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=St12mod50515093t>0

The four numbers Nt=(S4t4,S4t3,S4t2,S4t1) 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 3x9 and 20y47, and would include 196 clocks.

For each t (t>0), the clocks within the range represented by Nt are moved to the next hour 1212.

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=St12mod50515093t>0

故我们可以用一个四元组 Nt=(S4t4,S4t3,S4t2,S4t1) 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 3x920y47 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 1212 的规则走时 1 小时。

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)=30621295449583788C(1)=30613048345941659C(10)=21808930308198471C(100)=16190667393984172

C(105)


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