[莫比乌斯反演 约数个数] BZOJ 3994 [SDOI2015]约数个数和

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#define P 20101009
using namespace std;
typedef long long ll;

inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
return *p1++;
}

inline void read(ll &x){
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

inline int write(ll x,char c)
{
if (x==0) return putchar('0'),putchar(c),0;
if (x<0) putchar('-'),x=-x;
char s[50]={0},len=0;
while (x) s[++len]=x%10+'0',x/=10;
for (int i=len;i;i--) putchar(s[i]); putchar(c);
}

const int maxn=50000;
int vst[maxn+5],prime[maxn+5],num=0;
int e[maxn+5];
ll f[maxn+5];
int mobius[maxn+5];
int sum[maxn+5];

inline void Pre()
{
mobius[1]=1; f[1]=1;
for (int i=2;i<=maxn;i++)
{
if (!vst[i])
{
mobius[i]=-1,prime[++num]=i;
f[i]=2; e[i]=1;
}
for (int j=1;j<=num && prime[j]*i<=maxn;j++)
{
vst[i*prime[j]]=1;
if (i%prime[j]==0)
{
mobius[i*prime[j]]=0;
f[i*prime[j]]=f[i]/(e[i]+1)*(e[i]+2);
e[i*prime[j]]=e[i]+1;
break;
}
else
{
mobius[i*prime[j]]=-mobius[i];
f[i*prime[j]]=f[i]*f[prime[j]];
e[i*prime[j]]=1;
}
}
}
for (int i=1;i<=maxn;i++)
f[i]+=f[i-1],
sum[i]=sum[i-1]+mobius[i];
}

ll n,m;
ll ans;

int main()
{
ll Q;
Pre();
read(Q);
while (Q--)
{
read(n); read(m);
if (n>m) swap(n,m);
ans=0;
for (ll j,i=1;i<=n;i=j+1)
{
j=min(n/(n/i),m/(m/i));
ans+=(ll)(sum[j]-sum[i-1])*f[n/i]*f[m/i];
}
write(ans,'\n');
}
return 0;
}