[Luogu] 余数求和

question：

$$\sum_{i=1}^{n} k \bmod i$$
$$\sum_{i=1}^{n} k - \lfloor \frac{k}{i} \rfloor i$$
$$\sum_{i=1}^{n}k - \sum_{i=1}^{n}\lfloor \frac{k}{i} \rfloor i$$

直接数论分块

#include <iostream>
int main() {
    int n, k;
    long long Answer = ;
    std:: cin >> n >> k;
    for(int i = , r; i <= n; i = r + ) {
        if(k / i) r = std:: min(n, (int)k / (k / i));
        else r = n;
        Answer += (1LL * (r - i + ) * k - 1LL * (k / i) * (i + r) * (r - i + ) / );
    }
    std:: cout << Answer;
    return ;
}