【链接】 我是链接,点我呀:)
【题意】

在这里输入题意

【题解】

x>=y => $\frac{1}{x}<=\frac{1}{y}$ => $\frac{1}{x}=\frac{1}{k}-\frac{1}{y}$ 结合两个式子可以得到 y<=2*k 则枚举y,然后根据式子得到x,判断合法性就ok

【代码】

/*
  	1.Shoud it use long long ?
  	2.Have you ever test several sample(at least therr) yourself?
  	3.Can you promise that the solution is right? At least,the main ideal
  	4.use the puts("") or putchar() or printf and such things?
  	5.init the used array or any value?
  	6.use error MAX_VALUE?
*/

#include <bits/stdc++.h>
using namespace std;

int k;
vector <pair<int,int> > v;

int main(){
	#ifdef LOCAL_DEFINE
	    freopen("F:\\c++source\\rush_in.txt", "r", stdin);
	#endif
	ios::sync_with_stdio(0),cin.tie(0);
    while (cin >> k){
        v.clear();
        for (int y = k+1;y <= 2*k;y++){
            int fenzi = k*y;
            int fenmu = y-k;
            if (fenzi%fenmu==0){
                int x = fenzi/fenmu;
                if (x>=y) v.push_back({x,y});
            }
        }
        cout << v.size() << endl;
        for (int i = 0;i < (int)v.size();i++){
                int x = v[i].first,y = v[i].second;
            cout <<"1/"<<k<<" = 1/"<<x<<" + 1/"<<y<<endl;
        }
    }
	return 0;
}