# 2019 Multi-University,HDU Day 9

## Problem A

Solved by Kilo_5723. 02:16:39 (+)

\begin{aligned} &f_i(k)\\ =&\frac{2}{k} \cdot \sum_{j=0}^{k-1} f_i(j)\\ =&\frac{2}{k} \cdot (f_i(k-1)+\sum_{j=0}^{k-2}f_i(j))\\ =&\frac{2}{k} \cdot (\frac{2}{k-1} \cdot \sum_{j=0}^{k-2}f_i(j)+\sum_{j=0}^{k-2}f_i(j))\\ =&\frac{2}{k} \cdot (1+\frac{2}{k-1}) \cdot \sum_{j=0}^{k-2}f_i(j)\\ =&\frac{2}{k} \cdot \frac{k+1}{k-1} \cdot \frac{k-1}{2} \cdot( \frac{2}{k-1} \cdot \sum_{j=0}^{k-2}f_i(j))\\ =&\frac{k+1}{k} \cdot f_i(k-1) \;\;\;(k-1>i) \end{aligned}

\begin{aligned} &\sum_{i=m+1}^{n-1}\frac{2 \cdot (n+1) \cdot (i-1)}{(i+1)\cdot(i+2)}\\ =&\sum_{i=m+1}^{n-1}(\frac{2 \cdot (n+1) \cdot (i-1)}{i+1}-\frac{2 \cdot (n+1) \cdot (i-1)}{i+2})\\ =&\frac{2 \cdot (n+1)\cdot m }{m+2} + (\sum_{i=m+2}^{n-1} \frac{2 \cdot (n+1) \cdot (i-1)}{i+1})-\frac{2 \cdot (n+1) \cdot (n-2)}{n+1}-(\sum_{i=m+1}^{n-2}\frac{2 \cdot (n+1) \cdot (i-1)}{i+2}) \\ =&\frac{2 \cdot (n+1)\cdot m }{m+2}-\frac{2 \cdot (n+1) \cdot (n-2)}{n+1} + (\sum_{i=m+2}^{n-1} \frac{2 \cdot (n+1) \cdot (i-1)}{i+1})-(\sum_{i=m+2}^{n-1}\frac{2 \cdot (n+1) \cdot (i-2)}{i+1}) \\ =&\frac{2 \cdot (n+1)\cdot m }{m+2}-\frac{2 \cdot (n+1) \cdot (n-2)}{n+1} + \sum_{i=m+2}^{n-1} \frac{2 \cdot (n+1)}{i+1} \\ =&\frac{2 \cdot (n+1)\cdot m }{m+2}-\frac{2 \cdot (n+1) \cdot (n-2)}{n+1} + 2 \cdot (n+1) \cdot \sum_{i=m+3}^{n} \frac{1}{i} \;\;\; (n-m>3) \end{aligned}

## Problem B

Solved by Xiejiadong && Kilo_5723. 03:38:23 (+)

• + 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 $1$ 。

## 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)

Unsolved.

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Unsolved.