POJ 2406

运用KMP 的 next数组寻找串的循环节循环的最多次数

思路：KMP中的get_next()，对next数组的应用。next[len]是最后一个字符跳的步长，如果他有相同字符串，则该串长度是len-next[len]，如果整个长度len能分解成x个这种串（能整除），就得到ans了。否则不能分解。只能是由他自己组成串，长度为1。

Power Strings

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 45348 Accepted: 18927

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

abcdaaaaababab.

Sample Output

143

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

Source

Waterloo local 2002.07.01

#include<stdio.h>#include<string.h>char str[1123456];int next[1123456];void get_next(){    next[0]=-1;    int i=0;    int j=-1;    while(str[i]!='\0')    {        if(j==-1 || str[i]==str[j])        {            i++;            j++;            next[i]=j;        }        else        {            j=next[j];        }    }}int main(){    while(~scanf("%s",str) && strcmp(str,".")!=0)    {        get_next();        int len=strlen(str);        int ans=1;        if(len%(len-next[len])==0)        {            ans=len/(len-next[len]);        }        printf("%d\n",ans);    }    return 0;}