莫比乌斯反演经典例题回顾

P2522 [HAOI2011]Problem b

$$\sum _{i=a}^b\sum _{j=c}^d[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=k]$$

•         \ \ \ \ \ \,      反演过程：

$$F(n,m)=\sum _{i=1}^n\sum _{j=1}^m[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=k]$$

$$ans=F(a,b)-F(a-1,d)-F(b,c-1)+F(a-1,c-1)$$

$$F(n,m)=$$

$$\sum _{i=1}^{\lfloor \frac{n}{k}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{k}\rfloor }[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=1]$$

$$\sum _{i=1}^{\lfloor \frac{n}{k}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{k}\rfloor }ϵ(\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=1)$$

$$\sum _{d=1}^{\lfloor \frac{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}{k}\rfloor }\sum _{d|T}\mu (d)\lfloor \frac{n}{T}\rfloor \lfloor \frac{m}{T}\rfloor$$

$$\sum _{d=1}^{\lfloor \frac{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}{k}\rfloor }\mu (d)\lfloor \frac{n}{kd}\rfloor \lfloor \frac{m}{kd}\rfloor$$

•         \ \ \ \ \ \,       $O(n)$预处理$\mu$，每次询问分块，复杂度$O(\sqrt{n})$，总复杂度为$O(n+T4\sqrt{n})$

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int inf=0x7fffffff;
const double eps=1e-10;
const double pi=acos(-1.0);
const int N=50010;
inline int read(){
  int x=0,f=1;char ch;ch=getchar();
  while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
  while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
  return x*f;
}
bool vis[N];
int prim[N],mu[N],sum[N],k;
void get_mu(int n){
  mu[1]=1;
  for(int i=2;i<=n;i++){
    if(!vis[i]){mu[i]=-1;prim[++prim[0]]=i;}
    for(int j=1;j<=prim[0]&&i*prim[j]<=n;j++){
      vis[i*prim[j]]=1;
      if(i%prim[j]==0)break;
      else mu[i*prim[j]]=-mu[i];
    }
  }
  for(int i=1;i<=n;i++)
  sum[i]=sum[i-1]+mu[i];
}
long long Ans(int a,int b){
    a/=k;b/=k;
  int max_rep=min(a,b);
  long long ans=0;
  for(int l=1,r;l<=max_rep;l=r+1){
    r=min(a/(a/l),b/(b/l));
    ans+=(long long)(a/l)*(long long)(b/l)*(long long)(sum[r]-sum[l-1]);
  }
  return ans;
}
int main()
{
 	int T=read();
 	get_mu(N-10);
 	while(T--){
 		int a=read(),b=read(),c=read(),d=read();k=read();
 		printf("%lld\n",Ans(b,d)-Ans(b,c-1)-Ans(a-1,d)+Ans(a-1,c-1));
  }
  return 0;
}

---

P2257 YY的GCD

$$\sum _{i=1}^n\sum _{j=1}^m[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)\mathrm{i}\mathrm{s}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{e}]$$

•         \ \ \ \ \ \,      反演过程：

$$\sum _{p=1,[p\mathrm{i}\mathrm{s}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{e}]}\sum _{i=1}^n\sum _{j=1}^m[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=p]$$

$$\sum _{p=1,[p\mathrm{i}\mathrm{s}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{e}]}\sum _{i=1}^{\lfloor \frac{n}{p}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{p}\rfloor }[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=1]$$

$$\sum _{p=1,[p\mathrm{i}\mathrm{s}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{e}]}\sum _{d=1}^{\lfloor \frac{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}{p}\rfloor }\mu (d)\lfloor \frac{n}{pd}\rfloor \lfloor \frac{m}{pd}\rfloor$$

$$\sum _{T=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\sum _{d|T}\mu (d)\lfloor \frac{n}{T}\rfloor \lfloor \frac{m}{T}\rfloor \left[\frac{T}{d}\mathrm{i}\mathrm{s}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{e}\right]$$

$$\sum _{T=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\lfloor \frac{n}{T}\rfloor \lfloor \frac{m}{T}\rfloor \sum _{d|T,\left[d\mathrm{i}\mathrm{s}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{e}\right]}\mu \left(\frac{T}{d}\right)$$

$$F(x)=\sum _{d|x,\left[d\mathrm{i}\mathrm{s}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{e}\right]}\mu \left(\frac{x}{d}\right)$$

$$\sum _{T=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\lfloor \frac{n}{T}\rfloor \lfloor \frac{m}{T}\rfloor F(T)$$

•         \ \ \ \ \ \,       $O(n)$预处理$\mu$，$O(n\log n)$预处理$F$，每次询问分块，复杂度$O(\sqrt{n})$，总复杂度为$O(n\log n+T\sqrt{n})$

