1 条题解
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#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; }
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信息
- ID
- 11537
- 时间
- 2000ms
- 内存
- 256MiB
- 难度
- 10
- 标签
- 递交数
- 5
- 已通过
- 1
- 上传者