正题

题目链接:https://www.luogu.com.cn/problem/P4449

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题目大意

$T$组询问给出$n,m$求$$\sum _{i=1}^n\sum _{j=1}^{m}gcd(i,j{)}^k$$

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解题思路

$$\sum _{i=1}^n\sum _{j=1}^{m}gcd(i,j{)}^k$$

$$\sum _{d=1}^n{d}^k\sum _{i=1}^n\sum _{j=1}^m[gcd(i,j)==d]$$

$$f(x)=\sum _{i=1}^n\sum _{j=1}^m[gcd(i,j)==x]$$

$$F(x)=\sum _{x|i}f(i)$$

$$F(x)=\lfloor \frac{n}{x}\rfloor \lfloor \frac{m}{x}\rfloor$$

$$f(x)=\sum _{x|i}F(x)=\sum _{x|i}\mu (\frac{i}{x})\lfloor \frac{n}{i}\rfloor \lfloor \frac{m}{i}\rfloor$$

$$\sum _{x=1}^n\lfloor \frac{n}{x}\rfloor \lfloor \frac{m}{x}\rfloor \sum _{x|i}\mu (\frac{i}{x})i^k$$

然后预处理${\sum }_{x|i}\mu (\frac{i}{x})i^k$的前缀和即可。

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$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const ll N=5e6+10,XJQ=1e9+7;
ll T,k,n,m,cnt,mu[N],g[N],pw[N],pri[N];
bool vis[N];
ll power(ll x,ll b){ll ans=1;while(b){if(b&1)ans=ans*x%XJQ;x=x*x%XJQ;b>>=1;}return ans;
} 
void prime(){g[1]=1;for(ll i=2;i<N;i++){if(!vis[i]){pri[++cnt]=i;pw[cnt]=power(i,k);g[i]=(pw[cnt]-1+XJQ)%XJQ;}for(ll j=1;i*pri[j]<N&&j<=cnt;j++){vis[i*pri[j]]=1;if(i%pri[j]==0){g[i*pri[j]]=g[i]*pw[j]%XJQ;break;}g[i*pri[j]]=g[i]*g[pri[j]]%XJQ;}}for(ll i=1;i<N;i++)g[i]=(g[i-1]+g[i])%XJQ;return;
}
int main()
{scanf("%lld%lld",&T,&k);prime();while(T--){scanf("%lld%lld",&n,&m);if(n>m)swap(n,m);ll ans=0;for(ll l=1,r;l<=n;l=r+1){r=min(n/(n/l),m/(m/l));(ans+=(n/l)*(m/l)%XJQ*(g[r]-g[l-1]+XJQ)%XJQ)%=XJQ;}printf("%lld\n",(ans+XJQ)%XJQ);}
}