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int inf=0x7fffffff;
const double eps=1e-10;
const double pi=acos(-1.0);
const int N=1e7+10;
inline int read(){
    int x=0,f=1;char ch;ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
bool vis[N];
int prim[N],mu[N];
long long F[N];
void get_mu(int n){
  mu[1]=1;
  for(int i=2;i<=n;i++){
    if(!vis[i]){mu[i]=-1;prim[++prim[0]]=i;}
    for(int j=1;j<=prim[0]&&i*prim[j]<=n;j++){
      vis[i*prim[j]]=1;
      if(i%prim[j]==0)break;
      else mu[i*prim[j]]=-mu[i];
    }
  }
  for(int i=1;i<=prim[0];i++)
  for(int j=1;j*prim[i]<=n;j++)F[j*prim[i]]+=(long long)mu[j];
  for(int i=1;i<=n;i++)F[i]+=F[i-1];
}
long long Ans(int n,int m){
  int lim=min(n,m);
  long long ans=0;
  for(int l=1,r;l<=lim;l=r+1){
    r=min(n/(n/l),m/(m/l));
    ans+=(long long)(n/l)*(long long)(m/l)*(F[r]-F[l-1]);
  }
  return ans;
}
int main()
{
  get_mu(N-10);
  int T=read();
  while(T--){
    int n=read(),m=read();
    printf("%lld\n",Ans(n,m));
  }
  return 0;
}

---

P3312 [SDOI2014]数表

$$\sum _{i=1}^n\sum _{j=1}^m\sum _{k|\mathrm{g}\mathrm{c}\mathrm{d}(i,j)}k\left[\sum _{k|\mathrm{g}\mathrm{c}\mathrm{d}(i,j)}k\le a\right]$$

•         \ \ \ \ \ \,      反演过程：

$$F(x)=\sum _{i|x}i$$

$$\sum _{i=1}^n\sum _{j=1}^{m}F(\mathrm{g}\mathrm{c}\mathrm{d}(i,j))[F(\mathrm{g}\mathrm{c}\mathrm{d}(i,j))\le a]$$

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\sum _{i=1}^n\sum _{j=1}^{m}F(d)[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=d][F(d)\le a]$$

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}F(d)[F(d)\le a]\sum _{i=1}^n\sum _{j=1}^m[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=d]$$

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}F(d)[F(d)\le a]\sum _{d^{\prime }=1}^{\lfloor \frac{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}{d}\rfloor }\mu (d^{\prime })\lfloor \frac{n}{d^{\prime }d}\rfloor \lfloor \frac{m}{d^{\prime }d}\rfloor$$

$$\sum _{T=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\sum _{d|T}F(d)\mu \left(\frac{T}{d}\right)\lfloor \frac{n}{T}\rfloor \lfloor \frac{m}{T}\rfloor [F(d)\le a]$$

$$\sum _{T=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\lfloor \frac{n}{T}\rfloor \lfloor \frac{m}{T}\rfloor \sum _{d|T}F(d)\mu \left(\frac{T}{d}\right)[F(d)\le a]$$

$$f(x)=\sum _{d|T}F(d)\mu \left(\frac{T}{d}\right)[F(d)\le a]$$

$$\sum _{T=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\lfloor \frac{n}{T}\rfloor \lfloor \frac{m}{T}\rfloor f(T)$$

•         \ \ \ \ \ \,       $O(n\log n)$预处理$F$和$\mu$，把每次询问的$a$值离线排序，用树状数组维护$f$，花费时间$O(n{\log }^{2}n)$，询问分块，复杂度$O(\sqrt{n}\log n)$，总时间复杂度$O(n{\log }^{2}n+T\sqrt{n}\log n)$

