POJ 2406 Power Strings

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

143

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

Source

Waterloo local 2002.07.01

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KMP~

这里的字符串是从0开始的，所以求next[i]的时候k要-1~

很巧妙的用了next数组~

（输入字符串s可以不加&~）

#include<cstdio>#include<cstring>#define maxn 1000006int len,k,next[maxn];char s[maxn];void g(char u[],int v){for(int i=1;i<v;i++){k=next[i-1];while(k>0 && u[i]!=u[k]) k=next[k-1];if(u[i]==u[k]) next[i]=k+1;else next[i]=0;}}int main(){scanf("%s",s); while(s[0]!='.'){len=strlen(s);next[0]=0;g(s,len);if(len%(len-next[len-1])==0) printf("%d\n",len/(len-next[len-1]));else printf("1\n");scanf("%s",s);}return 0;}