How to Compute the Clumsy Factorials using Iterative Algorithm?

  • 时间:2020-09-28 16:28:51
  • 分类:网络文摘
  • 阅读:171 次

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

Clumsy Factorial Algorithms

Numbers can be proccessed in batches of four. Special cases are the four numbers from N to N-3 and the rest are stored in intermediate variable num, which is (a*b/c-d). Then the num is subtracted from the result.

The remainder items (up to 3) should be also subtracted. The following C++ implements the iterative algorithm that runs at O(N) complexity, and requires O(1) constant space.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
class Solution {
public:
    int clumsyFactorial(int N) {
        int r = 0, c = 0, num = 0;
        for (int i = N; i >= 1; -- i, ++ c) {
            switch (c % 4) {
                case 0:                    
                    num = i; break;
                case 1:
                    num *= i; break;
                case 2:
                    num /= i; break;
                case 3:
                    if (N - i <= 4) r = num + i; else r = r - num + i; break;
            }
        }
        if (N <= 3) return num;
        if (c % 4 != 0) r -= num;
        return r;
    }
};
class Solution {
public:
    int clumsyFactorial(int N) {
        int r = 0, c = 0, num = 0;
        for (int i = N; i >= 1; -- i, ++ c) {
            switch (c % 4) {
                case 0:                    
                    num = i; break;
                case 1:
                    num *= i; break;
                case 2:
                    num /= i; break;
                case 3:
                    if (N - i <= 4) r = num + i; else r = r - num + i; break;
            }
        }
        if (N <= 3) return num;
        if (c % 4 != 0) r -= num;
        return r;
    }
};

–EOF (The Ultimate Computing & Technology Blog) —

推荐阅读:
数正方形个数的数学题:5X5方格共有36个格点  数学题:小兔晴天可以采12个蘑菇  数学题:老师说我像你这么大的时候你才刚刚1岁  数学题:甲、丙两组人比乙组人数多2倍还多2人  数学题:22名家长和老师陪同学生参加某次数学竞赛  数学题:求X的长度是多少厘米  正好可以把它平均切成2个相等的正方体  数学题:求三角形ABC中阴影正方形的边长是多少厘米  数学题:三角形ABF的面积比三角形CEF的面积大8平方厘米  数学题:这个式子是不是某个数的平方 
评论列表
添加评论