#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int inf=0x7fffffff;
const double eps=1e-10;
const double pi=acos(-1.0);
//char buf[1<<15],*S=buf,*T=buf;
//char getch(){return S==T&&(T=(S=buf)+fread(buf,1,1<<15,stdin),S==T)?0:*S++;}
inline int read(){
  int x=0,f=1;char ch;ch=getchar();
  while(ch<'0'||ch>'9'){if(ch=='-') f=0;ch=getchar();}
  while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch&15);ch=getchar();}
  if(f)return x;else return -x;
}
const int N=1e5+10;
const int Q=2e4+10;
int sum[N];
bool vis[N];
int prim[N],mu[N];
int S=1;
struct Sum{int data,id;}F[N];
bool operator <(Sum a,Sum b){return a.data<b.data;}
int ans[Q];
#define lowbit(x) ((x)&(-x))
void add(int i,int c)
{for(;i<=N-5;i+=lowbit(i))sum[i]+=c;}
int query(int i){
  int ans=0;
  for(;i;i-=lowbit(i))ans+=sum[i];
  return ans;
}
void Add(int i,int data){
  int ed=((N-5)/i);
  for(register int j=1;j<=ed;++j)
  add(i*j,data*mu[j]);
}
void init(){
  mu[1]=1;
  for(register int i=2;i<=N-5;++i){
    if(!vis[i]){mu[i]=-1;prim[++prim[0]]=i;}
    for(register int j=1;j<=prim[0]&&i*prim[j]<=N-5;++j){
      vis[i*prim[j]]=1;
      if(i%prim[j]==0)break;
      else mu[i*prim[j]]=-mu[i];
    }
  }
  for(register int i=1;i<=N-5;++i){
    F[i].id=i;
    int ed=((N-5)/i);
    for(register int j=1;j<=ed;++j)
    F[i*j].data+=i;
  }
  sort(F+1,F+1+N-5);
}
struct Query{int n,m,a,id;}q[Q];
inline bool operator <(const Query &a,const Query &b){return a.a<b.a;}
int main()
{
  int T=read();
  init();
  for(register int i=1;i<=T;i++){
    q[i].n=read();q[i].m=read();q[i].a=read();q[i].id=i;
    if(q[i].n>q[i].m)swap(q[i].n,q[i].m);
  }
  sort(q+1,q+T+1);
  for(register int i=1;i<=T;i++){
    int n=q[i].n,m=q[i].m,a=q[i].a,id=q[i].id;
    while(F[S].data<=a){Add(F[S].id,F[S].data);S++;}
    for(register int l=1,r;l<=n;l=r+1){
      r=min(n/(n/l),(m/(m/l)));
      ans[id]+=(n/l)*(m/l)*(query(r)-query(l-1));
    }
  }
  for(register int i=1;i<=T;i++)printf("%d\n",ans[i]&0x7fffffff);
  return 0;
}

---

P3704 [SDOI2017]数字表格

$$\prod _{i=1}^n\prod _{j=1}^{m}f\left(\mathrm{g}\mathrm{c}\mathrm{d}(i,j)\right)$$

        \ \ \ \ \ \,       $f$为斐波拉契序列。

•         \ \ \ \ \ \,      反演过程：

$$\prod _{i=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\prod _{i=1}^n\prod _{j=1}^{m}f(d)\left[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=d\right]$$

$$\prod _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}f(d{)}^{{\sum }_{i=1}^n{\sum }_{j=1}^m\left[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=d\right]}$$

$$\prod _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}f(d{)}^{{\sum }_{d^{\prime }=1}^{\lfloor \frac{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}{d}\rfloor }\mu (d^{\prime })\lfloor \frac{n}{d^{\prime }d}\rfloor \lfloor \frac{m}{d^{\prime }d}\rfloor }$$

$$\prod _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\prod _{d^{\prime }=1}^{\lfloor \frac{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}{d}\rfloor }f(d{)}^{\mu (d^{\prime })\lfloor \frac{n}{d^{\prime }d}\rfloor \lfloor \frac{m}{d^{\prime }d}\rfloor }$$

$$\prod _{T=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\prod _{d|T}f(d{)}^{\mu \left(\frac{T}{d}\right)\lfloor \frac{n}{T}\rfloor \lfloor \frac{m}{T}\rfloor }$$

$$C(x)=\prod _{d|T}f(d{)}^{\mu \left(\frac{T}{d}\right)}$$

$$\prod _{T=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}C(T{)}^{\lfloor \frac{n}{T}\rfloor \lfloor \frac{m}{T}\rfloor }$$

•         \ \ \ \ \ \,       $O(n)$预处理$f$和$\mu$，$O(n\log n)$预处理$C$，每次询问分块，复杂度$O(\sqrt{n}\log n)$，总复杂度为$O(n\log n+T\sqrt{n}\log n)$

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int inf=0x7fffffff;
const double eps=1e-10;
const double pi=acos(-1.0);
const int N=1e6+10;
const long long mod=1e9+7;
inline int read(){
    int x=0,f=1;char ch;ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
long long power(long long a,long long b){  
  long long ans=1ll;  
  a=a%mod;
  while(b!=0){  
    if(b&1) ans=(ans*a)%mod;
    b>>=1;a=(a*a)%mod;
  }  
  return ans;  
}
long long f[N];
bool vis[N];
int prim[N],mu[N],sum[N];
long long C[N],g[N];
void get_mu(int n){
  mu[1]=1;
  g[1]=C[0]=C[1]=1ll;
  for(int i=2;i<=n;i++){
  	g[i]=power(f[i],mod-2);C[i]=1;
    if(!vis[i]){mu[i]=-1;prim[++prim[0]]=i;}
    for(int j=1;j<=prim[0]&&i*prim[j]<=n;j++){
      vis[i*prim[j]]=1;
      if(i%prim[j]==0)break;
      else mu[i*prim[j]]=-mu[i];
    }
  }
  for(int i=1;i<=n;++i){
    if(!mu[i])continue;
  	for(int j=i;j<=n;j+=i)
    C[j]=1ll*C[j]*(mu[i]==1?f[j/i]:g[j/i])%mod;
  }
  for(int i=2;i<=n;++i)C[i]=1ll*C[i]*C[i-1]%mod;
}
int n,m;
int main()
{
  f[1]=f[2]=1ll;
  for(int i=3;i<=N-10;i++)f[i]=f[i-1]+f[i-2],f[i]%=mod;
  get_mu(N-10);
  int T=read();
  while(T--){
    n=read();m=read();
    if(n>m)swap(n,m);
    long long ans=1ll;
    for(int i=1,lim=0;i<=n;i=lim+1){
      lim=min(n/(n/i),m/(m/i));
      ans=1ll*ans*power(C[lim]*power(C[i-1],mod-2),1ll*(n/i)*(m/i)%(mod-1));
      ans%=mod;
    }
    printf("%lld\n",ans);
  }
  return 0;
}

