POJ 2406 Power Strings

F - Power Strings

Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2406

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<iostream>
 #include<cstring>
 #include<cstdio>
 #include<map>
 #define maxn 10010000
 using namespace std;
 char p[maxn];
 int next[maxn],len;
 int kmp(){
     int i=,j=-;
     next[]=-;
     while(i<len){
         if(j==-||p[i]==p[j]) next[++i]=++j;
         else j=next[j];
     }
 }
 int main()
 {
     while(scanf("%s",p)!=EOF&&p[]!='.'){
         len=strlen(p);
         kmp();
         if(len%(len-next[len]))printf("1\n");
         else printf("%d\n",len/(len-next[len]));
     }
     return ;
 }