求\(\sum_{i=1}^n\sum_{j=1}^m\sigma(gcd(i,j))[\sigma(gcd(i,j)<=a)]\)
\(f(d)=\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)=d]\)
\(F(d)=\sum_{i=1}^n\sum_{j=1}^m [|dgcd(i,j)]=\sum_{d|k}f(k)\)
\(f(d)=\sum_{d|k}\mu(\dfrac{k}{d})F(k)=\sum_{d|k}\mu(\left\lfloor\frac{k}{d}\right\rfloor)\left\lfloor\frac{n}{k}\right\rfloor\left\lfloor\frac{m}{k}\right\rfloor\)
\(Ans=\sum_{i=1}^{min(n,m)}\sigma(i)f(i)\)
\(=\sum_{i=1}^{min(n,m)}\sigma(i)\sum_{i|k}\mu(\left\lfloor\frac{k}{i}\right\rfloor)\left\lfloor\frac{n}{k}\right\rfloor\left\lfloor\frac{m}{k}\right\rfloor\)
\(=\sum_{k=1}^{min(n,m)}\left\lfloor\frac{n}{k}\right\rfloor\left\lfloor\frac{m}{k}\right\rfloor\sum_{i|k}\mu(\left\lfloor\frac{k}{i}\right\rfloor)\sigma(i)\)
后半部分预处理筛一遍\(sum_{i|k}\mu(\left\lfloor\frac{k}{i}\right\rfloor)\sigma(i)\)，前面分块一下，别忘了还有限制条件\([\sigma(gcd(i,j)<=a)]\)，丢到树状数组里就行了
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=1e5+9;
inline int Read(){
int 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 mu[maxn],prime[maxn];
bool visit[maxn];
inline void F_phi(int N){
mu[1]=1; int tot(0);
for(int i=2;i<=N;++i){
if(!visit[i]){
prime[++tot]=i,
mu[i]=-1;
}
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];
}
}
}
struct _F{
int id;
LL val;
}F[maxn];
inline bool cmp_F(_F x,_F y){
return x.val<y.val;
}
struct Qy{
int n,m,a,id;
}q[maxn];
inline bool cmp_Q(Qy x,Qy y){
return x.a<y.a;
}
int sum[maxn];
inline int Lowbit(int i){
return i&(-i);
}
inline void Add(int i,int val){
for(;i<=100000;i+=Lowbit(i))
sum[i]+=val;
}
inline int Query(int i){
int num(0);
for(;i;i-=Lowbit(i))
num+=sum[i];
return num;
}
inline int Calc(int n,int m){
int num(0);
if(n>m)
swap(n,m);
for(int l=1,r;l<=n;l=r+1){
r=min(n/(n/l),(m/(m/l)));
num+=(n/l)*(m/l)*(Query(r)-Query(l-1));
}
return num;
}
int ans[maxn];
int main(){
F_phi(100000);
for(int i=1;i<=100000;++i){
F[i].id=i;
for(int j=1;j*i<=100000;++j)
F[j*i].val+=i;
}
sort(F+1,F+1+100000,cmp_F);
int T=Read();
for(int i=1;i<=T;++i){
int n=Read(),m=Read(),a=Read();
q[i]=(Qy){n,m,a,i};
}
sort(q+1,q+1+T,cmp_Q);
int now(0);
for(int i=1;i<=T;++i){
while(now<100000&&F[now+1].val<=q[i].a){
++now;
for(int j=1;j*F[now].id<=100000;++j)
Add(j*F[now].id,mu[j]*F[now].val);
}
ans[q[i].id]=Calc(q[i].n,q[i].m);
}
for(int i=1;i<=T;++i)
printf("%d\n",ans[i]&0x7fffffff);
return 0;
}/*
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*/原文地址：https://www.cnblogs.com/y2823774827y/p/10227373.html