链接：https://www.nowcoder.com/acm/contest/141/H
来源：牛客网
 

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

题目描述

Eddy has solved lots of problem involving calculating the number of coprime pairs within some range. This problem can be solved with inclusion-exclusion method. Eddy has implemented it lots of times. Someday, when he encounters another coprime pairs problem, he comes up with diff-prime pairs problem. diff-prime pairs problem is that given N, you need to find the number of pairs (i, j), where and are both prime and i ,j ≤ N. gcd(i, j) is the greatest common divisor of i and j. Prime is an integer greater than 1 and has only 2 positive divisors.

Eddy tried to solve it with inclusion-exclusion method but failed. Please help Eddy to solve this problem.

Note that pair (i1, j1) and pair (i2, j2) are considered different if i1 ≠ i2 or j1 ≠ j2.

输入描述:

Input has only one line containing a positive integer N.

1 ≤ N ≤ 107

输出描述:

Output one line containing a non-negative integer indicating the number of diff-prime pairs (i,j) where i, j ≤ N

示例1

输入

3

输出

2

示例2

输入

5

输出

6

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL MAXN = 1e7+10;
const LL N = 1e6+10;
bool pri[MAXN];
LL a[N];
LL num,ans;
 
void get_prime(LL n){
    memset(pri,true,sizeof(pri));
    pri[0]=pri[1]=false;
    for (LL i = 2; i <=n ; ++i) {
        if(pri[i]) {
            for (LL j = 2; j*i <=n ; ++j) {
                pri[j*i]=false;
            }
        }
    }
    num = 0;
    for (LL i=1;i<=n;++i){
        if(pri[i]){
            a[++num] = i;
        }
    }
}
int main(){
    LL n;
    scanf("%lld",&n);
    get_prime(n);
    for(LL j=2;j<=num;j++)
    {
        long long  k;
        for( k=1;;k++){
            if(a[j]*k > n)
                break;
        }
        ans+=(j-1)*(k-1);
    }
    printf("%lld\n",ans*2);
    return 0;
}