UVA - 10976:Fractions Again?!

Fractions Again?!
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タイトル
It is easy to see that for every fraction in the form 1 k (k > 0), we can always find two positive integers x and y, x ≥ y, such that: 1/k = 1/x + 1/y Now our question is: can you write a program that counts how many such pairs of x and y there are for any given k?
入力
Input contains no more than 100 lines, each giving a value of k (0 < k ≤ 10000).
しゅつりょく
For each k, output the number of corresponding (x, y) pairs, followed by a sorted list of the values of x and y, as shown in the sample output.
入力サンプル
2 12
出力サンプル
2 1/2 = 1/6 + 1/3 1/2 = 1/4 + 1/4 8 1/12 = 1/156 + 1/13 1/12 = 1/84 + 1/14 1/12 = 1/60 + 1/15 1/12 = 1/48 + 1/16 1/12 = 1/36 + 1/18 1/12 = 1/30 + 1/20 1/12 = 1/28 + 1/21 1/12 = 1/24 + 1/24
リファレンスコード

#include
#include
const int maxn=10000;
int main()
{
	int k;
	while(scanf("%d",&k)!=EOF)
	{
		double x,y;
		int ax[maxn],ay[maxn];
		int cnt=0;
		for(y=1;y<=2*k;y++)
		{
			if(y==k)continue;
			x=k*y/(y-k);
			if(x>0 && x==floor(x))
			{
				ax[cnt]=(int)x;
				ay[cnt]=(int)y;
				cnt++;
			}
		}
		printf("%d
",cnt);
		for(int i=0;i<cnt;i++)
			printf("1/%d = 1/%d + 1/%d
",k,ax[i],ay[i]);
	}
	return 0;
}