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 4 3
就是问一个字符串写成(a)^n的形式,求最大的n.
根据KMP的next函数的性质,已知字符串t第K个字符的next[k],那么d=k-next[k],如果k%d==0,那么t[1……k]最多可均匀的分成k/d份。也就是可以生成一个长度为d的重复度为k/d的字串
根据KMP的next函数的性质,已知字符串t第K个字符的next[k],那么d=k-next[k],如果k%d==0,那么t[1……k]最多可均匀的分成k/d份。也就是可以生成一个长度为d的重复度为k/d的字串
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int maxn=1000005; int m,next[maxn]; char s[maxn]; void getnext() { int i=0,j=0; next[0]=-1; j=next[i]; while(i<m) { if(j==-1||s[i]==s[j]) { next[++i]=++j; } else { j=next[j]; } } } int main() { while(scanf("%s",s)!=EOF) { if(s[0]=='.') break; m=strlen(s); getnext(); int n=1; if(m%(m-next[m])==0) n=m/(m-next[m]); printf("%d\n",n); } return 0; }