确定可分解为 2^p5^q 的数字集的算法

Question

我正在尝试编写一种算法，该算法可以返回小于实例 n 的正整数集，并且可以分解为 2^p5^q。我的数学不是最好的，所以我不知道如何确定一个数字是否可以分解为这种特定形式...

任何帮助将不胜感激:)

Answer 1

I have no idea how I can determine whether a number can be factorised in this specific form

与其测试给定数字是否可以因式分解并遍历所有小于 n 的数字，不如直接生成它们，例如:

void findNumbers(int n)
{
    int p = 0;
    int power2 = 1; 
    while(power2 < n)
    {
        int q = 0;
        int power5 = 1;
        while(power5 * power2 < n)
        {
            printf("%d p = %d q = %d\n", power5 * power2, p, q);
            power5 *= 5;
            q++;
        }
        power2 *= 2;
        p++;
    }
}

n = 500 的输出:

1 p = 0 q = 0
5 p = 0 q = 1
25 p = 0 q = 2
125 p = 0 q = 3
2 p = 1 q = 0
10 p = 1 q = 1
50 p = 1 q = 2
250 p = 1 q = 3
4 p = 2 q = 0
20 p = 2 q = 1
100 p = 2 q = 2
8 p = 3 q = 0
40 p = 3 q = 1
200 p = 3 q = 2
16 p = 4 q = 0
80 p = 4 q = 1
400 p = 4 q = 2
32 p = 5 q = 0
160 p = 5 q = 1
64 p = 6 q = 0
320 p = 6 q = 1
128 p = 7 q = 0
256 p = 8 q = 0

它只是循环遍历 p 和 q 的每个组合，直到 n。

如果要排除 p = 0 和 q = 0，只需从 1 开始循环并设置 power2 = 2 和 power5 = 5。