Description
Train
Of
Thought
首先
1/k=1/x+1/y
可以变形为
(k∗y)/(y−k)
然后我们就可以枚举
y,然后只要满足
(k∗y)为0时,就是一个满足条件的数对
Code
#include<iostream>
#include<cstdio>
using namespace std;
long long k,ans;
int main()
{
scanf("%lld",&k);
for (long long y=k+1; (k*y)/(y-k)>=y; ++y)
if ((k*y)%(y-k)==0) ans++;
printf("%lld",ans);
return 0;
}