Difference between revisions of "2019 Multi-University,HDU Day 9"
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Xiejiadong (talk | contribs) |
Xiejiadong (talk | contribs) |
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Solved by Kilo_5723. 02:16:39 (+) | Solved by Kilo_5723. 02:16:39 (+) | ||
− | + | 题意:给出\begin{equation} | |
g_m(i)=\left\{ | g_m(i)=\left\{ | ||
\begin{aligned} | \begin{aligned} | ||
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求 $g_m(n)$。 | 求 $g_m(n)$。 | ||
− | 题解:可以把答案看作每个 $g_m(i)$ 的 $i-1$ | + | 题解:可以把答案看作每个 $g_m(i)$ 的 $i-1$ 部分的贡献总和,试求每一个 $i-1$ 对 $g_m(n)$ 有多少贡献。 |
+ | |||
+ | 我们发现,假设 $i-1$ 对 $k$ 的贡献是 $f_i(k) \cdot (i-1) $,则 $f_i(i)=1,f_i(k)= \frac{2}{k} cdot \sum_j=i^k-1 f_i(j) (k>i)$。 | ||
== Problem B == | == Problem B == |
Revision as of 14:34, 19 August 2019
Problem A
Solved by Kilo_5723. 02:16:39 (+)
题意:给出\begin{equation} g_m(i)=\left\{ \begin{aligned} &0&(1 \le i \le m) \\ &i-1+\frac{1}{i}\sum_{j=1}^i(g_m(j-1)+g_m(i-j))&(m<i) \end{aligned} \right. \end{equation} 求 $g_m(n)$。
题解:可以把答案看作每个 $g_m(i)$ 的 $i-1$ 部分的贡献总和,试求每一个 $i-1$ 对 $g_m(n)$ 有多少贡献。
我们发现,假设 $i-1$ 对 $k$ 的贡献是 $f_i(k) \cdot (i-1) $,则 $f_i(i)=1,f_i(k)= \frac{2}{k} cdot \sum_j=i^k-1 f_i(j) (k>i)$。
Problem B
Solved by Xiejiadong && Kilo_5723. 03:38:23 (+)
题意:给出一些从边界出发的直线,求平面被分割成了几块。
题解:可以对交点分类讨论,因为保证了出发点不会同行也不会同列,所以只有两种交点是有贡献的:
- + 交点,且贡献为 $1$ ;
- 」 交点,贡献为 $0.5$ ,且只出现在边界上,共 $4$ 个,贡献为 $1$ 。
所以统计 + 交点个数,+1 就好了,需要区间加,单点询问,直接上树状数组。
Problem C
Upsolved by Kilo_5723 && Xiejiadong. (-5)
Problem D
Solved by Weaver_zhu. 00:15:48 (+)
Problem E
Solved by Kilo_5723. 04:07:47 (+9)
Problem F
Unsolved.
Problem G
Unsolved.
Problem H
Unsolved.
Problem I
Unsolved.
Problem J
Unsolved.
Problem K
Unsolved.