求dalao解释，为什么 $O(n\sqrt{n})$ 代码能过，在本机上输入 $n, m = 10^7$ 最多只跑了 900ms，还是说这个复杂度跑不满吗？

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

const int N = 1e7 + 5;
const ll mod = 20101009;

int n, m;
ll prime[N], vis[N], mob[N]; int cnt;
ll sum[N], a[N];

ll calc(int n, int m) {
    ll ans = 0;
    for (int l = 1, r; l <= n; l = r + 1) {
        r = min(n / (n / l), m / (m / l));
        ans = (ans + ((sum[r] - sum[l - 1] + mod) % mod) % mod * a[n / l] % mod * a[m / l] % mod) % mod;
    }
    return ans;
}

int main() {
    sum[1] = mob[1] = 1;
    for (int i = 2; i <= 1e7; i++) {    
        if (!vis[i]) {
            vis[i] = i;
            prime[++cnt] = i;
            mob[i] = -1;
        }
        for (int j = 1; j <= cnt && prime[j] <= 1e7 / i; j++) {
            vis[prime[j] * i] = prime[j];
            if (i % prime[j] == 0) {
                mob[prime[j] * i] = 0;
                break;
            } else {
                mob[prime[j] * i] = -mob[i];
            }
        }
        sum[i] = (sum[i - 1] + mob[i] * i % mod * i % mod) % mod;
    }

    for (int i = 1; i <= 1e7; i++) {
        a[i] = (a[i - 1] + i) % mod;
    }

    cin >> n >> m;
    if (n > m) swap(n, m);
    ll ans = 0;
    for (int d = 1; d <= n; d++) {
        ans = (ans + d * calc(n / d, m / d) % mod) % mod;
    }
    cout << ans << '\n';
    return 0;
}