Project Euler：Problem 64 Odd period square roots

All square roots are periodic when written as continued fractions and can be written in the form:

√N = a0 +	1
	a1 +	1
		a2 +	1
			a3 + ...

For example, let us consider √23:

√23 = 4 + √23 — 4 = 4 +	1	= 4 +	1
	1 √23—4		1 +	√23 – 3 7

If we continue we would get the following expansion:

√23 = 4 +	1
	1 +	1
		3 +	1
			1 +	1
				8 + ...

The process can be summarised as follows:

a0 = 4,		1 √23—4	=	√23+4 7	= 1 +	√23—3 7
a1 = 1,		7 √23—3	=	7(√23+3) 14	= 3 +	√23—3 2
a2 = 3,		2 √23—3	=	2(√23+3) 14	= 1 +	√23—4 7
a3 = 1,		7 √23—4	=	7(√23+4) 7	= 8 +	√23—4
a4 = 8,		1 √23—4	=	√23+4 7	= 1 +	√23—3 7
a5 = 1,		7 √23—3	=	7(√23+3) 14	= 3 +	√23—3 2
a6 = 3,		2 √23—3	=	2(√23+3) 14	= 1 +	√23—4 7
a7 = 1,		7 √23—4	=	7(√23+4) 7	= 8 +	√23—4

It can be seen that the sequence is repeating. For conciseness, we use the notation √23 = [4;(1,3,1,8)], to indicate that the block (1,3,1,8) repeats indefinitely.
The first ten continued fraction representations of (irrational) square roots are:

√2=[1;(2)], period=1

√3=[1;(1,2)], period=2

√5=[2;(4)], period=1

√6=[2;(2,4)], period=2

√7=[2;(1,1,1,4)], period=4

√8=[2;(1,4)], period=2

√10=[3;(6)], period=1

√11=[3;(3,6)], period=2

√12= [3;(2,6)], period=2

√13=[3;(1,1,1,1,6)], period=5
Exactly four continued fractions, for N ≤ 13, have an odd period.
How many continued fractions for N ≤ 10000 have an odd period?

#include <iostream>
using namespace std;

int getPeriod(int n)
{
	int pd = 0;
	int m = 0, d = 1;
	int a0 = int(sqrt(n));
	int a = a0;
	while (a != 2 * a0)
	{
		m = a*d - m;
		d = (n - m*m) / d;
		a = int((sqrt(n) + m) / d);
		pd++;
	}
	return pd;
}

int main()
{
	int count = 0;
	for (int i = 2; i <= 10000; i++)
	{
		if (int(sqrt(i))*int(sqrt(i)) == i)
			continue;
		if (getPeriod(i) % 2 != 0)
			count++;
		//cout << i << " " << getPeriod(i) << endl;
	}
	cout << count << endl;
	system("pause");
	return 0;
}