牛客多校第三场H题

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

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 $\frac{i}{gcd(i, j)}$ and $\frac{j}{gcd(i,j)}$ 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

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输出

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示例2

输入

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输出

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一直没找到规律。。。比赛的时候发现越找越多了心态爆炸

正解思路：（我猜是打表找的规律？我队有人打表，打了两个小时，最后在赛后过了emmmmm）

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×⌊n/y⌋2×⌊n/y⌋ 个数对；

例如，n=7，取到素数对(2,3)，那么 ⌊n/3⌋=⌊7/3⌋=2⌊n/3⌋=⌊7/3⌋=2，就有 2×2=42×2=4 个数对：(1*2,1*3) = (2,3)、(3,2)、(2*2,2*3) = (4,6)、(6,4)；

#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;
}