题目大意：
就是说输入一个k，然后求出满足1/k = 1/x+1/y的所有形式，使得x>=y。
解题思路：
拿到这个题后，看到题目仅仅给了x>=y.这个条件，感觉无从下手，但是仔细想想就不难发现，只要我们
首先确定了枚举的范围，那么我们就能够在这个区间内,找到符合我们要求的解了。
推导：
因为x>=y,所以1/x<=1/y，由1/k-1/x=1/y －＞ 1/k-1/y <= 1/y -> y<=2k,
然后，由等式1/k = 1/x+1/y －> x = (y*k)/(y-k)
这样一来，我们只需要在[k+1,2k]上枚举y的值，然后找到 (y*k)%(y-k)==0的所有x的值就好了。
代码：
# include<cstdio> # include<iostream> # include<algorithm> # include<functional> # include<cstring> # include<string> # include<cstdlib> # include<iomanip> # include<numeric> # include<cctype> # include<cmath> # include<ctime> # include<queue> # include<stack> # include<list> # include<set> # include<map> using namespace std; const double PI=4.0*atan(1.0); typedef long long LL; typedef unsigned long long ULL; # define inf 999999999 # define MAX 10000 int k; struct node { int x, y; }a[MAX]; int main(void) { while ( cin>>k ) { int num = 0; for ( int yy = k+1;yy <= 2*k;yy++ ) { int fenzi = yy*k; int fenmu = yy-k; if ( fenzi%fenmu == 0 ) { int xx = fenzi/fenmu; a[num].x = xx; a[num++].y = yy; } } cout<<num<<endl; for ( int i = 0;i < num;i++ ) { printf("1/%d = 1/%d + 1/%d ",k,a[i].x,a[i].y); } } return 0; }