【POJ2406】Power Strings

                                          Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 58102		Accepted: 24134

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.

 

解析：

       从小到大枚举子串长度，第一个合法长度即为答案子串，用字符串哈希来判断即可，小优化必不可少。

 

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
using namespace std;

const int Max=1000010;
int n,m,ans;
unsigned long long Pow[Max],hash[Max];
char ch[Max];

inline void pre()
{
	Pow[0]=1;
	for(int i=1;i<=n;i++)
	{
	  hash[i] = hash[i-1] * 131 + (ch[i] - 'a' + 1);
	  Pow[i] = Pow[i-1] * 131;
	}
}

inline bool check(int len)
{
	unsigned long long x=0;
	for(int i=len;i<=n;i+=len)
	{
	  x = x * Pow[len] + hash[len];
	  if(hash[i] != x) return 0;
	}
	return 1;
}

int main()
{
	while(1)
	{
	  scanf("%s",ch+1);
	  if(ch[1]=='.') break;
	  n=strlen(ch+1);
	  pre();
	  for(int i=1;i<=n;i++)
	  {
	  	if(n%i) continue;   //不能被整除显然不可能是答案串
	  	if(check(i)) {ans=i;break;}  //满足条件直接结束
	  }
	  cout<<n/ans<<"\n";
	}

	return 0;
}