UVa 10976 Fractions Again?!

题意:给出正整数k,找到所有整数使得满足1/k=1/x+1/y

根据基本不等式可以求出k<=y<=2k，在这个范围内枚举即可

#include<iostream>  
#include<cstdio>  
#include<cstring> 
#include <cmath> 
#include<stack>
#include<vector>
#include<map> 
#include<queue> 
#include<algorithm>  
#define mod=1e9+7;
using namespace std;

typedef long long LL;

int main(){
    int k,i,j,a,b,x,y;
    while(cin>>k){
        int ans=0;
        for(y=1;y<=2*k;y++){
            if(y!=k) x=(k*y)/(y-k);//注意分母为0的情况 
            int m=k*(x+y)-x*y;
            if(x>=y&&m==0) ans++;
        }
        printf("%d
",ans);
        
        for(y=1;y<=2*k;y++){
            if(y!=k) x=(k*y)/(y-k);
            int m=k*(x+y)-x*y;
            if(x>=y&&(m==0)) 
            printf("1/%d = 1/%d + 1/%d
",k,x,y);
        }
    }
    return 0;
}