my solution of Atcoder Beginner Contest 293 E

Правка en4, от 1xx55, 2023-03-14 06:39:52

The goal is to find $$$\sum^{X-1}_{i=0}A^{i}$$$ $$$ mod $$$ $$$m$$$ , and a simple thought is that we can calculate $$$\frac{A^{X}-1}{A-1}$$$. But when we divide $$$A^{X}-1$$$ by $$$A-1$$$ then $$$mod$$$ $$$m$$$, problem occurs: when we calculate $$$A^{X}-1$$$ , it's already $$$mod$$$ $$$m$$$ ,and there's maybe no reverse to it.

To solve this problem , I use a simple method: store $$$A^X$$$ with two numbers : $$$k$$$ and $$$r$$$ ,satisfying

$$$A^X \equiv k(A-1)+r \pmod {m(A-1)}$$$ , $$$0 \leq r \le A-1$$$

When we calculate $$$A^{x}$$$,we refresh $$$k$$$ and $$$r$$$ by multiplication ,then refresh $$$k$$$ by modulo $$$m$$$ , and refresh $$$r$$$ when $$$r \ge A-1 $$$ : we do $$$k$$$ $$$+=$$$ $$$r/(A-1)$$$ and $$$r$$$ $$$=$$$ $$$r$$$ % $$$(A-1)$$$

История

 
 
 
 
Правки
 
 
  Rev. Язык Кто Когда Δ Комментарий
en14 Английский 1xx55 2023-03-14 10:10:25 0 (published)
en13 Английский 1xx55 2023-03-14 10:04:56 1205 (saved to drafts)
en12 Английский 1xx55 2023-03-14 08:46:09 0 (published)
en11 Английский 1xx55 2023-03-14 08:45:53 31 (saved to drafts)
en10 Английский 1xx55 2023-03-14 08:37:00 0 (published)
en9 Английский 1xx55 2023-03-14 08:35:02 4 Tiny change: ' , we can NOT use $k$ a' -> ' , we can **NOT** use $k$ a'
en8 Английский 1xx55 2023-03-14 08:33:58 13 Tiny change: 'nCode here:\n~~~~~\n#' -> 'nCode here (GNU C11)\n~~~~~\n#'
en7 Английский 1xx55 2023-03-14 08:29:50 1155 Tiny change: 'n\nCode:\n ' -> 'n\nCode:\n~~~~~\nincl\n~~~~~\n\n\n '
en6 Английский 1xx55 2023-03-14 08:16:01 135 Tiny change: '} \equiv \left\{k_{X} & ' -> '} \equiv \{k_{X} & '
en5 Английский 1xx55 2023-03-14 07:59:14 337 Tiny change: ' $r=$\left{ 01A=2A>20A=21A>2\begin{arr' -> ' $r=$\left\{ \begin{arr'
en4 Английский 1xx55 2023-03-14 06:39:52 4 Tiny change: ')+r \pmod m(A-1)$ , $0 \le' -> ')+r \pmod {m(A-1)}$ , $0 \le'
en3 Английский 1xx55 2023-03-14 06:38:22 80 Tiny change: ': we do $k% %+=% %r/(A-1)$ a' -> ': we do $k$ $+=$ $r/(A-1)$ a'
en2 Английский 1xx55 2023-03-14 06:16:49 524 Tiny change: '{i=0}A^{i} mod m$' -> '{i=0}A^{i}$ $ mod $ $m$'
en1 Английский 1xx55 2023-03-14 05:56:14 94 Initial revision (saved to drafts)