您如何找到所有数字的列表，这些数字是2、3和5的幂的倍数? [英] How do you find the list of all numbers that are multiples of only powers of 2, 3, and 5?

问题描述

我正在尝试生成所有可以用，其中a，b和c是整数.我尝试了以下方法，

I am trying to generate a list of all multiples which can be represented by the form , where a, b, and c are whole numbers. I tried the following,

[ a * b * c | a <- map (2^) [0..], b <- map (3^) [0..], c <- map (5^) [0..] ] 

但它仅列出5的幂，而从不继续增加2或3.

but it only lists powers of 5 and never goes on to 2 or 3.

对不起，似乎我没有足够清楚地阐明这个问题.我想要的是一个有序的无限列表，尽管我可以对一个有限列表进行排序，但我觉得好像有一种更有效的解决方案.

My apologies, it seems that I did not clarify the question enough. What I want is an ordered infinite list, and while I could sort a finite list, I feel as if there may be a solution that is more efficient.

推荐答案

只有5的幂的原因是Haskell尝试评估a = 2 ^ 0和b = 3 ^ 0的所有可能c，并且仅当完成后，结果为a = 2 ^ 0和b = 3 ^ 1.因此，这种方式只能构造如下这样的有限列表:
[a * b * c |a<-地图(2 ^)[0..n]，b<-地图(3 ^)[0..n]，c<-地图(5 ^)[0..n]]
给定n.

The reason why there are only powers of 5 is that Haskell tries to evaluate every possible c for a = 2^0 and b = 3^0 and only when it is finished it goes for a = 2^0 and b = 3^1. So this way you can only construct a finite list like this:
[ a * b * c | a <- map (2^) [0..n], b <- map (3^) [0..n], c <- map (5^) [0..n] ]
for a given n.

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