---

P3327 [SDOI2015]约数个数和

$$\sum _{i=1}^n\sum _{j=1}^m\sum _{k|\mathrm{g}\mathrm{c}\mathrm{d}(i,j)}1$$

•         \ \ \ \ \ \,      反演过程：

$$F(x)=\sum _{i|x}1$$

$$\sum _{i=1}^n\sum _{j=1}^{m}F(\mathrm{g}\mathrm{c}\mathrm{d}(i,j))$$

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\sum _{i=1}^n\sum _{j=1}^{m}F(d)[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=d]$$

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}F(d)\sum _{i=1}^n\sum _{j=1}^m[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=d]$$

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}F(d)\sum _{d^{\prime }=1}^{\lfloor \frac{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}{d}\rfloor }\mu (d^{\prime })\lfloor \frac{n}{d^{\prime }d}\rfloor \lfloor \frac{m}{d^{\prime }d}\rfloor$$

$$\sum _{T=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\sum _{d|T}F(d)\mu \left(\frac{T}{d}\right)\lfloor \frac{n}{T}\rfloor \lfloor \frac{m}{T}\rfloor$$

$$\sum _{T=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\lfloor \frac{n}{T}\rfloor \lfloor \frac{m}{T}\rfloor \sum _{d|T}F(d)\mu \left(\frac{T}{d}\right)$$

$$f(x)=\sum _{d|T}F(d)\mu \left(\frac{T}{d}\right)$$

$$\sum _{T=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\lfloor \frac{n}{T}\rfloor \lfloor \frac{m}{T}\rfloor f(T)$$

•         \ \ \ \ \ \,       $O(n\log n)$预处理$F$，$f$和$\mu$，询问分块，复杂度$O(\sqrt{n}\log n)$，总时间复杂度$O(n\log n+T\sqrt{n}\log n)$，和P3312 [SDOI2014]数表这道题差不多

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int inf=0x7fffffff;
const double eps=1e-10;
const double pi=acos(-1.0);
const int N=50010;
inline int read(){
  int x=0,f=1;char ch;ch=getchar();
  while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
  while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
  return x*f;
}
bool vis[N];
int prim[N],mu[N],sum[N];
long long g[N];
void get_mu(int n){
  mu[1]=1;
  for(int i=2;i<=n;i++){
    if(!vis[i]){mu[i]=-1;prim[++prim[0]]=i;}
    for(int j=1;j<=prim[0]&&i*prim[j]<=n;j++){
      vis[i*prim[j]]=1;
      if(i%prim[j]==0)break;
      else mu[i*prim[j]]=-mu[i];
    }
  }
  for(int i=1;i<=n;i++)sum[i]=sum[i-1]+mu[i];
  for(int i=1;i<=n;i++){
    long long ans=0;
    for(int l=1,r;l<=i;l=r+1){
      r=(i/(i/l));
      ans+=1ll*(r-l+1)*1ll*(i/l);
    }
    g[i]=ans;
  }
}
long long Ans(int n,int m){
  int max_rep=min(n,m);
  long long ans=0;
  for(int l=1,r;l<=max_rep;l=r+1){
    r=min(n/(n/l),m/(m/l));
  	ans+=(sum[r]-sum[l-1])*1ll*g[n/l]*1ll*g[m/l];
  }
  return ans;
}
int main()
{
  int T=read();
 	get_mu(N-10);
 	while(T--){
 		int a=read(),b=read();
 		printf("%lld\n",Ans(a,b));
  }
  return 0;
}

---

P3455 [POI2007]ZAP-Queries

$$\sum _{i=1}^a\sum _{j=1}^b[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=c]$$

