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
14
3
这题就是求最小的循环节，直接用len-next[len] && next[len]!=0就行，，对于len%(len-next[len])!=0要输出0.
#include<stdio.h>#include<string.h>char s[1000006];int len,next[1000006];void nextt(){int i,j;i=0;j=-1;memset(next,-1,sizeof(next));while(i<len){if(j==-1 || s[i]==s[j]){i++;j++;next[i]=j;}else j=next[j];}}int main(){int n,m,i,j;while(scanf("%s",s)!=EOF){if(strcmp(s,".")==0)break;len=strlen(s);//printf("%d\n",len);nextt();if(len%(len-next[len])==0 && next[len]!=0){printf("%d\n",len/(len-next[len]));}else printf("1\n");}}
人生没有彩排，只有现场直播，所以每一件事都要努力做得最好
