余数求和
对于任意整数 \(x\in[1,n]\) 设 \(g(x)=\lfloor\frac{k}{\lfloor\frac{k}{x}\rfloor}\rfloor\)
\(\because f(x)=\frac{k}{x}\) 单调递减
又 \((x)=\lfloor\frac{k}{\lfloor\frac{k}{x}\rfloor}\rfloor\geq\lfloor\frac{k}{(\frac{k}{x})}\rfloor=x\)
\(\therefore \lfloor\frac{k}{g(x)}\rfloor\leq\lfloor\frac{k}{x}\rfloor\)
\(\because \lfloor\frac{k}{g(x)}\rfloor\geq\lfloor\frac{k}{(\frac{k}{\lfloor\frac{k}{x}\rfloor})}\rfloor=\lfloor\frac{k}{k}\lfloor\frac{k}{x}\rfloor\rfloor=\lfloor\frac{k}{x}\rfloor\)
\(\therefore \lfloor\frac{k}{g(x)}\rfloor=\lfloor\frac{k}{x}\rfloor\)
综上得，对于 \(i\in[x,\lfloor\frac{k}{\lfloor\frac{k}{x}\rfloor}\rfloor]\),\(\lfloor\frac{k}{i}\rfloor\)的值相等
莫比乌斯函数
\(\mu(n) = \begin{cases} 0, & \exists i \in [1,m],c_i>1 \\ 1, & m \equiv 0 \pmod 2,\forall i \in [1,m],c_i=1 \\ -1, & m \equiv 1 \pmod 2,\forall i \in [1,m],c_i=1 \end{cases}\)
线性筛法
$ Code:$
mu[1]=1;
for(re int i=2;i<N;i++){
if(!mark[i]){
p[++cnt]=i;
mu[i]=-1;
}
for(re int j=1;j<=cnt&&i*p[j]<N;j++){
mark[i*p[j]]=1;
if(!(i%p[j])){
mu[i*p[j]]=0;
break;
}else mu[i*p[j]]=-mu[i];
}
}
莫比乌斯反演
狄利克雷卷积
对于 \(h(n)=\sum_{d|n} f(d)*g(\frac{n}{d})\)
记作 \(h=f\otimes g\) ,若 \(f,g\) 为积性函数,则 \(h\) 也为积性函数
对于给定数论函数 $ f,g$
\(f(n)=\sum_{d|n} g(d) \iff g(n)=\sum_{d|n} \mu(d)*f(\frac{n}{d})\)
性质：
-
\(\sum_{d|n} \mu(d) = [n=1]\)
-
\(\sum_{d|n} \phi(d) =n\)
杜教筛
应用范围： 对于积性函数 $ f(x)$ ,求 \(S(n)=\sum_{i=1}^n f(i)\)
若 \(h=f\otimes g\)
则 \(\sum_{i=1}^{n} h(i)\)
\(=\sum_{i=1}^n \sum_{d|i} f(d)·g(\frac{i}{d})\)
\(=\sum_{i=1}^n g(i)\sum_{d=1}^{\lfloor\frac{n}{i}\rfloor} f(d)\)
\(=\sum_{i=1}^n g(i) S(\lfloor\frac{n}{i}\rfloor)\)
\(=g(1)S(n)+\sum_{i=2}^n g(i)S(\lfloor\frac{n}{i}\rfloor)\)
\(\therefore S(n)=\frac{\sum_{i=1}^n h(i)-\sum_{i=2}^ng(i)S(\lfloor\frac{n}{i}\rfloor)}{g(1)}\)
\(Code:\)
#pragma GCC optimize("O2")
#include<stdio.h>
#include<map>
using namespace std;
#define ll long long
#define re register
#define N 6000001
template<class T>
inline void read(T &x){
x=0;char c=getchar();T flag=1;
while(c<'0'||c>'9'){if(c=='-')flag=-1;c=getchar();}
while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-48;c=getchar();}
x*=flag;
}
map<ll,ll> mp1,mp2;
ll phi[N],mu[N];
int p[N],cnt=0;
bool mark[N];
inline ll dfs1(ll x){
if(x<N) return phi[x];
if(mp1[x]) return mp1[x];
ll ans=1LL*(1+x)*x/2;
for(re ll l=2,r;l<=x;l=r+1){
r=(x/(x/l));
ans-=(r-l+1)*dfs1(x/l);
}
return mp1[x]=ans;
}
inline ll dfs2(ll x){
if(x<N) return mu[x];
if(mp2[x]) return mp2[x];
ll ans=1;
for(re ll l=2,r;l<=x;l=r+1){
r=(x/(x/l));
ans-=(r-l+1)*dfs2(x/l);
}
return mp2[x]=ans;
}
int T;
ll n;
int main(){
mark[1]=phi[1]=mu[1]=1;
for(re int i=2;i<N;i++){
if(!mark[i]){
p[++cnt]=i;
phi[i]=i-1;
mu[i]=-1;
}
for(re int j=1;j<=cnt&&i*p[j]<N;j++){
mark[i*p[j]]=1;
if(!(i%p[j])){
mu[i*p[j]]=0;
phi[i*p[j]]=phi[i]*p[j];
break;
}else{
mu[i*p[j]]=-mu[i];
phi[i*p[j]]=phi[i]*(p[j]-1);
}
}
}
for(re int i=1;i<N;i++)
phi[i]+=phi[i-1],mu[i]+=mu[i-1];
read(T);
while(T--){
read(n);
printf("%lld %lld\n",dfs1(n),dfs2(n));
}
}