[杜教筛] 51Nod 1237 最大公约数之和 V3

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ans = ∑1<=i<=n ∑1<=j<=n ∑d|(i,j) phi(d)

= ∑1<=i<=n phi(i) * (n / i) * (n / i)

直接用杜教筛跑欧拉函数前缀和就好了

#include<cstdio>#include<cstdlib>#include<algorithm>#include <tr1/unordered_map>typedef long long ll;using namespace std;using namespace std::tr1;const int maxn=10000000;int prime[1000000],num;int vst[maxn+5],phi[maxn+5];const ll P=1e9+7;const ll inv=500000004;inline void Pre(){  phi[1]=1;  for (int i=2;i<=maxn;i++){    if (!vst[i]) prime[++num]=i,phi[i]=i-1;    for (int j=1;j<=num && (ll)i*prime[j]<=maxn;j++){      vst[i*prime[j]]=1;      if (i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;      }elsephi[i*prime[j]]=phi[i]*phi[prime[j]];    }  }  for (int i=1;i<=maxn;i++) (phi[i]+=phi[i-1])%=P;}unordered_map<ll,int> S;inline int Sum(ll n){  if (n<=maxn) return phi[n];  if (S.find(n)!=S.end()) return S[n];  int tem=(ll)(n%P)*((n+1)%P)%P*inv%P; ll l,r;  for (l=2;l*l<=n;l++) (tem+=P-Sum(n/l))%=P;  for (ll t=n/l;l<=n;l=r+1,t--)    r=n/t,(tem+=P-(ll)(r-l+1)*Sum(t)%P)%=P;  return S[n]=tem;}int main(){  ll n; int Ans=0;  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  Pre();  scanf("%lld",&n);  for (ll i=1,j;i<=n;i=j+1){    j=n/(n/i);    (Ans+=((n/i)%P)*((n/i)%P)%P*(Sum(j)+P-Sum(i-1))%P)%=P;  }  printf("%d\n",Ans);  return 0;}

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