Difference between revisions of "2019 Multi-University,HDU Day 9"
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求 $g_m(n)$。 | 求 $g_m(n)$。 | ||
| − | + | 题解：函数中 $i>m$ 的部分可以改写成 $g_m(i)=i-1+\frac{2}{i}\sum_{j=0}^{i-1} g_m(j)$，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。 | |
| − | 假设 $g_m(i)$ 的 $i-1$ 对 $g_m(k)$ 的贡献是 $f_i(k) \cdot (i-1) $，则 $f_i(k)=0(k<i),f_i(i)=1,f_i(k)= \frac{2}{k | + | 假设 $g_m(i)$ 的 $i-1$ 对 $g_m(k)$ 的贡献是 $f_i(k) \cdot (i-1) $，则 $f_i(k)=0(k<i),f_i(i)=1,f_i(k)= \frac{2}{k} \cdot \sum_{j=1}^{k-1} f_i(j) (k>i)$。我们发现， |
\begin{equation} \begin{aligned} | \begin{equation} \begin{aligned} | ||
| − | &f_i(k | + | &f_i(k)\\ |
| − | =&\frac{2}{k} \cdot \sum_{j=1}^{k} f_i(j)\\ | + | =&\frac{2}{k} \cdot \sum_{j=1}^{k-1} f_i(j)\\ |
| − | =&\frac{2}{k} \cdot (f_i(k)+\sum_{j=1}^{k-1}f_i(j))\\ | + | =&\frac{2}{k} \cdot (f_i(k-1)+\sum_{j=1}^{k-1}f_i(j))\\ |
| − | =&\frac{2}{k} \cdot (\frac{2}{k | + | =&\frac{2}{k} \cdot (\frac{2}{k} \cdot \sum_{j=1}^{k-1}f_i(j)+\sum_{j=1}^{k-1}f_i(j))\\ |
| − | =&\frac{2}{k} \cdot (1+\frac{2}{k | + | =&\frac{2}{k} \cdot (1+\frac{2}{k}) \cdot \sum_{j=1}^{k-1}f_i(j)\\ |
| − | =&\frac{2}{k} \cdot \frac{k+ | + | =&\frac{2}{k} \cdot \frac{k+2}{k} \cdot \frac{k}{2} \cdot( \frac{2}{k} \cdot \sum_{j=1}^{k-1}f_i(j))\\ |
=&\frac{k+2}{k} \cdot f_i(k) | =&\frac{k+2}{k} \cdot f_i(k) | ||
\end{aligned} \end{equation} | \end{aligned} \end{equation} | ||
| − | |||
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== Problem B == | == Problem B == | ||
Revision as of 15:11, 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：函数中 $i>m$ 的部分可以改写成 $g_m(i)=i-1+\frac{2}{i}\sum_{j=0}^{i-1} g_m(j)$，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。
假设 $g_m(i)$ 的 $i-1$ 对 $g_m(k)$ 的贡献是 $f_i(k) \cdot (i-1) $，则 $f_i(k)=0(k<i),f_i(i)=1,f_i(k)= \frac{2}{k} \cdot \sum_{j=1}^{k-1} f_i(j) (k>i)$。我们发现，
\begin{equation} \begin{aligned} &f_i(k)\\ =&\frac{2}{k} \cdot \sum_{j=1}^{k-1} f_i(j)\\ =&\frac{2}{k} \cdot (f_i(k-1)+\sum_{j=1}^{k-1}f_i(j))\\ =&\frac{2}{k} \cdot (\frac{2}{k} \cdot \sum_{j=1}^{k-1}f_i(j)+\sum_{j=1}^{k-1}f_i(j))\\ =&\frac{2}{k} \cdot (1+\frac{2}{k}) \cdot \sum_{j=1}^{k-1}f_i(j)\\ =&\frac{2}{k} \cdot \frac{k+2}{k} \cdot \frac{k}{2} \cdot( \frac{2}{k} \cdot \sum_{j=1}^{k-1}f_i(j))\\ =&\frac{k+2}{k} \cdot f_i(k) \end{aligned} \end{equation}
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.