[国家集训队]Crash的数字表格(莫比乌斯反演)
这道题在bzoj有多组数据,那么洛谷题解里的双重分块在bzoj就会T
那我就讲一下只有一重分块的做法
时间复杂度(大概) O(\(N_{max}+T\sqrt{N_i}\)) 其中T为数据组数
\[\sum\limits_{i=1}^n\sum\limits_{j=1}^m\operatorname{lcm(i,j)}
\]
\[=\sum\limits_{i=1}^n\sum\limits_{j=1}^m\frac{i\cdot j}{\gcd(i,j)}
\]
\[=\sum\limits_{i=1}^n
\sum\limits_{i=1}^m
\sum\limits_{d=1}^{min(n,m)}\frac{i\cdot j}d[gcd(i,j)=d]
\]
\[=\sum\limits_{d=1}^{min(n,m)}
\sum\limits_{i=1}^{\lfloor\frac{n}d\rfloor}
\sum\limits_{i=1}^{\lfloor\frac{m}d\rfloor}
dij\cdot [gcd(i,j)=1]
\]
\[=\sum\limits_{d=1}^{min(n,m)}
\sum\limits_{i=1}^{\lfloor\frac{n}d\rfloor}
\sum\limits_{i=1}^{\lfloor\frac{m}d\rfloor}
dij
\sum\limits_{x|gcd(i,j)}^{}μ(x)
\]
\[=\sum\limits_{d=1}^{min(n,m)}
\sum\limits_{i=1}^{\lfloor\frac{n}d\rfloor}
\sum\limits_{i=1}^{\lfloor\frac{m}d\rfloor}
dij
\sum\limits_{x|i,x|j}μ(x)
\]
然后把μ(x)和d调到前面去,i变成ix,j变成jx
\[=\sum\limits_{d=1}^{min(n,m)}d
\sum\limits_{x=1}^{min(\lfloor\frac{n}d\rfloor,\lfloor\frac{m}d\rfloor)}μ(x)
\sum\limits_{i=1}^{\lfloor\frac{n}{dx}\rfloor}ix
\sum\limits_{j=1}^{\lfloor\frac{m}{dx}\rfloor}jx
\]
然后把x也调到前面去
\[=\sum\limits_{d=1}^{min(n,m)}d
\sum\limits_{x=1}^{min(\lfloor\frac{n}d\rfloor,\lfloor\frac{m}d\rfloor)}x^2μ(x)
\sum\limits_{i=1}^{\lfloor\frac{n}{dx}\rfloor}
\sum\limits_{j=1}^{\lfloor\frac{m}{dx}\rfloor}ij
\]
然后等差数列
\[=\sum\limits_{d=1}^{min(n,m)}d
\sum\limits_{x=1}^{min(\lfloor\frac{n}d\rfloor,\lfloor\frac{m}d\rfloor)}x^2\mu(x)
\frac{(1+\lfloor\frac{n}{dx}\rfloor)\lfloor\frac{n}{dx}\rfloor}{2}\frac{(1+\lfloor\frac{m}{dx}\rfloor)\lfloor\frac{m}{dx}\rfloor}{2}
\]
在洛谷的话推到这一步就够了
但是想要在bzoj过这道题还要接着优化
然后设T=dx,所以x=\(\frac{T}{d}\)
\[=\sum\limits_{T=1}^{min(n,m)}
\frac{(1+\lfloor\frac{n}{T}\rfloor)\lfloor\frac{n}{T}\rfloor}{2}\frac{(1+\lfloor\frac{m}{T}\rfloor)\lfloor\frac{m}{T}\rfloor}{2}
\sum\limits_{d|T}d\frac{T^2}{d^2}\mu(\frac{T}{d})
\]
由于对称性
\[=\sum\limits_{T=1}^{min(n,m)}
\frac{(1+\lfloor\frac{n}{T}\rfloor)\lfloor\frac{n}{T}\rfloor}{2}\frac{(1+\lfloor\frac{m}{T}\rfloor)\lfloor\frac{m}{T}\rfloor}{2}
\sum\limits_{d|T}\frac{T}{d}d^2\mu({d})
\]
\[=\sum\limits_{T=1}^{min(n,m)}
\frac{(1+\lfloor\frac{n}{T}\rfloor)\lfloor\frac{n}{T}\rfloor}{2}\frac{(1+\lfloor\frac{m}{T}\rfloor)\lfloor\frac{m}{T}\rfloor}{2}
\sum\limits_{d|T}Td\mu({d})
\]
然后我们设
\[g(T)=\sum\limits_{d|T}Td\mu({d})
\]
然后可以发现g是一个积性函数
所以有
\[g(\Pi P_i^{ai})=\Pi g(P_i^{ai})
\]
\[=\Pi (T*P_i^{0}*\mu(1)+T*P_i*\mu(1))
\]
\[=\Pi (T-T*P_i)
\]
\[=\Pi (T*(1-P_i))
\]
因此
\[g(p)=1-p
\]
\[g(p^k)=g(p^{k-1})
\]
当np互质时,有
\[g(np)=g(n)*g(p)
\]
然后筛一筛就出来了
代码：
#pragma GCC optimize(3)
#include<bits/stdc++.h>
#define int long long
#define N 10000010
using namespace std;
const int mod=20101009;
int vis[N],prim[N/10],mu[N],low[N],sum[N],cnt;
void pre(int n){
mu[1]=1;
sum[1]=low[1]=1;
for(int i=2;i<=n;i++){
if(!vis[i])mu[i]=-1,prim[++cnt]=i,low[i]=i,sum[i]=(1-i+mod)%mod;
for(int j=1;j<=cnt&&i*prim[j]<=n;j++){
vis[i*prim[j]]=1;
if(i%prim[j]==0){
low[i*prim[j]]=low[i]*prim[j];
if(low[i]==i)sum[i*prim[j]]=sum[i];
else sum[i*prim[j]]=sum[i/low[i]]*sum[prim[j]*low[i]]%mod;
break;
}else{
low[i*prim[j]]=prim[j];
sum[i*prim[j]]=sum[i]*sum[prim[j]]%mod;
mu[i*prim[j]]=-mu[i];
}
}
}
for(int T=1;T<=n;T++)sum[T]=(T*sum[T]%mod+sum[T-1])%mod;
}
int ask(int x,int y){return (x+y)*(y-x+1)/2%mod;}
int query(int n,int m){
int re=0;
int mx=min(n,m);
for(int l=1,r;l<=mx;l=r+1){
r=min(n/(n/l),m/(m/l));
int tmp=ask(1,n/l)*ask(1,m/l)%mod;
tmp=tmp*(sum[r]-sum[l-1]+mod)%mod;
re=(re+tmp)%mod;
}
return re;
}
int t;
signed main(){
pre(10000000);
int n,m;
scanf("%lld%lld",&n,&m);
printf("%lld\n",query(n,m));
return 0;
}
