NC17247 H、Diff-prime Pairs

题目链接

题目

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.

\(1 ≤ N ≤ 10^7\)

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

题解

知识点：组合数学，枚举，筛法。

考虑枚举每一对素数 \((i,j)\) 产生的贡献，其中显然 \(i\neq j\) 。

不妨先假设 \(i<j\) ，那么形如 \((ki,kj)\) 的一对数结果会是 \((i,j)\) ，其中 \(k \leq \left\lfloor \dfrac{n}{j} \right\rfloor\) 。

我们发现，对于任意的素数 \(i\) 满足 \(i<j\) ，答案都是 \(\left\lfloor \dfrac{n}{j} \right\rfloor\) ，因此我们可以只枚举 \(j\) ，那么以 \(j\) 较大数时产生的贡献，答案是 \((\pi(j)-1) \cdot \left\lfloor \dfrac{n}{j} \right\rfloor\) 。

最后不要忘记乘 \(2\) ，因为我们一开始设 \(i<j\) ，为了计数方便。

时间复杂度 \(O(n)\)

空间复杂度 \(O(n)\)

代码

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

const int N = 1e7 + 7;
bool vis[N];
vector<int> prime;
void get_prime(int n) {
    for (int i = 2;i <= n;i++) {
        if (!vis[i]) prime.push_back(i);
        for (auto j : prime) {
            if (i * j > n) break;
            vis[i * j] = 1;
            if (!(i % j)) break;
        }
    }
}

int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n;
    cin >> n;
    get_prime(n);
    ll ans = 0;
    for (int i = 0;i < prime.size();i++) {
        ans += i * (n / prime[i]);
    }
    cout << ans * 2 << '\n';
    return 0;
}