题意：
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 $b$.
思路：
一开始对答案打表，发现毫无规律，原来是打表的打开方式不对~~
堆GCD打表
如下
3
3
4
3 2
5
3 2 5
6
3 2 5 1
7
3 2 5 1 7
8
3 2 5 1 8 2
…
发现，如果对于 $GCD(n)$，如果 $n$，有一个质因子，那么 $GCD（n）$ = 这个质因子，否则 $GCD（n） = 1$。预处理输出即可。试除法预处理的时间复杂度为 $n\sqrt{n}$.

$AC\ Code:$

#include<iostream>
#include<cstring>
#include<queue>
#include<map>
#include<cmath>
#include<set>
#include<stack>
#include<cstdio>
#include<sstream>
#include<vector>
#include<bitset>
#include<algorithm>

using namespace std;
#define read(x) scanf("%d",&x)
#define Read(x,y) scanf("%d%d",&x,&y)
#define gc(x)  scanf(" %c",&x);
#define mmt(x,y)  memset(x,y,sizeof x)
#define write(x) printf("%d\n",x)
#define pii pair<int,int>
#define INF 0x3f3f3f3f
#define ll long long
const int N = 100000 + 100;
const int M = 3e6 + 1005;
typedef long long LL;
int gcd(int a,int b){
    if(b) return gcd(b,a%b);
    else return a;
}
ll f[M];
int prim[M];
int tot = 0;
void init(){
    int x;
    for(int i = 3;i <= 1000000;++i){
        x = i;
        tot = 0;
        for(int j = 2;j * j<=x;++j){
            if(!(x%j)){
                prim[++tot] = j;
                while(!(x%j)) x /= j;
            }
        }
        if(x > 1) prim[++tot] = x;
        if(tot > 1) f[i] = 1;
        else f[i] = prim[1];
        f[i] += f[i-1];//递推预处理
    }
}
int main(){
    int n;
    init();
    while(scanf("%d",&n)==1){
        printf("%lld\n",f[n]);
    }
}