BZOJ_2820_YY的GCD_莫比乌斯反演

BZOJ_2820_YY的GCD_莫比乌斯反演

题意&分析：

$sumlimits_pis[p]sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}[gcd(i,j)=p]$

$=sumlimits_pis[p]sumlimits_{i=1}^{lfloor frac{n}{p} floor}sumlimits_{j=1}^{lfloor frac{m}{p} floor}[gcd(i,j)=1]$

$=sumlimits_pis[p]sumlimits_{i=1}^{lfloor frac{n}{p} floor}sumlimits_{j=1}^{lfloor frac{m}{p} floor}[gcd(i,j)=1]$

$=sumlimits_pis[p]sumlimits_{i=1}^{lfloor frac{n}{p} floor}sumlimits_{j=1}^{lfloor frac{m}{p} floor}sumlimits_{d|gcd(i,j)}mu(d)$

$=sumlimits_pis[p]sumlimits_{d=1}^{lfloor frac{n}{p} floor}mu(d)sumlimits_{i=1}^{lfloor frac{n}{dp} floor}sumlimits_{j=1}^{lfloor frac{m}{dp} floor}$

$=sumlimits_{Q=1}^{n}lfloor frac{n}{Q} floorlfloorfrac{m}{Q} floorsumlimits_{p|Q}is[p]mu(lfloorfrac{Q}{p} floor)$

$f(n)=sumlimits_{p|n}is[p]mu(lfloorfrac{n}{p} floor)$

首先$f[i]$非积性，但可以通过μ处理，所以我们考虑线筛

1.当$i$为质数时$f[i]=1$;

2.当$i$%$p==0$时

$f(i*p)=sumlimits_{d|i}is[d]mu(i*p/d)$

当$d!=p$时$i*p/d$有两个以上的$p$，贡献为$0$，因此此时$f(i*p)=mu(i)$

3.当$i$%$p!=0$时$i$与$p$互质

$f(i*p)=sumlimits_{d|i}is[d]mu(i*p/d)+sumlimits_{d|p}is[d]mu(i*p/d)$
$=f(i)*mu(p)+f(p)*mu(i)$
$=mu(i)-f(i)$

再记录下f[i]的前缀和，分块计算

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define LL long long
int prime[4000010],vis[10000100],miu[10000100],f[10000100],sum[10000100],cnt;
int T,n,m;
inline void init()
{
    miu[1]=1;
    for(int i=2;i<=10000000;i++)
    {
        if(!vis[i])
        {
            miu[i]=-1;
            f[i]=1;
            prime[++cnt]=i;    
        }
        for(int j=1;j<=cnt&&i*prime[j]<=10000000;j++)
        {
            vis[i*prime[j]]=1;
            if(i%prime[j]==0)
            {
                miu[i*prime[j]]=0;
                f[i*prime[j]]=miu[i];
                break;
            }
            miu[i*prime[j]]=-miu[i];
            f[i*prime[j]]=miu[i]-f[i];
        }
        sum[i]=sum[i-1]+f[i];
    }
}
int main()
{
    init();
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        if(n>m)swap(n,m);
        int lst;
        LL ans=0;
        for(int i=1;i<=n;i=lst+1)
        {
            lst=min(n/(n/i),m/(m/i));
            ans+=1ll*(sum[lst]-sum[i-1])*(n/i)*(m/i);
        }
        printf("%lld
",ans);
    }
}