[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;
}

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