HDU-5780 gcd(欧拉函数)

$Description$
$Calculate \quad \sum_{i=1}^{n}\sum_{j=1}^{n}gcd(x^i-1,x^j-1).$

$Solution$
$\sum_{i=1}^{n}\sum_{j=1}^{n}gcd(x^i-1,x^j-1) \\=\sum_{i=1}^{n}\sum_{j=1}^{n}(x^{gcd(i,j)}-1) \\=\sum_{d=1}^{n}(x^d-1)\sum_{i=1}^{n}\sum_{j=1}^{n}[gcd(i,j)=d] \\=\sum_{d=1}^{n}(x^d-1)\sum_{i=1}^{\frac{n}{d}}\sum_{i=j}^{\frac{n}{d}}[gcd(i,j)=1] \\=\sum_{d=1}^{n}(x^d-1)\cdot (2\sum_{i=1}^{\frac{n}{d}}\varphi(i)-1)$

$Code$

#include<bits/stdc++.h>
#define me(a,x) memset(a,x,sizeof(a))
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
#define sc scanf
#define itn int
#define STR clock_t startTime = clock();
#define END clock_t endTime = clock();cout << double(endTime - startTime) / CLOCKS_PER_SEC *1000<< "ms" << endl;
using namespace std;
const int N=1e6+5;
const long long mod=1e9+7;
const long long mod2=998244353;
const int oo=0x7fffffff;
const int sup=0x80000000;
typedef long long ll;
typedef unsigned long long ull;
template <typename it>void db(it *begin,it *end){while(begin!=end)cout<<(*begin++)<<" ";puts("");}
template <typename it>
string to_str(it n){string s="";while(n)s+=n%10+'0',n/=10;reverse(s.begin(),s.end());return s;}
template <typename it>int o(it a){cout<<a<<endl;return 0;}
inline ll mul_64(ll x,ll y,ll c){return (x*y-(ll)((long double)x/c*y)*c+c)%c;}
inline ll ksm(ll a,ll b,ll c){ll ans=1;for(;b;b>>=1,a=a*a%c)if(b&1)ans=ans*a%c;return ans;}
inline void exgcd(ll a,ll b,ll &x,ll &y){if(!b)x=1,y=0;else exgcd(b,a%b,y,x),y-=x*(a/b);}
int prime[N],tot=0;
bool vis[N]={0};
ll phi[N];
void pre(){
    phi[1]=1;
    for(int i=2;i<N;i++){
        if(!vis[i])prime[++tot]=i,phi[i]=i-1;
        for(int j=1;j<=tot&&i*prime[j]<N;j++){
            vis[i*prime[j]]=1;
            if(i%prime[j]==0){
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }else phi[i*prime[j]]=phi[i]*(prime[j]-1);
        }
    }
    for(int i=1;i<N;i++)phi[i]=(phi[i-1]+phi[i])%mod;
}
ll cal(ll x,ll n){
    if(n==0)return 0;
    ll inv=ksm(x-1,mod-2,mod);
    ll k=ksm(x,n,mod);
    ll ans=x*(k-1)%mod*inv%mod;
    ans=(ans-n+mod)%mod;
    return ans;
}
int main(){
    pre();
    int t;cin>>t;
    while(t--){
        int x,n;
        sc("%d%d",&x,&n);
        if(x==1){
            o(0);
            continue;
        }
        ll ans=0;
        for(ll i=1,last;i<=n;i=last+1){
            last=n/(n/i);
            ans+=(cal(x,last)-cal(x,i-1))*(2LL*phi[n/i]-1)%mod;
            if(ans>=mod)ans-=ans/mod*mod;
            if(ans<0)ans=(ans%mod+mod)%mod;
        }
        printf("%lld\n",ans);
    }   
}