北大ACM poj2406 Power Strings

Power Strings

 

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

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

 

 

/*本题还可以用字符串匹配的经典算法KMP来解决。我觉得KMP很难理解，不过的确是个经典算法。这个问题，只需要用到KMP的构造，也就是对待搜索字符串的自身匹配就可以了KMP是一种在长字符串中匹配短字符串的无回溯算法，它特别适合待搜索字符串自身有重复的情况，就是像这道题中的abababab之类的串。使用一个辅助数组next[i]记录当下标为i的字符匹配失败时，下一次匹配开始的下标（这里下标从0开始）。如此完成next的构造，并多计算一位，设s的长度为len，则a的长度就是len-next[len]，只需再做一个除法就可以完成n的求解了*/#include <stdio.h>#include <string.h>char s[1000001];int next[1000001];int main(void){    int i,j,len,n;    while(scanf("%s",s),s[0]!='.'){        len=strlen(s);        //KMP构造        i=0,j=-1;        next[0]=-1;        while(i<len){            if(j==-1||s[i]==s[j]){                i++,j++;                if(s[i]!=s[j]) next[i]=j;                else next[i]=next[j];                printf("next[%d]=%d\n",i,next[i]);            }else j=next[j];        }        //判断是否能整除，决定n的值        //比如abababa，不能整除，n就应该为1，否则就成了错误的答案3        if(len%(i=len-next[len])==0)            n=len/i;        else n=1;        printf("%d\n",n);    }    return 0;}