2694: Lcm 莫比乌斯反演

题解：

莫比乌斯反演复习。
题目即求：

$$\sum_{i=1}^A\sum_{j=1}^B[\mu(gcd(i,j))!=0]lcm(i,j)$$
枚举gcd：
$$\sum_{d=1}^{min(A,B)}[\mu(d)==1]d\sum_{i=1}^{\lfloor {A\over d} \rfloor}\sum_{j=1}^{\lfloor {B\over d} \rfloor}[gcd(i,j)==1]ij$$
$$\sum_{d=1}^{min(A,B)}[\mu(d)==1]d\sum_{i=1}^{\lfloor {A\over d} \rfloor}\sum_{j=1}^{\lfloor {B\over d} \rfloor}ij\sum_{k|i且k|j}\mu(k)$$
$$\sum_{d=1}^{min(A,B)}[\mu(d)==1]d\sum_{k=1}^{min({\lfloor {A\over d} \rfloor},\lfloor {B\over d} \rfloor)}\mu(k)k^2 \frac{(1+\lfloor{A\over dk}\rfloor)\lfloor{A\over dk}\rfloor}{2}\frac{(1+\lfloor{B\over dk}\rfloor)\lfloor{B\over dk}\rfloor}{2}$$
设T=dk，枚举T
$$\sum_{T=1}^{min(A,B)} \frac{(1+\lfloor{A\over T}\rfloor)\lfloor{A\over T}\rfloor}{2}\frac{(1+\lfloor{B\over T}\rfloor)\lfloor{B\over T}\rfloor}{2}\sum_{d|T}[\mu(d)==1]d\mu(\frac{T}{d})(\frac{T}{d})^2$$
然后右边那堆可以 $O(nlogn)$预处理，左边可以分块。

代码：

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define pa pair<int,int>
const int Maxn=4000010;
const int inf=4000000;
const LL mod=(1LL<<30);
int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
    return x*f;
}
int prime[400000],len=0,mu[Maxn];
LL f[Maxn],sum[Maxn];
bool mark[Maxn];
void pre()
{
    memset(mark,false,sizeof(mark));
    mu[1]=1;
    for(int i=2;i<=inf;i++)
    {
        if(!mark[i])prime[++len]=i,mu[i]=-1;
        for(int j=1;j<=len&&prime[j]*i<=inf;j++)
        {
            mark[prime[j]*i]=true;
            if(i%prime[j]==0)
            {
                mu[prime[j]*i]=0;
                break;
            }
            mu[prime[j]*i]=-mu[i];
        }
    }
    for(int d=1;d<=inf;d++)
    {
        if(mu[d]==0)continue;
        for(int T=d;T<=inf;T+=d)
        f[T]=(f[T]+(d*((T/d)*(T/d)%mod)%mod)*mu[T/d]+mod)%mod;
    }
    sum[0]=0;
    for(int i=1;i<=inf;i++)sum[i]=(sum[i-1]+f[i])%mod;
} 
int main()
{
    pre();
    int T=read();
    while(T--)
    {
        int n=read(),m=read();
        int pos;
        LL ans=0;
        for(int T=1;T<=min(n,m);T=pos+1)
        {
            pos=min(n/(n/T),m/(m/T));
            ans=(ans+((sum[pos]-sum[T-1]+mod)%mod)*((1+n/T)*(n/T)/2LL%mod)*((1+m/T)*(m/T)/2LL%mod))%mod;
        }
        printf("%lld\n",ans);
    }
}