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}+Tsqrt{N_i})) 其中T为数据组数
[sumlimits_{i=1}^nsumlimits_{j=1}^moperatorname{lcm(i,j)}
]
[=sumlimits_{i=1}^nsumlimits_{j=1}^mfrac{icdot j}{gcd(i,j)}
]
[=sumlimits_{i=1}^n
sumlimits_{i=1}^m
sumlimits_{d=1}^{min(n,m)}frac{icdot j}d[gcd(i,j)=d]
]
[=sumlimits_{d=1}^{min(n,m)}
sumlimits_{i=1}^{lfloorfrac{n}d
floor}
sumlimits_{i=1}^{lfloorfrac{m}d
floor}
dijcdot [gcd(i,j)=1]
]
[=sumlimits_{d=1}^{min(n,m)}
sumlimits_{i=1}^{lfloorfrac{n}d
floor}
sumlimits_{i=1}^{lfloorfrac{m}d
floor}
dij
sumlimits_{x|gcd(i,j)}^{}μ(x)
]
[=sumlimits_{d=1}^{min(n,m)}
sumlimits_{i=1}^{lfloorfrac{n}d
floor}
sumlimits_{i=1}^{lfloorfrac{m}d
floor}
dij
sumlimits_{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
[=sumlimits_{d=1}^{min(n,m)}d
sumlimits_{x=1}^{min(lfloorfrac{n}d
floor,lfloorfrac{m}d
floor)}μ(x)
sumlimits_{i=1}^{lfloorfrac{n}{dx}
floor}ix
sumlimits_{j=1}^{lfloorfrac{m}{dx}
floor}jx
]
然后把x也调到前面去
[=sumlimits_{d=1}^{min(n,m)}d
sumlimits_{x=1}^{min(lfloorfrac{n}d
floor,lfloorfrac{m}d
floor)}x^2μ(x)
sumlimits_{i=1}^{lfloorfrac{n}{dx}
floor}
sumlimits_{j=1}^{lfloorfrac{m}{dx}
floor}ij
]
然后等差数列
[=sumlimits_{d=1}^{min(n,m)}d
sumlimits_{x=1}^{min(lfloorfrac{n}d
floor,lfloorfrac{m}d
floor)}x^2mu(x)
frac{(1+lfloorfrac{n}{dx}
floor)lfloorfrac{n}{dx}
floor}{2}frac{(1+lfloorfrac{m}{dx}
floor)lfloorfrac{m}{dx}
floor}{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})
[=sumlimits_{T=1}^{min(n,m)}
frac{(1+lfloorfrac{n}{T}
floor)lfloorfrac{n}{T}
floor}{2}frac{(1+lfloorfrac{m}{T}
floor)lfloorfrac{m}{T}
floor}{2}
sumlimits_{d|T}dfrac{T^2}{d^2}mu(frac{T}{d})
]
由于对称性
[=sumlimits_{T=1}^{min(n,m)}
frac{(1+lfloorfrac{n}{T}
floor)lfloorfrac{n}{T}
floor}{2}frac{(1+lfloorfrac{m}{T}
floor)lfloorfrac{m}{T}
floor}{2}
sumlimits_{d|T}frac{T}{d}d^2mu({d})
]
[=sumlimits_{T=1}^{min(n,m)}
frac{(1+lfloorfrac{n}{T}
floor)lfloorfrac{n}{T}
floor}{2}frac{(1+lfloorfrac{m}{T}
floor)lfloorfrac{m}{T}
floor}{2}
sumlimits_{d|T}Tdmu({d})
]
然后我们设
[g(T)=sumlimits_{d|T}Tdmu({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
",query(n,m));
return 0;
}