[洛谷4450]双亲数

题目链接：https://www.luogu.org/problemnew/show/P4450

题目要求（之后令\(n<m\)）

\[\sum\limits_{i=1}^n\sum\limits_{j=1}^m[gcd(i,j)==d] \]

同时除上\(d\)，有

\[\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}[gcd(i,j)==1] \]

然后发现最后面其实是莫比乌斯反演的形式，继续改写式子

\[\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}\sum\limits_{x|i,x|j}\mu(x) \]

然后我们把枚举改一下顺序，得到

\[\sum\limits_{x=1}^{\lfloor\frac{n}{d}\rfloor}\mu(x)\lfloor\dfrac{n}{dx}\rfloor\lfloor\dfrac{m}{dx}\rfloor \]

然后我们只要线筛求出\(\mu\)函数之后就可以做这题了(当然可以使用整除分块)

/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
inline int frd(){
	int x=0,f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline int read(){
	int x=0,f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=1e6;
int prime[N+10],miu[N+10],sum[N+10];
bool inprime[N+10];
void prepare(){
	miu[1]=sum[1]=1; int tot=0;
	for (int i=2;i<=N;i++){
		if (!inprime[i])	miu[prime[++tot]=i]=-1;
		for (int j=1;j<=tot&&i*prime[j]<=N;j++){
			inprime[i*prime[j]]=1;
			if (i%prime[j]==0){
				miu[i*prime[j]]=0;
				break;
			}miu[i*prime[j]]=-miu[i];
		}
		sum[i]=sum[i-1]+miu[i];
	}
}
int main(){
	prepare();
	int n=read(),m=read(),d=read();
	ll Ans=0; n/=d,m/=d;
	if (n>m)	swap(n,m);
	for (int i=1,pos;i<=n;i=pos+1){
		pos=min(n/(n/i),m/(m/i));
		Ans+=1ll*(sum[pos]-sum[i-1])*(n/i)*(m/i);
	}
	printf("%lld\n",Ans);
	return 0;
}