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 43
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"); } }