[exercise-7] (UVA 10976) fractions again?! (fraction split)

translate ：

Enter a positive integer k, Find all positive integers x≥y, bring
$$\frac{1}{k}=\frac{1}{x}+\frac{1}{y}$$
This question is also very interesting , Do some advanced calculation ：
Since it is required to find all x,y, The enumerated object is naturally x,y 了 . But the problem is , How about the scope of enumeration ？ from 1/12=1/156+1/13 It can be seen that ,x Comparable y Much larger . Should we enumerate endlessly ？ Of course not. . because x≥y, Yes 1/x ≤1/y, therefore 1/k - 1/y ≤ 1/y, namely y≤2k. This only needs to be done in [k+1,2k] Enumeration in scope y, And then according to y Try to work out x that will do .
（ Content from Purple Book ）
That's great ！

AC Code ：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5+5;
const ll mod = 1e9+7;
int main()
{
    
	int n;
	while(cin>>n && n)
	{
    
		int cnt = 0;
		for(int i=n+1;i<=2*n;i++)
			if((i*n)%(i-n)==0)
				cnt++;
		cout<<cnt<<endl;
		for(int i=n+1;i<=2*n;i++)
			if((i*n)%(i-n)==0)
				printf("1/%d = 1/%d + 1/%d\n",n,(i*n)/(i-n),i);
	}
}

Write directly twice and calculate cnt One side of the output is because I'm too lazy ... There may be simple ways ？？ Well, it should be about the same .
The middle judgment idea is like this ：1/k = 1/x +1/y, in other words 1/k - 1/y = 1/x Is to meet the conditions .
The left side of the equation is simplified to y/ky - k/ky ==》 (y-k)/yk That is to say, as long as this formula can be reduced to 1/？ Is to meet the conditions .
That is to say 1/(yk/(y-k)) The denominator of is an integer . That is to say yk%(y-k) ==0
So we get the formula in the code .

summary ：

The problem of this formula , We should use unknown numbers to deduce several new formulas to simplify the problem .
Mainly push formula , If you can't push it out, you'll be finished ...