POJ2406——经典哈希（求最大循环次数）

题目链接：poj.org/problem?id=2406

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.

题目理解：

给一个串s，求出它的最大循环次数。

 

这个题用kmp也可以做，哈希也行。

先预处理求出能被整除的长度，并且哈希处理字符串

之后一个个枚举长度，找到符合条件的。

#include 
#include
#include
#include
#define ull unsigned long long
using namespace std;
const int maxn=1e6+10;
const ull P=999983;
ull hash[maxn],p[maxn];
char s[maxn];
int main(int argc, char** argv) {
	p[0]=1;
	for(int i=1;i<=maxn;++i)
		p[i]=p[i-1]*P;
	while(~scanf("%s",s+1)){
		int len=strlen(s+1);
		if(len==1&&s[1]=='.') break;
		for(int i=1;i<=len;++i)
			hash[i]=hash[i-1]*P+(s[i]-'a'+1);
		vector v;
		for(int i=1;i<=len;++i)
			if(len%i==0)
				v.push_back(i);
		int times=1;
		for(int i=0;i