51Nod-1363 最小公倍数之和(积性函数)

$Description$
$Calculate \quad \sum_{i=1}^{n}lcm(i,n)$

$Solution$
$\sum_{i=1}^{n}lcm(i,n) \\=\sum_{i=1}^{n}\frac{i\cdot n}{gcd(i,n)} \\=\sum_{d|n}\sum_{i=1}^{n}\frac{i\cdot n}{d}[gcd(i,n)=d] \\ =\sum_{d|n}n\sum_{i=1}^{\frac{n}{d}}i[gcd(i,\frac{n}{d})=1] \\=n\sum_{d|n}\frac{d\cdot \varphi(d)+[d=1]}{2} \\=\frac{n}{2}\cdot (\prod_{p^k||n}(1+\frac{p^{2\cdot k+1}-p}{p+1})+1)$

$Code$

#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 int inv2=mod+1>>1;
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(ll a,ll b,ll c){ll ans=0;for(;b;b>>=1,a=(a+a)%c)if(b&1)ans=(ans+a)%c;return ans;}
inline ll ksm(ll a,ll b,ll c){ll ans=1;for(;b;b>>=1,a=mul(a,a,c))if(b&1)ans=mul(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]={};
void pre(){
    for(int i=2;i<N;i++){
        if(!vis[i])prime[++tot]=i;
        for(int j=1;j<=tot&&i*prime[j]<N;j++){
            vis[i*prime[j]]=1;
            if(i%prime[j]==0)break;
        }
    }
}
int main(){
    pre();
    int t;cin>>t;
    while(t--){
        int n;
        sc("%d",&n);
        ll ans=1,m=n;
        for(int i=1;i<=tot&&1LL*prime[i]*prime[i]<=n;i++){
            if(n%prime[i]==0){
                int k=0;
                while(n%prime[i]==0)n/=prime[i],k++;
                ll res=1LL+(ksm(prime[i],2*k+1,mod)-prime[i]+mod)%mod*ksm(prime[i]+1,mod-2,mod);
                res%=mod;
                ans=ans*res%mod;
            }
        }
        if(n>1)ans=ans*((1LL*n*n-n+1)%mod)%mod;
        printf("%lld\n",(ans+1)*m%mod*inv2%mod);
    }
}