YbtOJ LCM计数

Description

题目链接：YbtOJ

，求

sum_{i=1}^nsum_{j=1}^moperatorname{lcm}(i,j)[forall n>1,n^2not gcd(i,j)]

Solution

由常见套路可得：

，则原式等同于：

sum_{T=1}^n S(frac nT)S(frac mT)F(T)

显然

总时间复杂度为 mathcal O(Nlog N Tsqrt N)

Code

代码语言：javascript复制

#include<bits/stdc  .h>
#define Tp template<typename Ty>
#define Ts template<typename Ty,typename... Ar>
#define W while
#define I inline
#define RI register int
#define LL long long
#define Cn const
#define CI Cn int&
#define gc getchar
#define D isdigit(c=gc())
#define pc(c) putchar((c))
#define min(x,y) ((x)<(y)?(x):(y))
#define max(x,y) ((x)>(y)?(x):(y))
using namespace std;
namespace Debug{
    Tp I void _debug(Cn char* f,Ty t){cerr<<f<<'='<<t<<endl;}
    Ts I void _debug(Cn char* f,Ty x,Ar... y){W(*f!=',') cerr<<*f  ;cerr<<'='<<x<<",";_debug(f 1,y...);}
    Tp ostream& operator<<(ostream& os,Cn vector<Ty>& V){os<<"[";for(Cn auto& vv:V) os<<vv<<",";os<<"]";return os;}
    #define gdb(...) _debug(#__VA_ARGS__,__VA_ARGS__)
}using namespace Debug;
namespace FastIO{
    Tp I void read(Ty& x){char c;int f=1;x=0;W(!D) f=c^'-'?1:-1;W(x=(x<<3) (x<<1) (c&15),D);x*=f;}
    Ts I void read(Ty& x,Ar&... y){read(x),read(y...);}
    Tp I void write(Ty x){x<0&&(pc('-'),x=-x,0),x<10?(pc(x '0'),0):(write(x/10),pc(x '0'),0);}
    Tp I void writeln(Cn Ty& x){write(x),pc('n');}
}using namespace FastIO;
Cn int N=4e6 10,P=(1<<30);
int Tt,n,m,p[N],v[N],mu[N],tot,F[N],Inv2,b[N];
I int HB(RI n,CI x){
    RI S=0;W(!(n%x)){
        S  ,n/=x;
        if(S>2) return 0;
    }return -1LL*x*x%P*x%P*F[n]%P;
}
I void GM(){
    RI i,k;LL j;for(mu[1]=1,i=2;i<N;i  ) for(!v[i]&&(mu[p[  tot]=i]=-1),j=1;j<=tot&&i*p[j]<N;j  )
    if(v[i*p[j]]=1,i%p[j]) mu[i*p[j]]=-mu[i];else{mu[i*p[j]]=0;break ;}
    for(i=2;i<N;i  ) for(j=1LL*i*i;j<N;j =i*i) b[j]=1;
    for(i=1;i<N;i  ) if(!b[i]) for(j=i;j<N;j =i) F[j]=(1LL*mu[j/i]*(j/i)%P*j%P F[j])%P; 
    for(i=1;i<N;i  ) F[i]=(1LL*F[i] F[i-1])%P;
}
I int S(CI n,CI m){
    #define Sum(x) (1LL*(1 (x))*(x)/2%P)
    RI i,j;LL X=0;
    for(i=1;i<=min(n,m);i=j 1) j=min(n/(n/i),m/(m/i)),X=(1LL*X 1LL*Sum(n/i)*Sum(m/i)%P*(F[j]-F[i-1])%P)%P,X=(X P)%P;
    return X;
}
I int QP(RI a,RI b){RI s=1;W(b) b&1&&(s=1LL*s*a%P),a=1LL*a*a%P,b>>=1;return s;}
int main(){
    GM(),read(Tt);
    W(Tt--) read(n,m),writeln(S(n,m));
    return 0;
}

ode

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