•         \ \ \ \ \ \,      反演过程：

$$\sum _{i=1}^a\sum _{j=1}^b[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=1]$$

$$\sum _{i=1}^{\lfloor \frac{a}{c}\rfloor }\sum _{j=1}^{\lfloor \frac{b}{c}\rfloor }ϵ(\mathrm{g}\mathrm{c}\mathrm{d}(i,j))$$

$$\sum _{i=1}^{\lfloor \frac{a}{c}\rfloor }\sum _{j=1}^{\lfloor \frac{b}{c}\rfloor }(\mu \ast 1)(\mathrm{g}\mathrm{c}\mathrm{d}(i,j))$$

$$\sum _{i=1}^{\lfloor \frac{a}{c}\rfloor }\sum _{j=1}^{\lfloor \frac{b}{c}\rfloor }\sum _{d|\mathrm{g}\mathrm{c}\mathrm{d}(i,j)}\mu (d)\times 1$$

$$\sum _{x=1}^{\lfloor \frac{\mathrm{m}\mathrm{i}\mathrm{n}(a,b)}{c}\rfloor }\sum _{d|x}\mu (d)\lfloor \frac{a}{xc}\rfloor \lfloor \frac{b}{xc}\rfloor$$

$$\sum _{d=1}^{\lfloor \frac{\mathrm{m}\mathrm{i}\mathrm{n}(a,b)}{c}\rfloor }\mu (d)\lfloor \frac{a}{dc}\rfloor \lfloor \frac{b}{dc}\rfloor$$

•         \ \ \ \ \ \,       裸题啊！模板啊！$O(n)$预处理$\mu$，询问分块，复杂度$O(\sqrt{n})$，总时间复杂度$O(n+T\sqrt{n})$

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int inf=0x7fffffff;
const double eps=1e-10;
const double pi=acos(-1.0);
const int N=50010;
inline int read(){
  int x=0,f=1;char ch;ch=getchar();
  while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
  while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
  return x*f;
}
bool vis[N];
int prim[N],mu[N],sum[N],k;
void get_mu(int n){
  mu[1]=1;
  for(int i=2;i<=n;i++){
    if(!vis[i]){mu[i]=-1;prim[++prim[0]]=i;}
    for(int j=1;j<=prim[0]&&i*prim[j]<=n;j++){
      vis[i*prim[j]]=1;
      if(i%prim[j]==0)break;
      else mu[i*prim[j]]=-mu[i];
    }
  }
  for(int i=1;i<=n;i++)
  sum[i]=sum[i-1]+mu[i];
}
long long Ans(int a,int b){
  int max_rep=min(a,b);
  long long ans=0;
  for(int l=1,r;l<=max_rep;l=r+1){
    r=min(a/(a/l),b/(b/l));
    ans+=(long long)(a/(l*k))*(long long)(b/(l*k))*(long long)(sum[r]-sum[l-1]);
  }
  return ans;
}
int T;
int main()
{
 	T=read();
 	get_mu(N-10);
 	while(T--){
 		int a=read(),b=read();k=read();
 		printf("%lld\n",Ans(a,b));
  }
  return 0;
}

---

P1829 [国家集训队]Crash的数字表格 / JZPTAB

$$\sum _{i=1}^n\sum _{j=1}^m\mathrm{l}\mathrm{c}\mathrm{m}(i,j)$$

•         \ \ \ \ \ \,      反演过程：

$$\sum _{i=1}^n\sum _{j=1}^m\frac{ij}{\mathrm{g}\mathrm{c}\mathrm{d}(i,j)}$$

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\sum _{i=1}^n\sum _{j=1}^m\frac{ij}{d}[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=d]$$

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\frac{1}{d}\sum _{i=1}^n\sum _{j=1}^{m}ij[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=d]$$

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\frac{1}{d}\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{d}\rfloor }ij\cdot d^2[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=1]$$

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}d\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{d}\rfloor }ij[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=1]$$

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}d\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{d}\rfloor }ijϵ(\mathrm{g}\mathrm{c}\mathrm{d}(i,j))$$

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}d\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{d}\rfloor }ij\sum _{k|\mathrm{g}\mathrm{c}\mathrm{d}(i,j)}\mu (k)$$

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}d\sum _{k=1}^{\lfloor \frac{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}{d}\rfloor }\mu (k)\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{d}\rfloor }ij[k|\mathrm{g}\mathrm{c}\mathrm{d}(i,j)]$$

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}d\sum _{k=1}^{\lfloor \frac{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}{d}\rfloor }\mu (k)\sum _{i=1}^{\lfloor \frac{n}{dk}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{dk}\rfloor }ij\cdot k^2$$

