51Nod 1239 欧拉函数之和
Description
求\(\sum_{i=1}^n\varphi(i),n\leqslant 10^{10}\)
Solution
杜教筛...贴代码...
Code
#include <bits/stdc++.h> using namespace std; typedef long long ll; typedef long double ld; const int N = 2000050; const ll p = 1000000007; ll Pow(ll a,ll b,ll r=1) { for(;b;b>>=1,a=a*a%p) if(b&1) r=r*a%p;return r; } ll mul(ll a,ll b) { return (a*b-((ll)((ld)a/p*b+1e-3))*p+p)%p; } int b[N],pr[N],cp; ll phi[N],sp[N],inv2=Pow(2,p-2); void pre(int n) { for(int i=2;i<=n;i++) { if(!b[i]) pr[++cp]=i,phi[i]=i-1; for(int j=1;j<=cp && i*pr[j]<=n;j++) { b[i*pr[j]]=1; if(i%pr[j]) phi[i*pr[j]]=phi[i]*(pr[j]-1); else { phi[i*pr[j]]=phi[i]*pr[j];break; } } }phi[1]=1; for(int i=1;i<=n;i++) sp[i]=(sp[i-1]+phi[i])%p; } map<ll,ll> mp; ll S(ll n) { if(n<=2000000) return sp[n]; if(mp.count(n)) return mp[n]; ll fn=(n%p)*((n+1)%p)%p*inv2%p; for(ll i=2,j;i<=n;i=j+1) { j=n/(n/i); fn=(fn-(j-i+1)%p*S(n/i)%p+p)%p; }return mp[n]=fn; } int main() { pre(2000000); ll n; scanf("%lld",&n); printf("%lld\n",S(n)); return 0; }