790. Clock Grid

There is a grid of length and width 50515093 populated with a clock in each grid square. The clocks are all analogue showing a single hour hand initially pointing at 12.

A sequence St is created where:

S0=290797St=St−12mod50515093t>0

The four numbers Nt=(S4t−4,S4t−3,S4t−2,S4t−1) represent a range within the grid, with the first pair of numbers representing the x-bounds and the second pair representing the y-bounds. For example, if Nt=(3,9,47,20), the range would be 3≤x≤9 and 20≤y≤47, and would include 196 clocks.

For each t (t>0), the clocks within the range represented by Nt are moved to the next hour 12→1→2→⋯.

We define C(t) to be the sum of the hours that the clock hands are pointing to after timestep t. You are given C(0)=30621295449583788, C(1)=30613048345941659, C(10)=21808930308198471 and C(100)=16190667393984172.

Find C(105).

790. 时钟网格

现有一个长、宽均为 50515093 的网格。每一个小方格里均有一个初始指向 12 点的时钟。

定义序列 {St} 满足:

S0=290797St=St−12mod50515093t>0

故我们可以用一个四元组 Nt=(S4t−4,S4t−3,S4t−2,S4t−1) 来表示网格内的一个子矩形;其中,这个四元组的前两个数代表 x 坐标的范围,后两个数代表 y 坐标的范围。例如,若四元组 Nt=(3,9,47,20),则其表示网格中 3≤x≤9 且 20≤y≤47 的一部分;这一部分有 196 个时钟。

对于每一个时刻 t (t>0),在四元组 Nt 所表示的子矩形范围内的时钟将按 12→1→2→⋯ 的规则走时 1 小时。

记 C(t) 为在 t 时刻后,所有时钟所指向的小时数之和。已知 C(0)=30621295449583788,C(1)=30613048345941659,C(10)=21808930308198471,C(100)=16190667393984172。

求 C(105)。


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