正题

题目链接:https://www.luogu.org/problem/P5253

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题目大意

给一个$n$，求有多少对$x,y(x\le y)$使得
$$\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$$

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解题思路

$$\frac{x+y}{xy}=\frac{1}{n}$$
$$n(x+y)=xy$$
$$xy-n(x+y)=0$$
$$xy-n(x+y)+n^2=n^2$$
$$(x-n)(y-n)=n^2$$
设$a=x-n,b=y-n$，问题就变为了求有多少对$(a,b)$

然后对于$\sum p_i^{c_i}=n$

答案就是$\prod (2\ast c_i)$

注意要考虑本质不同

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$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
ll n,ans=1;
int main()
{scanf("%lld",&n);for(ll i=2;i*i<=n;i++){if(!(n%i)){int z=0;while(!(n%i)) n/=i,z++;ans*=2*z+1;}}if(n!=1) ans*=3; printf("%lld",ans/2+(ans&1));
}