2018牛客网暑期ACM多校训练营（第三场） H

题目链接：https://www.nowcoder.com/acm/contest/141/H

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 ≤ 10^7

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

题意：

给出一个数字 n (1 ≤ n ≤ 1e7)，求多少 数对(i, j) 满足 $frac{i}{{gcd left( {i,j} ight)}}$ 和 $frac{j}{{gcd left( {i,j} ight)}}$ 均为质数，且1 ≤ i, j ≤ n。

题解：

筛出[1,n]之间所有的素数，

不难知道，每次取到其中两个素数组成一个素数对(x, y)，不妨设 x < y，那么相应的就增加了 $2 imes leftlfloor {n/y} ight floor $ 个数对；

例如，n=7，取到素数对(2,3)，那么 $leftlfloor {n/3} ight floor = leftlfloor {7/3} ight floor = 2$，就有 $2 imes 2 = 4$ 个数对：(1*2,1*3) = (2,3)、(3,2)、(2*2,2*3) = (4,6)、(6,4)；

对欧拉筛法稍加改造，添加一行代码即可。时间复杂度O(n)。

AC代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e7+5;

int n;
ll ans;

bool isPrime[maxn];
int prime[maxn/10],cnt;
void screen()//欧拉筛法求素数
{
    cnt=0;
    memset(isPrime,1,sizeof(isPrime));
    isPrime[0]=isPrime[1]=0;
    for(int i=2;i<=n;i++)
    {
        if(isPrime[i])
        {
            prime[cnt++]=i;
            
            ans+=2*(n/i)*(cnt-1);
            //每找到一个素数i，其就可以与前面所有出现过的cnt-1个素数组成cnt-1个素数对，相应的就有2*(n/i)*(cnt-1)个数对
            
        }
        for(int j=0;j<cnt;j++)
        {
            if(i*prime[j]>n) break;
            isPrime[(i*prime[j])]=0;
            if(i%prime[j]==0) break;
        }
    }
}

int main()
{
    scanf("%d",&n);
    ans=0;
    screen();
    cout<<ans<<endl;
}