#define M 1000007
int next[M];
char pattern[M];
void pre(int len)
{
	int i = 0, j = -1;
	next[0] = -1;
	while(i != len)
	{
		if(j == -1 || pattern[i] == pattern[j])
			next[++i] = ++j;
		else
			j = next[j];
	}
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(scanf("%s",pattern)!=EOF)
    {
        if(strcmp(pattern,".")==0)break;
        int len=strlen(pattern);
        pre(len);
        if(len%(len-next[len])==0)
            printf("%d\n",len/(len-next[len]));
        else printf("1\n");
    }
    return 0;
}