【题意】给你一个数k，求所有使得1/k = 1/x + 1/y成立的x≥y的整数对。

【解题方法】数论，枚举。枚举所有在区间（k+1，2k）上的y即可，当1/k - 1/y的结果分子为1即为一组解。

【AC代码】

//
//Created by just_sort 2016/1/3
//Copyright (c) 2016 just_sort.All Rights Reserved
//

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <time.h>
#include <cstdlib>
#include <cstring>
#include <sstream> //isstringstream
#include <iostream>
#include <algorithm>
using namespace std;
using namespace __gnu_pbds;
typedef long long LL;
typedef pair<int, LL> pp;
#define REP1(i, a, b) for(int i = a; i < b; i++)
#define REP2(i, a, b) for(int i = a; i <= b; i++)
#define REP3(i, a, b) for(int i = a; i >= b; i--)
#define CLR(a, b)     memset(a, b, sizeof(a))
#define MP(x, y)      make_pair(x,y)
const int maxn = 100;
const int maxm = 2e5;
const int maxs = 10;
const int INF  = 1e9;
const int UNF  = -1e9;
const int mod  = 1e9+7;
int gcd(int x, int y) {return y == 0 ? x : gcd(y, x % y);}
//typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>order_set;
//head
int a[10010];
int main()
{
    int k;
    while(scanf("%d", &k) != EOF)
    {
        int cnt = 0;
        REP2(y, 1, 2 * k){
            if(y == k) continue;
            if((abs(k * y) % abs(y - k) == 0) && ((k * y) / (y - k) >= y)){
                a[++cnt] = y;
            }
        }
        printf("%d\n", cnt);
        REP2(i, 1, cnt){
            printf("1/%d = 1/%d + 1/%d\n", k, (k * a[i]) / (a[i] - k), a[i]);
        }
    }
    return 0;
}