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}d\sum _{k=1}^{\lfloor \frac{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}{d}\rfloor }\mu (k)\cdot k^2\sum _{i=1}^{\lfloor \frac{n}{dk}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{dk}\rfloor }ij$$

$$\begin{array}{rl}F(n,m)=& \sum _{k=1}^{\frac{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}{d}}\mu (k)\cdot k^2\sum _{i=1}^{\lfloor \frac{n}{k}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{k}\rfloor }ij\\ =& \sum _{k=1}^{\frac{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}{d}}\mu (k)\cdot k^2\cdot \frac{\lfloor \frac{n}{k}\rfloor \cdot (\lfloor \frac{n}{k}\rfloor +1)}{2}\times \frac{\lfloor \frac{m}{k}\rfloor \cdot (\lfloor \frac{m}{k}\rfloor +1)}{2}\end{array}$$

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}d\times F\left(\lfloor \frac{n}{k}\rfloor ,\lfloor \frac{m}{k}\rfloor \right)$$

•         \ \ \ \ \ \,       感觉很神奇，需要连续提取 $2$ 次公约数，$O(n)$预处理$\mu$，分块询问$F$,$O(\sqrt{\sqrt{n}})$,答案询问分块，复杂度$O(\sqrt{n})$，嵌套起来总时间复杂度$O(n+n^{\frac{3}{4}})$，代码比较丑

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int inf=0x7fffffff;
const double eps=1e-10;
const double pi=acos(-1.0);
inline int read(){
    int x=0,f=1;char ch;ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const int N=10010000;
const long long mod=20101009;
long long sum[N];
bool vis[N];
int prim[N],mu[N];
void get_mu(int n){
  mu[1]=1;
  for(int i=2;i<=n;i++){
    if(!vis[i]){mu[i]=-1;prim[++prim[0]]=i;}
    for(int j=1;j<=prim[0]&&i*prim[j]<=n;j++){
      vis[i*prim[j]]=1;
      if(i%prim[j]==0)break;
      else mu[i*prim[j]]=-mu[i];
    }
  }
  for(int i=1;i<=n;i++)
  sum[i]=(sum[i-1]+1ll*mu[i]*1ll*i%mod*1ll*i%mod)%mod;
}
int main()
{
 	int n=read(),m=read();
 	if(n>m)swap(n,m);
  get_mu(n);
  long long ans=0;
  long long inv2=(mod+1ll)/2ll;
  for(int d=1;d<=n;d++){
    int x=n/d,y=m/d,minn=min(x,y);
    long long Sum=0ll;
    for(int l=1,r;l<=minn;l=r+1ll){
    	r=min(x/(x/l),y/(y/l));
      Sum=(Sum+(sum[r]-sum[l-1])%mod*(((1ll+x/l)%mod*1ll*(x/l)%mod*inv2%mod)%mod)%mod*(((1ll+y/l)%mod*1ll*(y/l)%mod*inv2%mod)%mod)%mod)%mod;
    }
    ans=(ans+Sum*1ll*d)%mod;
  }
 	printf("%lld\n",(ans%mod+mod)%mod);
  return 0;
}

• 扩展：这个题其实是两道题的融合，上面的方法是过不了JZPTAB的，在bzoj上面是权限题，所以先挖坑，我们下面继续来化简：

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}d\sum _{k=1}^{\lfloor \frac{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}{d}\rfloor }\mu (k)\cdot k^2\sum _{i=1}^{\lfloor \frac{n}{dk}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{dk}\rfloor }ij$$

$$\sum _{d=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\sum _{k=1}^{\lfloor \frac{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}{d}\rfloor }d\cdot \mu (k)\cdot k^2\sum _{i=1}^{\lfloor \frac{n}{dk}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{dk}\rfloor }ij$$

$$\sum _{T=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\sum _{i=1}^{\lfloor \frac{n}{T}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{T}\rfloor }ij\sum _{d|T}\frac{T}{d}\cdot \mu (d)\cdot d^2$$

$$\sum _{T=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\sum _{i=1}^{\lfloor \frac{n}{T}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{T}\rfloor }ij\sum _{d|T}T\mu (d)d$$

$$F(T)=\sum _{d|T}T\mu (d)d$$

$$\sum _{T=1}^{\mathrm{m}\mathrm{i}\mathrm{n}(n,m)}\frac{\lfloor \frac{n}{T}\rfloor \cdot (\lfloor \frac{n}{T}\rfloor +1)}{2}\times \frac{\lfloor \frac{m}{T}\rfloor \cdot (\lfloor \frac{m}{T}\rfloor +1)}{2}\times F(T)$$

•         \ \ \ \ \ \,       显然，$F$是积性函数，我们最快可以$O(n)$地把他筛出来，具体操作可以看【积性函数的线性筛】，欧拉筛三步走，询问分块，复杂度$O(\sqrt{n})$,总时间复杂度$O(n+T\sqrt{n})$

