最大公约数

思路 :

\(gcd(x,y)=p,1\le x,y \le n \Rightarrow gcd(\frac{x}{p},\frac{y}{p})=1 \Rightarrow gcd(x′,y′)=1,1 \le x′,y′\le \frac{n}{p}\)

所以其实很经典的在矩形(n*n)坐标范围下求 \((x,y)\) 的互质对数.只不过这里的 \(n\) 也是变量.

本题扩展于可见的点

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);
inline int lowbit(int x) { return x & (-x); }
#define ll long long
#define ull unsigned long long
#define pb push_back
#define PII pair<int, int>
#define VIT vector<int>
#define x first
#define y second
#define inf 0x3f3f3f3f
const int N = 1e7 + 10;
int pri[N], cnt;
bool st[N];
ll phi[N];

void init() {
    phi[1] = 1;
    for (int i = 2; i < N; ++i) {
        if (!st[i]) {
            pri[cnt++] = i;
            phi[i] = i - 1;
        }
        for (int j = 0; pri[j] * i < N; ++j) {
            st[pri[j] * i] = true;
            if (i % pri[j] == 0) {
                phi[pri[j] * i] = phi[i] * pri[j];
                break;
            }
            phi[pri[j] * i] = phi[i] * (pri[j] - 1);
        }
    }
}

int main() {
    IO;
    init();
    int n;
    cin >> n;
    phi[1] = 0;
    for (int i = 2; i < N; ++i) phi[i] += phi[i - 1];
    ll ans = 0;
    for (int i = 0; pri[i] <= n; ++i) ans += phi[n / pri[i]] * 2 + 1;
    cout << ans << '\n'; 
    return 0;
}