又是个蓝莫反水题

首先发现对于位置(x,y)(x,y)，中间遮挡的点就是gcd⁡(x,y)−1\gcd(x, y) - 1个，那么题目就是求：

∑i=1n∑j=1m2×(gcd⁡(i,j)−1)+1\sum_{i = 1}^n \sum_{j = 1}^m 2 \times (\gcd(i, j) - 1) + 1

我们令f(x)=2×(x−1)+1=2x−1f(x) = 2 \times (x - 1) + 1 = 2x - 1，那么就是求：

∑i=1n∑j=1mf(gcd⁡(i,j))\sum_{i = 1}^n \sum_{j = 1}^m f(\gcd(i,j))

然后就是套路了，假设n≤mn \leq m：

=∑i=1n∑j=1m∑d∣gcd⁡(i,j)g(d)= \sum_{i = 1}^n \sum_{j = 1}^m \sum_{d | \gcd(i, j)}g(d)

=∑i=1n∑j=1m∑d∣i,d∣jg(d)= \sum_{i = 1}^n \sum_{j = 1}^m \sum_{d | i, d | j} g(d)

=∑d=1ng(d)⌊nd⌋⌊md⌋= \sum_{d = 1}^n g(d) \left \lfloor \frac{n}{d} \right \rfloor \left \lfloor \frac{m}{d} \right \rfloor

这里：

g(n)=∑d∣nμ(d)f(nd)g(n) = \sum_{d | n} \mu(d)f(\frac{n}{d})

所以我们可以直接预处理gg的前缀和，然后直接对这个式子数论分块，时间复杂度为O(nn)O(n\sqrt{n})。

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#include <bits/stdc++.h>
using namespace std;
#define int long long
int n, m;
vector<int> musieve(int n) {
    vector<int> primes, mu(n + 1);
    vector<bool> vis(n + 1, 0);
    mu[0] = 0;
    mu[1] = 1;
    for (int i = 2; i <= n; i++) {
        if (!vis[i]) {
            mu[i] = -1;
            primes.push_back(i);
        }
        for (int j = 0; j < primes.size() && i * primes[j] <= n; j++) {
            vis[i * primes[j]] = 1;
            if (i % primes[j]) {
                mu[i * primes[j]] = -mu[i];
            } else {
                mu[i * primes[j]] = 0;
                break;
            }
        }
    }
    return mu;
}
signed main() {
    cin >> n >> m;
    if (n > m) {
        swap(n, m);
    }
    auto mu = musieve(m);
    function<int(int)> f = [&](int x) -> int { return 2 * x - 1; };
    vector<int> g(m + 1, 0);
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j * i <= m; j++) {
            g[j * i] += f(j) * mu[i];
        }
    }
    for (int i = 2; i <= m; i++) {
        g[i] += g[i - 1];
    }
    int ans = 0;
    int l = 1;
    while (l <= n) {
        int d1 = n / l;
        int r1 = n / d1;
        int d2 = m / l;
        int r2 = m / d2;
        int r = min(r1, r2);
        ans += (g[r] - g[l - 1]) * (n / l) * (m / l);
        l = r + 1;
    }
    cout << ans;
    return 0;
}