BZOJ3994: [SDOI2015]约数个数和

$n leq 50000,m leq 50000$，求$sum_{i=1}^nsum_{j=1}^md(ij)$。

$d(ij)=sum_{a|i}sum_{b|j}[(a,b)=1]$，把$a$选中的质因数的次数加上$j$的质因数次数，就是$a$算“比$j$次数多的质因子”，$b$算“次数不超过$j$的质因子”。

$sum_{i=1}^nsum_{j=1}^md(ij)$

$=sum_{i=1}^nsum_{j=1}^msum_{a|i}sum_{b|j}[(a,b)=1]$

$=sum_{a=1}^{n}sum_{b=1}^{m}left lfloor frac{n}{a} ight floorleft lfloor frac{m}{b} ight floorsum_{t|awedge t|b}mu(t)$

$=sum_{t=1}^n mu(t)sum_{c=1}^{left lfloor frac{n}{t} ight floor}sum_{d=1}^{left lfloor frac{m}{t} ight floor}left lfloor frac{n}{tc} ight floorleft lfloor frac{m}{td} ight floor$

后面那东西预处理一下。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
//#include<math.h>
//#include<queue>
#include<algorithm>
//#include<iostream>
//#include<assert.h>
using namespace std;

int T,n,m;
#define maxn 50011
#define LL long long
LL f[maxn],g[maxn],sumf[maxn];
int miu[maxn],summiu[maxn],prime[maxn],lp; bool notprime[maxn];
void pre(int n)
{
    miu[1]=summiu[1]=1;
    for (int i=2;i<=n;i++)
    {
        if (!notprime[i]) prime[++lp]=i,miu[i]=-1;
        summiu[i]=summiu[i-1]+miu[i];
        for (int tmp,j=1;j<=lp && 1ll*i*prime[j]<=n;j++)
        {
            notprime[tmp=i*prime[j]]=1;
            if (i%prime[j]) miu[tmp]=-miu[i];
            else {miu[tmp]=0; break;}
        }
    }
    
    for (int i=1;i<=n;i++)
        for (int j=1,now,last;j<=i;j=last+1)
        {
            now=i/j,last=i/now;
            f[i]+=(last-j+1)*1ll*now;
        }
}

int main()
{
    scanf("%d",&T);
    pre(50000);
    while (T--)
    {
        scanf("%d%d",&n,&m);
        (n>m) && (n^=m^=n^=m);
        LL ans=0;
        for (int i=1,last;i<=n;i=last+1)
        {
            last=min(n/(n/i),m/(m/i));
            ans+=(summiu[last]-summiu[i-1])*f[n/i]*f[m/i];
        }
        printf("%lld
",ans);
    }
    return 0;
}