SkyWT

SkyWT

我们的征途是星辰大海

⚙️ 本网站主题尚未开发完善,可能偶现奇怪的 bug。
✨ 最近更新:添加全站 PJAX 无刷新(220930)。

Description

Yaroslav has n points that lie on the $Ox$ axis. The coordinate of the first point is $x_1$, the coordinate of the second point is $x_2$, ..., the coordinate of the n-th point is — $x_n$. Now Yaroslav wants to execute $m$ queries, each of them is of one of the two following types:

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In mathematical terms, the sequence $F_n$ of Fibonacci numbers is defined by the recurrence relation

$$F_1 = 1; F_2 = 1; F_n = F_{n - 1} + F_{n - 2} (n > 2)$$

DZY loves Fibonacci numbers very much. Today DZY gives you an array consisting of n integers: $a_1, a_2, \dots, a_n$. Moreover, there are $m$ queries, each query has one of the two types:

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矩阵乘法的很多应用都是围绕矩阵乘法的定义式展开的:

$$C[i,j]=\sum_{k=1}^{b} A[i,k]\ast B[k,j]$$

本质上是一种动态规划吧。

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$$ \begin{bmatrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \\ x_{31} & x_{32} & x_{33} \end{bmatrix} $$

每增加一个维度,世界便会增加无限的美感。

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坠痛苦的是,POJ 的辣鸡 G++ 编译器不支持 long long,害得我调试调了半天……

Description

For every pair of triplets, $T_a = (I_a, J_a, K_a)$ and $T_b = (I_b, J_b, K_b)$, we define the difference value between $T_a$ and $T_b$ as follows:

$$ D(T_a, T_b) = \max (I_a − I_b, J_a − J_b, K_a − K_b) − \min (I_a − I_b, J_a − J_b, K_a − K_b) $$

Now you are given $N$ triplets, could you write a program to calculate the sum of the difference values between every unordered pair of triplets?

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Topcoder Single Round Match 635 Div2 T3 LonglongestPathTree 题解

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Topcoder Single Round Match 638 Div2 T3 CandleTimerEasy 题解
每天被 XY 题困扰的我们怎么会去做 Topcoder 呢

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今天的 XY 题居然是递推专题,五道题目全都是递推,30+个人 AK 了……

递推是按照一定的规律来计算序列中的每个项,通常是通过计算前面的一些项来得出序列中的指定项的值。其思想是把一个复杂的庞大的计算过程转化为简单过程的多次重复,该算法利用了计算机速度快和不知疲倦的机器特点。

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Topcoder Single Round Match 640 DIv2 T3 TwoNumberGroupsEasy 题解

(TC SRM 从 640 到 616 持续施工~)
倒着开车,最为致命)(国庆之前怕是完不成了

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