#莫比乌斯反演，整除分块#bzoj 2154 bzoj 2693 jzoj 1938 洛谷 1829 Crash的数字表格 or JZPTAB

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题目

求 $\sum_{i=1}^n\sum_{j=1}^mlcm(i,j)$

分析

原式= $\sum_{i=1}^n\sum_{j=1}^m\frac{ij}{gcd(i,j)}$
那么原式= $\sum_{d=1}^{min(n,m)}d\times \sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}[gcd(i,j)=1]\times i\times j$
根据莫比乌斯反演可以得到原式= $\sum_{d=1}^{min(n,m)}d\sum_{t=1}^{min(\lfloor\frac{n}{d}\rfloor,\lfloor\frac{m}{d}\rfloor)}t^2\mu(t)\sum_{i=1}^{\lfloor\frac{n}{dt}\rfloor}i\sum_{j=1}^{\lfloor\frac{m}{dt}\rfloor}j$
就可以预处理 $t^2\mu(t)$ ,用整除分块求解，时间复杂度大概是 $O (n)$

代码

#include <cstdio>
#define rr register
#define min(a,b) ((a)<(b)?(a):(b))
#define res(n,m) ((1ll*n*(n+1)>>1)%mod*((1ll*m*(m+1)>>1)%mod)%mod)
using namespace std;
const int mod=20101009,M=10000010;
int n,m,v[M],prime[M],s[M],cnt;
inline void prepa(int N){s[1]=1;for (rr int i=2;i<=N;++i){if (!v[i]) s[i]=-1,v[i]=prime[++cnt]=i;for (rr int j=1;j<=cnt&&prime[j]*i<=N;++j){v[i*prime[j]]=prime[j];if (i%prime[j]) s[i*prime[j]]=-s[i];else break;}}for (rr int i=1;i<=N;++i) s[i]=1ll*i*i%mod*(s[i]+mod)%mod;for (rr int i=2;i<=N;++i) (s[i]+=s[i-1])%=mod;
}
inline signed answ(int n,int m){rr int ans=0,t=min(n,m);for (rr int l=1,r;l<=t;l=r+1){r=min(n/(n/l),m/(m/l));(ans+=1ll*(s[r]-s[l-1]+mod)*res(n/l,m/l)%mod)%=mod;}return ans;
}
inline signed solve(int n,int m){rr int ans=0,t;prepa(t=min(n,m)); for (rr int l=1,r;l<=t;l=r+1){r=min(n/(n/l),m/(m/l));(ans+=(1ll*(r-l+1)*(l+r)>>1)%mod*answ(n/l,m/l)%mod)%=mod;}return ans;
}
signed main(){scanf("%d%d",&n,&m);printf("%d",solve(n,m));return 0;
}