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75 lines
2.8 KiB
Markdown
75 lines
2.8 KiB
Markdown
# [1006. Clumsy Factorial](https://leetcode.com/problems/clumsy-factorial/)
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## 题目
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Normally, the factorial of a positive integer `n` is the product of all positive integers less than or equal to `n`. For example, `factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1`.
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We instead make a *clumsy factorial:* using the integers in decreasing order, we swap out the multiply operations for a fixed rotation of operations: multiply (*), divide (/), add (+) and subtract (-) in this order.
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For example, `clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1`. However, these operations are still applied using the usual order of operations of arithmetic: we do all multiplication and division steps before any addition or subtraction steps, and multiplication and division steps are processed left to right.
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Additionally, the division that we use is *floor division* such that `10 * 9 / 8` equals `11`. This guarantees the result is an integer.
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`Implement the clumsy` function as defined above: given an integer `N`, it returns the clumsy factorial of `N`.
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**Example 1:**
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```
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Input:4
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Output: 7
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Explanation: 7 = 4 * 3 / 2 + 1
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```
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**Example 2:**
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```
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Input:10
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Output:12
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Explanation:12 = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1
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```
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**Note:**
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1. `1 <= N <= 10000`
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2. `2^31 <= answer <= 2^31 - 1` (The answer is guaranteed to fit within a 32-bit integer.)
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## 题目大意
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通常，正整数 n 的阶乘是所有小于或等于 n 的正整数的乘积。例如，factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1。相反，我们设计了一个笨阶乘 clumsy：在整数的递减序列中，我们以一个固定顺序的操作符序列来依次替换原有的乘法操作符：乘法(*)，除法(/)，加法(+)和减法(-)。例如，clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1。然而，这些运算仍然使用通常的算术运算顺序：我们在任何加、减步骤之前执行所有的乘法和除法步骤，并且按从左到右处理乘法和除法步骤。另外，我们使用的除法是地板除法（floor division），所以 10 * 9 / 8 等于 11。这保证结果是一个整数。实现上面定义的笨函数：给定一个整数 N，它返回 N 的笨阶乘。
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## 解题思路
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- 按照题意，由于本题没有括号，所以先乘除后加减。4 个操作一组，先算乘法，再算除法，再算加法，最后算减法。减法也可以看成是加法，只是带负号的加法。
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## 代码
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```go
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package leetcode
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func clumsy(N int) int {
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res, count, tmp, flag := 0, 1, N, true
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for i := N - 1; i > 0; i-- {
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count = count % 4
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switch count {
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case 1:
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tmp = tmp * i
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case 2:
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tmp = tmp / i
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case 3:
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res = res + tmp
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flag = true
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tmp = -1
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res = res + i
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case 0:
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flag = false
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tmp = tmp * (i)
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}
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count++
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}
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if !flag {
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res = res + tmp
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}
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return res
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}
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``` |