CF2264F.Deranged Calculator

Hacks are disabled on this problem.

We are excited to introduce a new language for this Codeforces round: the Deranged Calculator language (DC).

The DC language is a simplified programming language with some weird constraints. It only supports a single expression on one line and only has a single positive integer $n$ as input. The syntax of the expression is similar to that of most programming languages (in fact, the expression is valid Python), but it is very limited:

• Expressions consist of the input parameter $n$, parenthesized expressions, and calls to the round function. These are joined together by the operators +, -, *, and /.
• Grouping with parentheses () can be used to change the order of operations, as usual.
• The operators * and / have equal precedence and take precedence over + and -, which also have equal precedence. Operators of equal precedence are evaluated from left to right.
• The round function takes a single expression as its argument and rounds its value to the nearest integer, rounding up when the fractional part is exactly $0.5$. (For example, round(n/(n-n-n-n)) will always evaluate to $0$, because $x/(x-x-x-x)=-\frac{1}{2}$ for a positive integer $x$. The fractional part is $0.5$, so it will be rounded up to $0$.)
• Only the input value $n$ can be used; no numeric constants are allowed in the program.
• All operations are performed using exact fractional arithmetic, so division does not round and can produce fractions.

A derangement of $n$ numbers is a permutation of those numbers in which none of the numbers appears in its original position. For example, the derangements of the sequence $[1, 2, 3]$ are $[2, 3, 1]$ and $[3, 1, 2]$.

You are given a single integer $k$. Your job is to make a valid DC program that, for every integer $n$ ($2 \le n \le k$), calculates the number of derangements of the sequence $[1, 2, \ldots, n]$, when $n$ is given as input to the DC program. The length of the program should not exceed $10\,000$ characters.

A local testing tool is provided to help you develop your solution. It can be found under Contest Materials.

本题禁用 hack。

我们很高兴在本次 Codeforces 比赛中引入一门新语言：错位计算器语言（Deranged Calculator language，简称 DC）。

DC 语言是一门简化版的编程语言，具有一些奇特的限制。它仅支持单行单表达式，且唯一输入为一个正整数 $n$。该表达式的语法与大多数编程语言类似（实际上，该表达式是合法的 Python 表达式），但功能非常受限：

• 表达式由输入参数 $n$、括号包围的子表达式以及对 round 函数的调用组成；这些成分通过运算符 +、-、* 和 / 连接。
• 可使用圆括号 () 进行分组以改变运算顺序，规则与常规一致。
• 运算符 * 和 / 优先级相同，且高于 + 和 -；而 + 和 - 的优先级也相同。同级运算符按从左到右顺序求值。
• round 函数接收一个表达式作为其唯一参数，并将其值四舍五入到最接近的整数；当小数部分恰好为 $0.5$ 时向上取整。（例如，round(n/(n-n-n-n)) 的结果恒为 $0$，因为对任意正整数 $x$，有 $x/(x-x-x-x)=-\frac{1}{2}$，其小数部分为 $0.5$，故向上取整得 $0$。）
• 程序中仅允许使用输入值 $n$，不得出现任何数值常量。
• 所有运算均采用精确的分数运算，因此除法不进行舍入，可能产生分数结果。

$n$ 个数的错位排列（derangement）是指这 $n$ 个数的一个排列，其中每个数都不出现在其原始位置上。例如，序列 $[1, 2, 3]$ 的错位排列为 $[2, 3, 1]$ 和 $[3, 1, 2]$。

你将获得一个整数 $k$。你的任务是编写一个合法的 DC 程序，使得：对每个整数 $n$（满足 $2 \le n \le k$），当以 $n$ 作为输入运行该 DC 程序时，程序输出序列 $[1, 2, \ldots, n]$ 的错位排列数目。该程序长度不得超过 $10\,000$ 个字符。

我们提供了一个本地测试工具以协助你开发解决方案。该工具位于“比赛资料”（Contest Materials）中。

The only line of input contains a single integer $k$ ($k \in {2,3,50}$).

Your DC program should output the correct number of derangements for all integers $n$ such that $2 \le n \le k$.

输入仅包含一行，其中有一个整数 $k$（$k \in \{2,3,50\}$）。

你的分治（DC）程序应输出所有满足 $2 \le n \le k$ 的整数 $n$ 对应的错排数。

Output a valid DC expression on a single line. The expression should only consist of the characters round()+-*/ and the length of the expression should not exceed $10\,000$ characters.

Your expression will be run on all integers $n$ such that $2 \le n \le k$. For each such run, the DC expression should not divide by $0$ at any time, and its value must be the number of derangements of $n$.

在单行内输出一个有效的 DC 表达式。该表达式仅可包含字符 round、+、-、*、/，且其长度不得超过 $10\,000$ 个字符。

您的表达式将在所有满足 $2 \le n \le k$ 的整数 $n$ 上运行。对每个这样的 $n$，DC 表达式在运行过程中不得出现除零错误，且其计算结果必须等于 $n$ 的错排数（即 $n$ 个元素的错排数目）。

For the first example, n-n/n is a valid DC program that calculates $n-1$. The number of derangements for $n=2$ is $1$, so this program correctly computes the answer for $n=2$.

In the second example, a slightly overcomplicated expression is used, which shows all the features of the language in action. For $n=3$, the value of the expression is: $3+\text{round}(3-3 \cdot(\frac{3}{3+3}))-3 = \text{round}(3 - \frac{3}{2}) = \text{round}(\frac{3}{2}) = \text{round}(1.5)=2$. It correctly calculates the number of derangements of the sequence $[1,2,3]$. For $n=2$ it can be verified that the value of the expression becomes $1$.

对于第一个例子，n-n/n 是一个有效的 DC 程序，用于计算 $n-1$。当 $n=2$ 时，错排数为 $1$，因此该程序正确地计算出了 $n=2$ 时的答案。

在第二个例子中，使用了一个略显复杂的表达式，它展示了该语言的所有特性。当 $n=3$ 时，该表达式的值为：
$3+\text{round}(3-3 \cdot(\frac{3}{3+3}))-3 = \text{round}(3 - \frac{3}{2}) = \text{round}(\frac{3}{2}) = \text{round}(1.5)=2$。
它正确地计算出了序列 $[1,2,3]$ 的错排数。当 $n=2$ 时，可以验证该表达式的值为 $1$。