hdu 2865 基礎polya問題

#include<cstdio>#include<iostream>using namespace std;const int MOD = 1000000007;typedef long long LL;LL n, m;int prime[37000],is[37000];void getprime(){int cnt=0;for(int i=2;i<37000;i++){if(!is[i]){prime[cnt++]=i;for(int j=i;j<37000;j+=i)is[j]=1;}}}LL pow(LL x,LL n){LL t=1;while(n){if(n&1)t=(t*x)%MOD;x=(x*x)%MOD;n >>= 1;}return t%MOD;}LL getptr(LL p,LL i){LL tmp=pow(p-1,i);if(i&1)tmp=(tmp+MOD-(p-1))%MOD;elsetmp=(tmp+p-1)%MOD;return tmp;}LL eular(LL x){if(x==1)return 1;LL rep=x;for(int i=0;prime[i]*prime[i]<=x;i++){if(x%prime[i]==0){rep-=rep/prime[i];x/=prime[i];}while(x%prime[i]==0)x/=prime[i];if(x==1)break;}if(x!=1)rep-=rep/x;return rep%MOD;}LL inv(LL n){return pow(n,MOD-2)%MOD;}int main(){getprime();while(cin>>n>>m){LL ans=0;int  i;m--;for(i = 1;i * i < n;i++){if(n%i==0){ans=(ans+(getptr(m,i)*eular(n/i))%MOD)%MOD;ans=(ans+(getptr(m,n/i)*eular(i))%MOD)%MOD;}}if(i*i==n)ans=(ans+(getptr(m,i)*eular(n/i))%MOD)%MOD;ans=((m+1)*ans)%MOD;ans=(ans*inv(n))%MOD;cout<<ans<<endl;}}