bzoj 1257

题解

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question：

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

计算 $n * k$ 然后数论分块
或者直接数论分块

code

#include <iostream>
int main() {
    int n, k; 
    long long Answer = 0;
    std:: cin >> n >> k;
    for(int i = 1, r; i <= n; i = r + 1) {
        if(k / i) r = std:: min(n, (int)k / (k / i));
        else r = n;
        Answer += (1LL * (r - i + 1) * k - 1LL * (k / i) * (i + r) * (r - i + 1) / 2);
    }
    std:: cout << Answer;
    return 0;
}