POJ-2406-Power Strings

Power Strings

Time Limit : 6000/3000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 96   Accepted Submission(s) : 34

Problem 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

题目分析：

意思是有多少个子串构成比如说  aaaa = a^4   ababab=(ab)^3    abcd=(abcd)^1    应该明白了吧！也就是这个字符串能被他的最短子串构成   由kmp ，next表的特点知如果满足  len%(len-p[len])==0  就能由子串构成且子串个数就是 len/(len-p[len])

#include<cstdio>
#include<cstring>
const int M = 1000000+10;
char str[M];
int p[M],a[M],len;
void getp()
{
  int i=0,j=-1;
  p[i]=j;
  while(i<len)
  {
    if(j==-1||str[i]==str[j])
    {
      i++;j++;
      p[i]=j;
    }
    else j=p[j];
  }
}
int main()
{
  while(~scanf("%s",str)&&str[0]!='.')
  {
    len=strlen(str);
    getp();
                if(len%(len-p[len])==0)
      printf("%d\n",len/(len-p[len]));
    else
      printf("1\n");
  }
  return 0;
}