[Luogu5106]dkw的lcm

https://minamoto.blog.luogu.org/solution-p5106

容易想到枚举质因子及其次数计算其贡献，容斥计算$\varphi(p^i)$的次方数。

#include<cstdio>
#include<algorithm>
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
typedef long long ll;
using namespace std;

const int N=1000010,mod=1e9+7;
int n,k,tot,ans,b[N],p[N];

int ksm(int a,int b,int mod){
    int res=1;
    for (; b; a=1ll*a*a%mod,b>>=1)
        if (b & 1) res=1ll*res*a%mod;
    return res;
}

void init(){
    rep(i,2,n){
        if (!b[i]) p[++tot]=i;
        for (int j=1; j<=tot && i*p[j]<=n; j++){
            b[i*p[j]]=1;
            if (i%p[j]==0) break;
        }
    }
}

int D(ll pi,ll pi1){
    int s1=ksm(n-n/pi,k,mod-1),s2=ksm(n-n/pi1,k,mod-1);
    return (s1-s2+mod-1)%(mod-1);
}

int main(){
    scanf("%d%d",&n,&k); init(); ans=1;
    rep(i,1,tot)
        for (ll t=p[i]; t<=n; t*=p[i])
            ans=1ll*ans*ksm(t-t/p[i],D(t*p[i],t),mod)%mod;
    printf("%d\n",ans);
    return 0;
}