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.
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<map> 5 #define maxn 10010000 6 using namespace std; 7 char p[maxn]; 8 int next[maxn],len; 9 int kmp(){ 10 int i=0,j=-1; 11 next[0]=-1; 12 while(i<len){ 13 if(j==-1||p[i]==p[j]) next[++i]=++j; 14 else j=next[j]; 15 } 16 } 17 int main() 18 { 19 while(scanf("%s",p)!=EOF&&p[0]!='.'){ 20 len=strlen(p); 21 kmp(); 22 if(len%(len-next[len]))printf("1 "); 23 else printf("%d ",len/(len-next[len])); 24 } 25 return 0; 26 }