LG P2257 YY的GCD

Description

给定 $N,M$，求 $1 \leq x \leq N$，$1 \leq y \leq M$ 且 $\gcd(x,y)$ 为质数的$(x,y)$ 有多少对。

Solution

题目要求$\sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(x,y) \in prime]$

设

$$f(d)=\sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)=d]$$

$$F(x)=\sum_{x|d}f(d)=\lfloor\frac{n}{x}\rfloor\lfloor\frac{m}{x}\rfloor$$

则有莫比乌斯反演

$$f(n)=\sum_{n|d}\mu(\lfloor\frac{d}{n}\rfloor)F(d)$$

化简原式：

\begin{align}
& \sum_{p\in prime}\sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)=p]\\
= & \sum_{p\in prime}f(p)\\
= & \sum_{p\in prime}\sum_{p|d}\mu(\frac{d}{p})F(d)\\
= & \sum_{p\in prime}\sum_{d=1}^{min(\lfloor\frac{n}{p}\rfloor,\lfloor\frac{m}{p}\rfloor)}\mu(d)F(dp)\\
= & \sum_{p\in prime}\sum_{d=1}^{min(\lfloor\frac{n}{p}\rfloor,\lfloor\frac{m}{p}\rfloor)}\mu(d)\lfloor\frac{n}{dp}\rfloor\lfloor\frac{m}{dp}\rfloor\\
= & \sum_{T=1}^{min(n,m)}\sum_{t|T,t\in prime}\mu(\frac{T}{t})\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor\\
= & \sum_{T=1}^{min(n,m)}\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor(\sum_{t|T}\mu(\frac{T}{t}))
\end{align}

数论分块即可

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int prime[10000005],mu[10000005],n,m,tot;
long long T,ans,sum[10000005];
bool vst[10000005];
inline int read()
{
    int f=1,w=0;
    char ch=0;
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')
            f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        w=(w<<1)+(w<<3)+ch-'0';
        ch=getchar();
    }
    return f*w;
}
inline void write(long long x)
{
    if(x<0)
    {
        putchar('-');
        x=-x;
    }
    if(x>9)
    {
        write(x/10);
    }
    putchar(x%10+'0');
}
int main()
{
    //freopen("P2257.in","r",stdin);
    //freopen("ans.txt","w",stdout);
    mu[1]=1;
    for(int i=2;i<=10000000;i++)
    {
        if(!vst[i])
        {
            prime[++tot]=i;
            mu[i]=-1;
        }
        for(int j=1;j<=tot&&i*prime[j]<=10000000;j++)
        {
            vst[i*prime[j]]=true;
            if(!(i%prime[j]))
            {
                break;
            }
            else
            {
                mu[i*prime[j]]=-mu[i];
            }
        }
    }
    for(int i=1;i<=5000000;i++)
    {
        for(int j=1;j<=tot&&i*prime[j]<=10000000;j++)
        {
            sum[i*prime[j]]+=mu[i];
        }
    }
    for(int i=1;i<=10000000;i++)
    {
        sum[i]+=sum[i-1];
    }
    T=read();
    for(;T;T--)
    {
        ans=0;
        n=read();
        m=read();
        if(n>m)
        {
            swap(n,m);
        }
        for(int i=1;i<=n;)
        {
            int j=min(n/(n/i),m/(m/i));
            ans+=1ll*(n/i)*(m/i)*(sum[j]-sum[i-1]);
            i=j+1;
        }
        write(ans);
        putchar(10);
    }
    return 0;
}