題目鏈接

\(Description\)

求\[\sum_{i=1}^n\sum_{j=1}^md(ij)\]

\(Solution\)

有結論：\[d(nm)=\sum_{i|d}\sum_{j|d}[\gcd(i,j)=1]\]
證明可以對質因子單獨考慮吧，不想寫了。

\[\begin{aligned}\sum_{i=1}^n\sum_{j=1}^md(ij)&=\sum_{i=1}^n\sum_{j=1}^m\sum_{a|i}\sum_{b|j}[\gcd(a,b)=1]\end{aligned}\]

轉為枚舉\(a,b\)，\[\sum_{i=1}^n\sum_{j=1}^md(ij)=\sum_{i=1}^n\sum_{j=1}^m\lfloor\frac{n}{i}\rfloor\lfloor\frac{m}{j}\rfloor[\gcd(i,j)=1]\]

然後反演，設\[F(d)=\sum_{i=1}^n\sum_{j=1}^m\lfloor\frac{n}{i}\rfloor\lfloor\frac{m}{j}\rfloor\left[d\mid(i,j)\right]=\sum_{i=1}^{\lfloor\frac nd\rfloor}\sum_{j=1}^{\lfloor\frac md\rfloor}\lfloor\frac{n}{id}\rfloor\lfloor\frac{m}{jd}\rfloor\\f(n)=\sum_{i=1}^n\sum_{j=1}^m\lfloor\frac{n}{i}\rfloor\lfloor\frac{m}{j}\rfloor\left[(i,j)=n\right]\]

則\[\begin{aligned}f(1)&=\sum_{d=1}^{\min(n,m)}\mu(d)F(d)\\&=\sum_{d=1}^{\min(n,m)}\mu(d)\sum_{i=1}^{\lfloor\frac nd\rfloor}\lfloor\frac{n}{id}\rfloor\sum_{j=1}^{\lfloor\frac md\rfloor}\lfloor\frac{m}{jd}\rfloor\end{aligned}\]

令\(g(n)=\sum_{i=1}\lfloor\frac ni\rfloor\)，則\(g(n)\)可以同樣用數論分塊\(O(n\sqrt n)\)

終於填了這個近半年前留下的坑了...

//1772kb    6448ms
#include <cstdio>
#include <cctype>
#include <algorithm>
//#define gc() getchar()
#define MAXIN 300000
#define gc() (SS==TT&&(TT=(SS=IN)+fread(IN,1,MAXIN,stdin),SS==TT)?EOF:*SS++)
const int N=5e4;

int cnt,P[N>>3],mu[N+3];
long long g[N+3];
bool not_P[N+3];

char IN[MAXIN],*SS=IN,*TT=IN;

inline int read()
{
    int now=0;register char c=gc();
    for(;!isdigit(c);c=gc());
    for(;isdigit(c);now=now*10+c-'0',c=gc());
    return now;
}
void Init()
{
    mu[1]=1;
    for(int i=2; i<=N; ++i)
    {
        if(!not_P[i]) P[++cnt]=i, mu[i]=-1;
        for(int j=1,v; j<=cnt&&(v=i*P[j])<=N; ++j)
        {
            not_P[v]=1;
            if(i%P[j]) mu[v]=-mu[i];
            else break;//mu[v]=0;
        }
    }
    for(int i=1; i<=N; ++i) mu[i]+=mu[i-1];
    for(int i=1; i<=N; ++i)
    {
        long long ans=0;
        for(int j=1,nxt; j<=i; j=nxt+1)
        {
            nxt=i/(i/j);
            ans+=1ll*(nxt-j+1)*(i/j);
        }
        g[i]=ans;
    }
}

int main()
{
    Init();
    for(int T=read(),n,m; T--; )
    {
        n=read(),m=read();
        long long ans=0;
        for(int i=1,nxt,lim=std::min(n,m); i<=lim; i=nxt+1)
        {
            nxt=std::min(n/(n/i),m/(m/i));
            ans+=1ll*(mu[nxt]-mu[i-1])*g[n/i]*g[m/i];
        }
        printf("%lld\n",ans);
    }
    return 0;
}

BZOJ.3994.[SDOI2015]約數個數和(莫比烏斯反演)