1006. Clumsy Factorial

https://leetcode.com/problems/clumsy-factorial/

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.)

Thoughts

求N的阶乘，只是乘要依次换成*/+-。观察到i * (i-1) / (i-2) = i+1，当 i >= 5时，因此

i * (i-1) / (i-2) + (i-3) - (i-4) * (i-5) / (i-6) + (i-7) - (i-8) * .... + rest elements=   (i+1) + "(i-3)" - "(i-4) * (i-5) / (i-6)" + "(i-7)" - "(i-8) * " .... + rest elements=   (i+1) + "(i-3) - (i-3)" + "(i-7) - (i-7)" +  ....  + rest elements=   (i+1) + rest elements

每四个元素看成一组，于是有以下四种情况：

1. when 0 element left: final result is (i+1) + ... + 5 - (4*3/2) + 1, which is i+1

2. when 1 element left: final result is (i+1) + ... + 6 - (5*4/3) + 2 - 1, which is i+2

3. when 2 element left: final result is (i+1) + ... + 7 - (6*5/4) + 3 - 2 * 1, which is i+2

4. when 3 element left: final result is (i+1) + ... + 8 - (7*6/5) + 4 - 3 * 2 / 1, which is i-1

因此分类讨论。参考自这。

Code

class Solution:
    def clumsy(self, N: int) -> int:
        if N <= 2: return N
        if N <= 4: return N + 3
        if N % 4 == 0: return N + 1
        elif N % 4 <= 2: return N + 2
        else: return N - 1
            

class Solution {
public:
    int clumsy(int N) {
        int i = N;
        int res = 0;
        for (; i >= 4; i -= 4) {
            int r = i * (i - 1) / (i - 2);
            res += i == N ? r : -r;
            res += (i - 3);
        }

        int r = 0;
        if (i >= 1) {
            int r = i * ((i - 1) >= 1 ? i - 1 : 1) / ((i - 2) >= 1 ? i - 2 : 1);
            res += i == N ? r : -r;
        }
        return res;
    }
};