  $\mathtt{s}\mathtt{t}\mathtt{e}\mathtt{p}1$. 如果$p$是素数：
  $f(p)=p-p^2$

  $\mathtt{s}\mathtt{t}\mathtt{e}\mathtt{p}2$. 如果$p$是素数，$i\%p̸=0$：$f(pi)=f(i)\times (p-p^2)$

  $\mathtt{s}\mathtt{t}\mathtt{e}\mathtt{p}3$. 如果$p$是素数，$i\%p=0$：我们把 $pi$ 分解成 $p^{c}x$ ：
  $f(pi)=f(p^{c}x)=f(x)\times (p^c-p^{c+1})$

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int inf=0x7fffffff;
const double eps=1e-10;
const double pi=acos(-1.0);
//char buf[1<<15],*S=buf,*T=buf;
//char getch(){return S==T&&(T=(S=buf)+fread(buf,1,1<<15,stdin),S==T)?0:*S++;}
inline int read(){
	int x=0,f=1;char ch;ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-') f=0;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch&15);ch=getchar();}
	if(f)return x;else return -x;
}
const int N=1e7+10;
const long long mod=20101009;
long long inv2=(mod+1ll)/2ll;
bool vis[N];
long long pri[N],cnt[N],power[N];
long long prime[N],f[N];
void Get_Shai(long long n){
  f[1]=1;
  for(long long i=1;i<=n;i++)power[i]=1;
  for(long long i=2;i<=n;i++){
    if(!vis[i]){
      cnt[i]=1;pri[i]=i;power[i]=i;
      prime[++prime[0]]=i;
      f[i]=(1ll*i-1ll*i*i%mod+mod)%mod;
    }
    for(long long j=1,v,pc;j<=prime[0]&&i*prime[j]<=n;j++){
      v=i*prime[j];
      vis[v]=1;
      if(i%prime[j]==0){
        cnt[v]=cnt[i]+1;pri[v]=pri[i];power[v]=(1ll*power[i]*pri[i])%mod;
        f[v]=f[v/power[v]]*(1ll*power[v]-1ll*power[v]*pri[i]%mod+mod)%mod;
        break;
      }
      cnt[v]=1;pri[v]=prime[j];power[v]=prime[j];
      f[v]=(f[i]*f[prime[j]])%mod;
    }
  }
  for(int i=1;i<=n;i++)f[i]=(1ll*f[i]+f[i-1])%mod;
}
long long Get_ans(int x,int y){
  long long Sum=0;
  for(int l=1,r;l<=x;l=r+1){
    r=min(x/(x/l),y/(y/l));
    Sum=(Sum+(1ll*f[r]-f[l-1]+mod)%mod*(((1ll+x/l)%mod*1ll*(x/l)%mod*inv2%mod)%mod)%mod*(((1ll+y/l)%mod*1ll*(y/l)%mod*inv2%mod)%mod)%mod)%mod;
  }
  return Sum;
}
int main()
{
	int n=read(),m=read();
	if(n>m)swap(n,m);
  Get_Shai(1ll*n);
  printf("%lld\n",(Get_ans(n,m)%mod+mod)%mod);
	return 0;
}

• 再扩展，这道题还可以用杜教筛加速的，具体戳这里，感觉和这道题差不多P3768，就不贴了。

---

P3768 简单的数学题

$$\sum _{i=1}^n\sum _{j=1}^{n}ij\mathrm{g}\mathrm{c}\mathrm{d}(i,j)$$

•         \ \ \ \ \ \,      反演过程：

$$\sum _{d=1}^n\sum _{i=1}^n\sum _{j=1}^{n}ijd[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=d]$$

$$\sum _{d=1}^{n}d\sum _{i=1}^n\sum _{j=1}^{n}ij[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=d]$$

$$\sum _{d=1}^{n}d\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }ij\cdot d^2[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=1]$$

$$\sum _{d=1}^n{d}^3\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }ij[\mathrm{g}\mathrm{c}\mathrm{d}(i,j)=1]$$

$$\sum _{d=1}^n{d}^3\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }ijϵ(\mathrm{g}\mathrm{c}\mathrm{d}(i,j))$$

$$\sum _{d=1}^n{d}^3\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }ij\sum _{k|\mathrm{g}\mathrm{c}\mathrm{d}(i,j)}\mu (k)$$

$$\sum _{d=1}^n{d}^3\sum _{k=1}^{\lfloor \frac{n}{d}\rfloor }\mu (k)\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }ij[k|\mathrm{g}\mathrm{c}\mathrm{d}(i,j)]$$

$$\sum _{d=1}^n{d}^3\sum _{k=1}^{\lfloor \frac{n}{d}\rfloor }\mu (k)\sum _{i=1}^{\lfloor \frac{n}{dk}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{dk}\rfloor }ij\cdot k^2$$

