bzoj1257 数论分块

求\(\sum_{i=1}^n k \% i\)
其中1 <= n <= 1e9, 1 <= k <= 1e9

把取模拆开

\[k \% i = k - \lfloor\frac{k}{i}\rfloor * i \]

注意到\(\lfloor\frac{k}{i}\rfloor\)总共有O(sqrt(k))种取值，可以数论分块。

代码：

#include <iostream>

using namespace std;

typedef long long LL;

LL sum(int l, int r) {
    return (LL)(l + r) * (r - l + 1) / 2;
}

int main() {
    int n, k;

    cin >> n >> k;
    LL ans = 0;
    for (int i = 1, r; i <= k && i <= n; i = r + 1) {
        int d = k / i;
        r = k / d;
        ans += d * sum(i, min(r, n));
    }
    ans = (LL)n * k - ans;
    cout << ans << endl;

    return 0;
}