N的阶乘二进制表示的最低位1的位置

《编程之美》中的一道题：

#include <iostream>
#include <string>
using std::endl;
using std::string;
using std::cout;
typedef unsigned long long ULLONG;
//给定整数N，求N!的二进制表示中最低位1的位置
template<class T>
string ToBinary(T n)
{
	if (n==0)
	{
		return string("0");
	}
	const int size=8*sizeof(n);
	string binstr(size,' ');
	for (int i=0;n;i++)
	{
		binstr.at(size-i-1)=(n&1)+'0';
		n>>=1;
	}
	if(binstr.at(0)==' ')
		binstr = binstr.substr(binstr.find_first_not_of(' '));
	return binstr;
}

ULLONG Fact(ULLONG n)
{
	ULLONG rst=1;
	while(n)
	{
		rst*=n--;
	}
	return rst;
}

int FactLowestOne1(int n)
{
	int nt=n;
	int cnt=0;
	while(n>1)
	{
		cnt++;
		n&=n-1;
	}
	return nt-cnt;
}

int FactLowestOne2(int n)
{
	int cnt=0;
	while(n)
	{
		n>>=1;
		cnt+=n;
	}
	return cnt;
}

int main()
{
	ULLONG n=12;
	ULLONG fact=Fact(n);
	cout<<n<<"!="<<fact<<endl;
	cout<<"binary form: "<<ToBinary(fact)<<endl;
	cout<<"Lowest one bit(1): "<<FactLowestOne1(n)<<endl;
	cout<<"Lowest one bit(2): "<<FactLowestOne2(n);
}