HDOJ 2582 f(n)

Discription

This time I need you to calculate the f(n) . (3<=n<=1000000) 

f(n)= Gcd(3)+Gcd(4)+…+Gcd(i)+…+Gcd(n). 
Gcd(n)=gcd(C[n][1],C[n][2],……,C[n][n-1]) 
C[n][k] means the number of way to choose k things from n some things. 
gcd(a,b) means the greatest common divisor of a and b.

Input

There are several test case. For each test case:One integer n(3<=n<=1000000). The end of the in put file is EOF.

Output

For each test case: 
The output consists of one line with one integer f(n).

Sample Input

3
26983

Sample Output

3
37556486


本来毫无思路，然而打了个表找了找规律，发现Gcd(x)无非两种情况：
1.当x=p^q时，其中p为质数，那么Gcd(x)=p
2.其他的时候Gcd(x)=1

然后就是个水题了

#include<bits/stdc++.h>
#define ll long long
#define maxn 1000000
using namespace std;
int zs[maxn/5],t=0,n;
ll f[maxn+5];
bool v[maxn+5];

inline void init(){
    for(int i=2;i<=maxn;i++){
        if(!v[i]) f[i]=i,zs[++t]=i;
        for(int j=1,u;j<=t&&(u=zs[j]*i)<=maxn;j++){
            v[u]=1;
            if(!(i%zs[j])){
                f[u]=f[i];
                break;
            }
            f[u]=1;
        }
    }
    
    for(int i=4;i<=maxn;i++) f[i]+=f[i-1];
}

int main(){
    init();
    while(scanf("%d",&n)==1) printf("%lld
",f[n]);
    return 0;
}