1006. Clumsy Factorial

# 1006. Clumsy Factorial#

## 题目 #

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`.

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.

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.

Additionally, the division that we use is floor division such that `10 * 9 / 8` equals `11`.  This guarantees the result is an integer.

`Implement the clumsy` function as defined above: given an integer `N`, it returns the clumsy factorial of `N`.

Example 1:

``````Input:4
Output: 7
Explanation: 7 = 4 * 3 / 2 + 1
``````

Example 2:

``````Input:10
Output:12
Explanation:12 = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1
``````

Note:

1. `1 <= N <= 10000`
2. `2^31 <= answer <= 2^31 - 1` (The answer is guaranteed to fit within a 32-bit integer.)

## 解题思路 #

• 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。

## 代码 #

``````package leetcode

func clumsy(N int) int {
res, count, tmp, flag := 0, 1, N, false
for i := N - 1; i > 0; i-- {
count = count % 4
switch count {
case 1:
tmp = tmp * i
case 2:
tmp = tmp / i
case 3:
res = res + tmp
flag = true
tmp = -1
res = res + i
case 0:
flag = false
tmp = tmp * (i)
}
count++
}
if !flag {
res = res + tmp
}
return res
}
``````

Apr 8, 2023