SkyWT

Fighting 2019!

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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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\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……

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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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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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