1 条题解

  • 0
    @ 2026-7-7 22:10:15
    #include<bits/stdc++.h>
    using namespace std;
    
    using ll=long long;
    
    const ll mod = 1e9+7;
    constexpr ll ksm(ll a,ll b){
        ll r = 1;
        while(b){
            if(b&1)r=r*a%mod;
            a=a*a%mod;
            b>>=1;
        }
        return r;
    }
    constexpr ll inv2 = ksm(2, mod-2);
    constexpr ll inv6 = ksm(6, mod-2);
    
    constexpr ll sq(ll n){n%=mod;return n*n%mod;}
    constexpr ll sum(ll n){n%=mod;return n*(n+1)%mod*inv2%mod;}
    constexpr ll sum_sq(ll n){n%=mod;return n*(n+1)%mod*(2*n+1)%mod*inv6%mod;}
    
    const int N = 2e5+5;
    
    
    ll p[N],tot; bool vis[N];
    void init_sieve(){
        for(int i=2;i<N;i++){
            if(!vis[i])p[++tot]=i;
            for(int j=1;j<=tot && i*p[j]<N;j++){
                vis[i*p[j]]=1;
                if(i%p[j]==0)break;
            }
        }
    }
    
    ll v[N],m;
    int id1[N],id2[N];
    ll g1[N],g2[N];
    ll n;
    ll get(ll x){
        return x<N?id1[x]:id2[n/x];
    }
    void init_sqrt(){
        for(ll l=1,r;l<=n;l=r+1){
            r=n/(n/l);
            v[++m]=n/l;
            if(v[m]<N)id1[v[m]]=m;
            else id2[n/v[m]]=m;
            g1[m]=(sum_sq(v[m])+mod-1)%mod, g2[m]=(sum(v[m])+mod-1)%mod;
        }
    }
    
    void calc_g(){
        for(int j=1;j<=tot;j++){
            for(int i=1;i<=m && p[j]<=v[i]/p[j];i++){
                g1[i] = (g1[i] - sq(p[j])*g1[get(v[i]/p[j])]%mod + sq(p[j])*g1[get(p[j-1])] + mod) % mod;
                g2[i] = (g2[i] - p[j]*g2[get(v[i]/p[j])]%mod + p[j]*g2[get(p[j-1])] + mod) % mod;
            }
        }
    }
    
    ll f(ll x){x%=mod;return (x*x%mod-x+mod)%mod;}
    ll S(ll x,int j){
        if(p[j] >= x) return 0;
        ll res = (g1[get(x)]-g2[get(x)]-g1[get(p[j])]+g2[get(p[j])]+mod+mod)%mod;
        for(int k=j+1;k<=tot && p[k]<=x/p[k];k++){
            ll w = p[k];
            for(;w<=x/p[k];w*=p[k]){
                res = (res + f(w) * S(x/w, k)%mod + f(w*p[k]))%mod;
            }
        }
        return res;
    }   
    
    int main(){
        ios::sync_with_stdio(0);cin.tie(0);
        cin>>n;
        init_sieve();
        init_sqrt();
        calc_g();
    
        cout << ((S(n, 0) + 1)%mod+mod)%mod << '\n';
        return 0;
    }
    

    #include<bits/stdc++.h>
    using namespace std;
    
    using ll=long long;
    
    const ll mod = 1e9+7;
    constexpr ll ksm(ll a,ll b){
        ll r = 1;
        while(b){
            if(b&1)r=r*a%mod;
            a=a*a%mod;
            b>>=1;
        }
        return r;
    }
    constexpr ll inv2 = ksm(2, mod-2);
    constexpr ll inv6 = ksm(6, mod-2);
    
    constexpr ll sq(ll n){n%=mod;return n*n%mod;}
    constexpr ll sum(ll n){n%=mod;return n*(n+1)%mod*inv2%mod;}
    constexpr ll sum_sq(ll n){n%=mod;return n*(n+1)%mod*(2*n+1)%mod*inv6%mod;}
    
    const int N = 2e5+5;
    
    
    ll p[N],tot; bool vis[N];
    void init_sieve(){
        for(int i=2;i<N;i++){
            if(!vis[i])p[++tot]=i;
            for(int j=1;j<=tot && i*p[j]<N;j++){
                vis[i*p[j]]=1;
                if(i%p[j]==0)break;
            }
        }
    }
    
    ll v[N],m;
    int id1[N],id2[N];
    ll g1[N],g2[N];
    ll n;
    ll get(ll x){
        return x<N?id1[x]:id2[n/x];
    }
    void init_sqrt(){
        for(ll l=1,r;l<=n;l=r+1){
            r=n/(n/l);
            v[++m]=n/l;
            if(v[m]<N)id1[v[m]]=m;
            else id2[n/v[m]]=m;
            g1[m]=(sum_sq(v[m])+mod-1)%mod, g2[m]=(sum(v[m])+mod-1)%mod;
        }   
    }
    
    void calc_g(){
        for(int j=1;j<=tot;j++){
            for(int i=1;i<=m && p[j]<=v[i]/p[j];i++){
                g1[i] = (g1[i] - sq(p[j])*g1[get(v[i]/p[j])]%mod + sq(p[j])*g1[get(p[j-1])] + mod) % mod;
                g2[i] = (g2[i] - p[j]*g2[get(v[i]/p[j])]%mod + p[j]*g2[get(p[j-1])] + mod) % mod;
            }
        }
    }
    
    ll f(ll x){x%=mod;return (x*x%mod-x+mod)%mod;}
    ll g[N],S[N];
    
    int main(){
        ios::sync_with_stdio(0);cin.tie(0);
        cin>>n;
        init_sieve();
        init_sqrt();
        calc_g();
    
        for(int i=1;i<=m;i++)S[i]=g[i]=(g1[i]-g2[i]+mod)%mod;
        for(int j=tot-1;j>=0;j--){
            for(int i=1;i<=m && p[j+1]<=v[i]/p[j+1];i++){
                for(ll w=p[j+1];w<=v[i]/p[j+1];w*=p[j+1]){
                    S[i] = (S[i] + f(w)*(S[get(v[i]/w)]-g[get(p[j+1])]+mod)%mod + f(w*p[j+1]))%mod;
                }
            }
        }
        cout << S[get(n)]+1 << '\n';
        return 0;
    }
    
    • 1

    信息

    ID
    11537
    时间
    2000ms
    内存
    256MiB
    难度
    10
    标签
    递交数
    5
    已通过
    1
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