又一道。。。分数和取模次数成正比$qwq$
求:$sum_{i=1}^Nsum_{j=1}^Mlcm(i,j)$
原式
$=sum_{i=1}^Nsum_{j=1}^Mfrac{i*j}{gcd(i.j)}$
$=sum_{d=1}^{N}sum_{i=1}^{lfloorfrac{N}{d} floor}sum_{j=1}^{lfloorfrac{M}{d} floor}dij[gcd(i,j)==1]$
$=sum_{d=1}^{N}sum_{i=1}^{lfloorfrac{N}{d} floor}sum_{j=1}^{lfloorfrac{M}{d} floor}dijsum_{k|gcd(i,j)}mu(k)$
$=sum_{d=1}^{N}sum_{i=1}^{lfloorfrac{N}{dk} floor}sum_{j=1}^{lfloorfrac{M}{dk} floor}dijk^2sum_{k=1}^{lfloorfrac{N}{d} floor} mu(k)$
$=sum_{d=1}^{N}dsum_{k=1}^{lfloorfrac{N}{d} floor} k^2 mu(k)sum_{i=1}^{lfloorfrac{N}{dk} floor}isum_{j=1}^{lfloorfrac{M}{dk} floor}j$
其中,$sum_{i=1}^{lfloorfrac{N}{dk} floor}isum_{j=1}^{lfloorfrac{M}{dk} floor}j$为等差数列,可以$O(1)$求,并且对于不同的$k$是可以整除分块的;
$sum_{d=1}^{N}dsum_{k=1}^{lfloorfrac{N}{d} floor} k^2mu(k)sum_{i=1}^{lfloorfrac{N}{dk} floor}isum_{j=1}^{lfloorfrac{M}{dk} floor}j$中的$lfloorfrac{N}{d} floor$对于不同的$d$也是可以整除分块的;然后对于$k^2mu(k)$先线性筛出来,再做个前缀和。
所以时间复杂度是$O(N)$的。
#include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<cctype> #include<cstdlib> #include<vector> #include<queue> #include<map> #include<set> #define ll long long #define R register int using namespace std; namespace Fread { static char B[1<<15],*S=B,*D=B; #define getchar() (S==D&&(D=(S=B)+fread(B,1,1<<15,stdin),S==D)?EOF:*S++) inline int g() { R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-1:fix; do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix; } }using Fread::g; const int M=20101009,N=10000010; int mu[N],pri[N/10],sum[N],n,m,cnt; bool v[N]; ll Inv; inline void MU(int n) { mu[1]=1; for(R i=2;i<=n;++i) { if(!v[i]) pri[++cnt]=i,mu[i]=-1; for(R j=1;j<=cnt&&i*pri[j]<=n;++j) { v[i*pri[j]]=true; if(i%pri[j]==0) break; mu[i*pri[j]]=-mu[i]; } } for(R i=1;i<=n;++i) sum[i]=(ll)(sum[i-1]+(ll)(mu[i]+M)*i%M*i)%M; } inline int S(int x,int y) {return (ll)x*(x+1)%M*Inv%M*y%M*(y+1)%M*Inv%M;} inline int F(int n,int m) {register ll ret=0; n>m?swap(n,m):void(0); for(R l=1,r;l<=n;l=r+1) { r=min(n/(n/l),m/(m/l)); ret=(ret+(ll)(sum[r]-sum[l-1]+M)*S(n/l,m/l)%M)%M; } return ret; } signed main() { #ifdef JACK freopen("NOIPAK++.in","r",stdin); #endif n=g(),m=g(); n>m?swap(n,m):void(0); MU(n); register ll ans=0;Inv=(M+1)/2; for(R l=1,r;l<=n;l=r+1) { r=min(n/(n/l),m/(m/l)); (ans+=(ll)(r-l+1)*(l+r)%M*Inv%M*F(n/l,m/l)%M)%=M; } printf("%lld ",ans); }
2019.06.09