「题解」洛谷 P2261 [CQOI2007]余数求和

题目

P2261 [CQOI2007]余数求和

简化题意

求 \(G(n,k) = \sum\limits_{i = 1}^{n} k\mod i\)

思路

整除分块。

\[\begin{aligned} G(n,k) &= \sum\limits_{i = 1}^{n} k\mod i \\ &= \sum_{i = 1} ^ {n} k - \left\lfloor\dfrac{k}{i}\right\rfloor \times i \\ &= nk - \sum_{i = 1} ^ {n}\left\lfloor\dfrac{k}{i}\right\rfloor \times i \end{aligned} \]

右边可以整除分块 \(O(2\sqrt n)\) 求出。

Code

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>

int n, k;
long long ans;

int main() {
    std::cin >> n >> k;
    ans = 1ll * n * k;
    for (int i = 1, j = 0; i <= n; i = j + 1) {
        if (k / i == 0) break;
        j = k / (k / i);
        j = j > n ? n : j;
        ans -= 1ll * (k / i) * (1ll * (i + j) * (j - i + 1) / 2);
    }
    std::cout << ans << '\n';
    return 0;
}

标签：lfloor,洛谷,int,题解,P2261,1ll,ans,include,sum
来源： https://www.cnblogs.com/poi-bolg-poi/p/13620610.html