紫书例题7-3 分数拆分

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It is easy to see that for every fraction in the form(k > 0), we can always find two positive integers x and y,x ≥ y, such that:

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?

Input

Input contains no more than 100 lines, each giving a value of k (0 < k ≤ 10000).

Output

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.

Sample Input
2

12

Sample Output

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

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#include<iostream>
#include<vector>
using namespace std;
int main(){
	int k,x;
	cin>>k;
	for(int y=k+1;y<2*k+1;y++){
		if(y*k%(y-k)==0)
		{
		  x=y*k/(y-k);
			cout<<"1/"<<k<<"="<<"1/"<<y<<"="<<"1/"<<x<<endl; 
}
}
	return 0;
}

参考链接
https://blog.csdn.net/CSDN_Liuzongyi/article/details/53501589