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Problem 790

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Clock Grid

There is a grid of length and width $50515093$ points. A clock is placed on each grid point. The clocks are all analogue showing a single hour hand initially pointing at $12$.

A sequence $S_t$ is created where:
$$
\begin{aligned}
S_0 & = 290797 &\\
S_t & = S_{t-1}^2 \bmod 50515093 & t>0
\end{aligned}
$$
The four numbers $N_t = (S_{4t-4}, S_{4t-3}, S_{4t-2}, S_{4t-1})$ 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 $N_t = (3,9,47,20)$, the range would be $3\le x\le 9$ and $20\le y\le47$, and would include $196$ clocks.

For each $t$ $(t>0)$, the clocks within the range represented by $N_t$ are moved to the next hour $12\rightarrow 1\rightarrow 2\rightarrow \cdots $.

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 $C(10^5)$.

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时钟格阵

考虑一个长和宽均包含$50515093$个点的格阵，每个格点上都放置有一台时钟。所有的时钟盘面上都只包含一根时针，且在开始时时针均指向$12$。

序列$S_t$按如下方式定义：
$$
\begin{aligned}
S_0 & = 290797 &\\
S_t & = S_{t-1}^2 \bmod 50515093 & t>0
\end{aligned}
$$
四个数$N_t = (S_{4t-4}, S_{4t-3}, S_{4t-2}, S_{4t-1})$代表了格阵的一部分，其中第一对数表示横轴的范围，第二对数表示纵轴的范围。例如，若$N_t = (3,9,47,20)$，相应的格阵范围满足$3\le x\le 9$和$20\le y\le47$，其中包含$196$个时钟。

对于每个$t$ $(t>0)$，包含在上述由$N_t$描述的格阵范围内的所有时钟均往后拨一小时，也即$12\rightarrow 1\rightarrow 2\rightarrow \cdots $。

记$C(t)$为经过第$t$步后，所有时钟时针所指向的数之和。
已知$C(0) = 30621295449583788$，$C(1) = 30613048345941659$，$C(10) = 21808930308198471$，$C(100) = 16190667393984172$。

求$C(10^5)$。

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