题目分析

$\begin{split} \sum\limits_{i=1}^n\sum\limits_{j=1}^mgcd(i,j)^k&=\sum\limits_{d=1}^nd^k\sum\limits_{i=1}^n\sum\limits_{j=1}^m[gcd(i,j)==d]\\ \end{split}$

$f(x)$表示$gcd(i,j)=x$$g(x)$表示$gcd(i,j)==kx,k\in Z$

$\begin{split} g(x)&=\sum\limits_{x|d}^nf(d)\\ &=\sum\limits_{i=1}^n\sum\limits_{j=1}^m[x|gcd(i,j)]\\ &=\sum\limits_{i=1}^{\lfloor\frac n x\rfloor}\sum\limits_{j=1}^{\lfloor\frac m x\rfloor}\lfloor\frac n x\rfloor\lfloor\frac m x\rfloor\\ f(x)&=\sum\limits_{x|d}^n\mu(\frac dx)g(d)=\sum\limits_{x|d}^n\mu(\frac dx)\lfloor\frac n d\rfloor\lfloor\frac m d\rfloor \end{split}$

$\begin{split} ans&=\sum\limits_{d=1}^nd^k\cdot f(d)\\ &=\sum\limits_{d=1}^nd^k\sum\limits_{d|T}^n\mu(\frac Td)\lfloor\frac n T\rfloor\lfloor\frac m T\rfloor\\ &=\sum\limits_{T=1}^n\lfloor\frac n T\rfloor\lfloor\frac m T\rfloor\sum\limits_{d|T}\mu(\frac Td)d^k \end{split}$

代码实现

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#define MAXN 0x7fffffff
typedef long long LL;
const int N=5000005,mod=1e9+7;
using namespace std;
inline int Getint(){register int x=0,f=1;register char ch=getchar();while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}return x*f;}
int g[N],mu[N],prime[N];
bool vis[N];
LL ksm(LL x,LL k){
LL ret=1;
while(k){
if(k&1)ret=ret*x%mod;
x=x*x%mod;
k>>=1;
}
return ret;
}
int low[N];
int main(){
int T=Getint(),K=Getint();

mu[1]=g[1]=1;
for(int i=2;i<=5e6;i++){
if(!vis[i]){
prime[++prime[0]]=i,mu[i]=-1;
low[i]=i,g[i]=ksm(i,K)-1;
}
for(int j=1;j<=prime[0]&&1ll*prime[j]*i<=5e6;j++){
vis[i*prime[j]]=1;
if(i%prime[j]==0){
low[i*prime[j]]=low[i]*prime[j];
if(low[i*prime[j]]==i*prime[j])
g[i*prime[j]]=g[i]*ksm(prime[j],K)%mod;
else
g[i*prime[j]]=(1ll*g[low[i*prime[j]]]*g[i*prime[j]/low[i*prime[j]]])%mod;
break;
}
low[i*prime[j]]=prime[j];
g[i*prime[j]]=(1ll*g[i]*g[prime[j]])%mod;
mu[i*prime[j]]=-mu[i];
}
}
for(int i=1;i<=5e6;i++)g[i]=(g[i]+g[i-1])%mod;

while(T--){
int n=Getint(),m=Getint();
if(n>m)swap(n,m);
int ans=0;
for(int l=1,r;l<=n;l=r+1){
r=min(n/(n/l),m/(m/l));
ans=(ans+1ll*(n/l)*(m/l)%mod*(g[r]-g[l-1])%mod+mod)%mod;
}
cout<<ans<<'\n';
}
return 0;
}

posted @ 2018-11-22 11:56  Emiya_2020  阅读(161)  评论(0编辑  收藏  举报