$$\sum _{d=1}^n{d}^3\sum _{k=1}^{\lfloor \frac{n}{d}\rfloor }\mu (k)\cdot k^2\sum _{i=1}^{\lfloor \frac{n}{dk}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{dk}\rfloor }ij$$

$$\sum _{T=1}^n\sum _{i=1}^{\lfloor \frac{n}{T}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{T}\rfloor }ij\sum _{d|T}{d}^3\cdot \mu \left(\frac{T}{d}\right)\cdot {\left(\frac{T}{d}\right)}^2$$

$$\sum _{T=1}^n\sum _{i=1}^{\lfloor \frac{n}{T}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{T}\rfloor }ij\sum _{d|T}{T}^2\cdot d\cdot \mu \left(\frac{T}{d}\right)$$

$$\sum _{T=1}^n\sum _{i=1}^{\lfloor \frac{n}{T}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{T}\rfloor }ij{T}^2\sum _{d|T}d\cdot \mu \left(\frac{T}{d}\right)$$

$$\sum _{T=1}^n\sum _{i=1}^{\lfloor \frac{n}{T}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{T}\rfloor }ij{T}^2(id\ast \mu )(T)$$

$$\sum _{T=1}^n\sum _{i=1}^{\lfloor \frac{n}{T}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{T}\rfloor }ij{T}^2\phi (T)$$

$$\sum _{T=1}^n\frac{\lfloor \frac{n}{T}\rfloor \cdot (\lfloor \frac{n}{T}\rfloor +1)}{2}\times \frac{\lfloor \frac{m}{T}\rfloor \cdot (\lfloor \frac{m}{T}\rfloor +1)}{2}\times T^2\phi (T)$$

•         \ \ \ \ \ \,       需要连续提取 $2$ 次公约数，和P1829有异曲同工之妙，杜教筛筛出 $T^2\phi (T)$ 的前缀和，分块询问答案询问，总时间复杂度$O(n^{\frac{2}{3}})$，杜教筛详见【莫比乌斯反演和杜教筛】

  $f(x)=x^2\phi (x)$

  $g(x)=x^2,{\sum }_{i=1}^{x}g(i)=\frac{x(x+1)(2x+1)}{6}$

  $(f\ast g)(x)=x^3,{\sum }_{i=1}^x(f\ast g)(i)=\frac{x^2(x+1{)}^2}{4}$

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
inline long long read(){
  long long x=0,f=1;char ch;ch=getchar();
  while(ch<'0'||ch>'9'){if(ch=='-') f=0;ch=getchar();}
  while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch&15);ch=getchar();}
  if(f)return x;else return -x;
}
const int N=8000010;
long long mod,n,inv6,inv4;
int prim[N];
long long f[N];
bool vis[N];
void get_f(int n){
  f[1]=1;
  for(int i=2;i<=n;i++){
    if(!vis[i]){f[i]=i-1;prim[++prim[0]]=i;}
    for(int j=1;j<=prim[0]&&i*prim[j]<=n;j++){
      vis[i*prim[j]]=1;
      if(i%prim[j]==0){f[i*prim[j]]=f[i]*prim[j];break;}
      else f[i*prim[j]]=f[i]*(prim[j]-1);
    }
  }
  for(int i=1;i<=n;++i)f[i]=(f[i-1]+1ll*f[i]*i%mod*i%mod)%mod;
}
map<long long,long long> mp;
long long power(long long a,long long b){
  long long ans=1;
  while(b){if(b&1)ans=ans*a%mod;a=a*a%mod;b>>=1;}
  return ans;
}
long long Sumfg(long long x){x%=mod;return x*(x+1)%mod*x%mod*(x+1)%mod*inv4%mod;}
long long Sumg(long long x){x%=mod;return x*(x+1)%mod*(x+x+1)%mod*inv6%mod;}
long long Sumf(long long x){
  if(x<=min(N-10,(int)n))return f[x];
  if(mp[x])return mp[x];
  long long ret=Sumfg(x);
  for(long long i=2,j;i<=x;i=j+1){
  	j=x/(x/i);
  	ret=(ret-(Sumg(j)-Sumg(i-1)+mod)%mod*Sumf(x/i)%mod+mod)%mod;
  }
  return mp[x]=(ret+mod)%mod;
}
int main()
{
  mod=read();n=read();
  inv4=power(4,mod-2);
  inv6=power(6,mod-2);
  get_f(min(N-10,(int)n));
  long long ans=0;
  for(long long i=1,j;i<=n;i=j+1){
    j=n/(n/i);
    ans=(ans+(Sumf(j)-Sumf(i-1))%mod*Sumfg(n/i)%mod)%mod;
  }
  printf("%lld\n",(ans+mod)%mod);
  return 0;
}

---

P4240 毒瘤之神的考验

$$\sum _{i=1}^n\sum _{j=1}^m\phi (ij)$$

•         \ \ \ \ \ \,