PKU 2406（KMP算法的靈活應用）

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 13798		Accepted: 5754

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 <stdio.h> #include <string.h> #define maxn 1000005 int next[maxn]; int len; void getnext(char str[]) { next[0] = -1; int i=0; int j=-1; while (i<len) { if (j==-1||str[i]==str[j]) { i++; j++; next[i] = j; } else { j=next[j]; } } } int main() { char str[maxn]; while (scanf("%s",str)) { if(strcmp(str,".") == 0) break; len = strlen(str); getnext(str); if(next[len]==0) { printf("1/n"); continue; } printf("%d/n",len%(len-next[len])==0?len/(len-next[len]):1); } return 0; }