poj 2406

http://poj.org/problem?id=2406

Power Strings Time Limit: 3000MS Memory Limit: 65536KTotal Submissions:66981 Accepted: 27644

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

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed. #include<cstdio> #include<cstring> const int maxn=1e7+3; int len; int next[maxn]; char str[maxn]; void getnext(){ int k=-1; int j=0; next[0]=-1; while(j<len){ if(k==-1||str[j]==str[k]){ k++; j++; next[j]=k; }else{ k=next[k]; } } return; } int main(){ while(scanf("%s",str)&&str[0]!='.'){ len=strlen(str); getnext(); int tmp=len-next[len]; if(len%tmp!=0){ printf("1\n"); }else{ printf("%d\n",len/tmp); } } return 0; }

 

 

转载于:https://www.cnblogs.com/qqshiacm/p/11443798.html

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