P3327 [SDOI2015]约数个数和
\(d(ij)=\sum_{x|i}\sum_{y|j}[gcd(x,y)=1]\)
则我们原式化为
\(Ans=\sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{x|i}\sum_{y|j}[gcd(x,y)=1]\)
莫比乌斯反演：
\(f(d)=\sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)=d]\)
\(F(n)=\sum_{n|d}f(d)\)
\(\Rightarrow f(n)=\sum_{n|d}\mu(\lfloor\frac{d}{n}\rfloor)F(d)\)
根据\(\mu\)函数性质
\(Ans=\sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{x|i}\sum_{y|j}\sum_{d|gcd(x,y)}\mu(d)\)
推(du)公(liu)式的地方来了：
\(Ans=\sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{x|i}\sum_{y|j}\sum_{d=1}^{min(n,m)}\mu(d)*[d|gcd(x,y)]\)
\(Ans=\sum_{d=1}^{min(n,m)}\mu(d)\sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{x|i}\sum_{y|j}[d|gcd(x,y)]\)
\(Ans=\sum_{d=1}^{min(n,m)}\mu(d)\sum_{x=1}^{n}\sum_{y=1}^{m}[d|gcd(x,y)]\lfloor\frac{n}{x}\rfloor\lfloor\frac{m}{y}\rfloor\)
\(Ans=\sum_{d=1}^{min(n,m)}\mu(d)\sum_{x=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{y=1}^{{\lfloor\frac{m}{d}\rfloor}}\lfloor\frac{n}{dx}\rfloor\lfloor\frac{m}{dy}\rfloor\)
\(Ans=\sum_{d=1}^{min(n,m)}\mu(d)(\sum_{x=1}^{\lfloor\frac{n}{d}\rfloor}\lfloor\frac{n}{dx}\rfloor)(\sum_{y=1}^{{\lfloor\frac{m}{d}\rfloor}}\lfloor\frac{m}{dy}\rfloor)\)
#include<cstring>
#include<iostream>
#include<algorithm>
#include<string>
#include<cstdio>
using namespace std;
typedef long long LL;
const LL maxn=50000+10;
inline int Read(){
LL x=0,f=1; char c=getchar();
while(c<'0'||c>'9'){
if(c=='-') f=-1; c=getchar();
}
while(c>='0'&&c<='9'){
x=(x<<3)+(x<<1)+c-'0',c=getchar();
}
return x*f;
}
int T;
int prime[maxn],mu[maxn],sum[maxn];
LL a[maxn];
bool visit[maxn];
inline void F_phi(LL N){
mu[1]=1;
int tot=0;
for(int i=2;i<=N;++i){
if(!visit[i]){
mu[i]=-1,
prime[++tot]=i;
}
for(int j=1;j<=tot&&i*prime[j]<=N;++j){
visit[i*prime[j]]=true;
if(i%prime[j]==0)
break;
else
mu[i*prime[j]]=-mu[i];
}
}
for(int i=1;i<=N;++i)
sum[i]=sum[i-1]+1ll*mu[i];
for(int i=1;i<=N;++i){
LL num(0);
for(int l=1,r;l<=i;l=r+1){
r=i/(i/l);
num=num+1ll*(r-l+1)*1ll*(i/l);
}
a[i]=num;
}
}
int main(){
T=Read();
F_phi(50000);
while(T--){
int n=Read(),m=Read();
if(n>m)
swap(n,m);
LL ans(0);
int N=n;
for(int l=1,r;l<=N;l=r+1){
r=min(n/(n/l),m/(m/l));
ans=ans+1ll*(sum[r]-sum[l-1])*(1ll*a[n/l])*(1ll*a[m/l]);
}
printf("%lld\n",ans);
}
return 0;
}/*
2
7 4
5 6
110
121
*/
