BZOJ 4805(欧拉函数求和-杜教筛)

Description

给出一个数字N，求sigma(phi(i)),1<=i<=N
Input

正整数N。N<=2*10^9
Output

输出答案。
Sample Input

10
Sample Output

32
HINT

杜教筛入门

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
#define MAXN (10000000+100) 
ll p[MAXN],tot,phi[MAXN];
bool b[MAXN]={0};
void make_prime(int n)
{
    tot=0; phi[1]=1;
    Fork(i,2,n)
    {
        if (!b[i]) p[++tot]=i,phi[i]=i-1;
        For(j,tot)
        {
            if (i*p[j]>n) break;
            b[i*p[j]]=1;
            phi[i*p[j]]=phi[i]*phi[p[j]];
            if (i%p[j]==0) {
                phi[i*p[j]]= phi[i]*p[j];
                break;
            }
        }
    }
}
map<ll,ll> h;
map<ll,ll>::iterator it;

ll calc_phi(ll n) {
    if (n<=1e7) return phi[n];
    if ((it=h.find(n))!=h.end()) return (*it).se;
    ll last=1,ans=n*(n+1)/2;
    for(int i=2;i<=n;i=last+1) {
        last=n/(n/i);
        ans-=calc_phi(n/i)*(last-i+1);
    }
    return h[n]=ans;

}
int main() 
{
//  freopen("bzoj4805.in","r",stdin);
//  freopen(".out","w",stdout);
    make_prime(1e7);
    Fork(i,2,1e7) phi[i]+=phi[i-1];
    ll n;
    cin>>n;
    cout<<calc_phi(n);  

    return 0;
}