2019南京网络赛 E-K Sum(杜教筛 反演)

K Sum

#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 int mod=1e9+7;
const int inv6=166666668;
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=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);}
itn prime[N],tot=0;
bool vis[N];
short int mu[N];
char s[N];
ll k,f[N]={0};
void f_mod(ll &x){
    if(x>=mod)x-=x/mod*mod;
}
ll inv_6(ll x){
    ll ans=x*(x+1);
    f_mod(ans);
    ans=ans*(2LL*x+1);
    f_mod(ans);
    ans=ans*inv6;
    f_mod(ans);
    return ans;
}
void pre(){
    mu[1]=1;
    for(itn i=2;i<N;i++){
        if(!vis[i])prime[++tot]=i,mu[i]=-1;
        for(int j=1;j<=tot&&i*prime[j]<N;j++){
            vis[i*prime[j]]=1;
            if(i%prime[j]==0){
                mu[i*prime[j]]=0;
                break;
            }else mu[i*prime[j]]=-mu[i];
        }
    }
    for(int i=1;i<N;i++){
        if(mu[i]!=0)
        for(int j=i;j<N;j+=i){
            f[j]+=1LL*(j/i)*(j/i)%mod*mu[i];
            f[j]=(f[j]%mod+mod)%mod;
        }
        f[i]+=f[i-1];
        f[i]=(f[i]%mod+mod)%mod;
    }
}
map<int,ll>p;
map<int,bool>q;
ll cal(int n){
    if(n<N)return f[n];
    if(q[n])return p[n];
    ll ans=inv_6(n);
    for(int i=2,last;i<=n;i=last+1){
        last=n/(n/i);
        ll x=1LL*(last-i+1)*cal(n/i);
        f_mod(x);
        if(x<0)x=(x%mod+mod)%mod;
        ans-=x;
        if(ans<0)ans=(ans%mod+mod)%mod;
    }
    q[n]=true;
    return p[n]=ans;
}
int main(){
    pre();
    int t;cin>>t;
    while(t--){
        itn n;
        sc("%d%s",&n,s);
        int mo=mod-1;
        k=0;
        ll pk=0;
        for(int i=0;s[i];i++){
            k=(1LL*k*10%mo+s[i]-'0')%mo;
            pk=(1LL*pk*10%mod+s[i]-'0')%mod;
        }
        ll m=(k+1)%mo;
        ll ans=0;
        for(ll i=1,last;i<=n;i=last+1){
            last=n/(n/i);
            ll q=n/i;
            if(q==1){
                ll x=1LL*(pk-1)*(cal(last)-cal(i-1));
                f_mod(x);
                if(x<0)x=(x%mod+mod)%mod;
                ans+=x;
                f_mod(ans);
                if(ans<0)ans=(ans%mod+mod)%mod;
            }
            ll inv=ksm(q-1,mod-2,mod);
            ll x=inv;
            ll y=(ksm(q,m+mod-1,mod)-q*q%mod);
            if(y<0)y=(y%mod+mod)%mod;
            x=x*y;
            f_mod(x);
            x=x*(cal(last)-cal(i-1));
            if(x<0)x=(x%mod+mod)%mod;
            f_mod(x);
            ans+=x;
            f_mod(ans);
            if(ans<0)ans=(ans%mod+mod)%mod;
        }
        printf("%lld\n",(ans%mod+mod)%mod);
    }
}