julia - 在 Julia 中使用 for 循环打印范围内的素数

Question

我正在学习 Julia 语法，而且我对一般编码不熟悉。我正在寻找一种使用 for 循环和 if 语句打印给定范围内所有素数的方法。(像这样)。

for num in 1:100
    if num > 1
        for i in 1:(num)
            if (num % i) == 0
                print(num)
            end
        end
    end
end

Answer 1

如果您有兴趣，这里有一个基于 Sieve of Eratosthenes 的替代解决方案:

UPPERLIMIT = 100

#=
Create an array of booleans called "is_prime" which is initially true
for all odd numbers, false for all even numbers
=#
is_prime = [(i % 2) > 0 for i in 1:UPPERLIMIT]
is_prime[2] = true  # 2 is the exception, an even prime

#=
Now iterate through the array up to the square root of UPPERLIMIT.
factors which are bigger than sqrt(UPPERLIMIT) must have co-factors
that are smaller, so there's no need to search above that limit.
=#
for num in 3:2:floor(Int,sqrt(UPPERLIMIT))
    #=
    Every time we find a prime, wipe out all of its multiples.
    We can start at 3*num since 2*num is already masked out
    because it's even.  Adding odd multiples would also yield
    even outcomes, so they can be stepped over as well.
    =#
    if is_prime[num]
        for i in 3*num:2*num:UPPERLIMIT
            is_prime[i] = false
        end
    end
end

# Everything that hasn't been masked out to false is now prime!
for i in 2:UPPERLIMIT
    if is_prime[i]
        println(i)
    end
end

请注意，这与蛮力方法的实际代码行数(去除注释后)几乎相同。两者产生相同的结果，但这个版本要快得多 - 在我的笔记本电脑上生成 664579 个素数到 10_000_000 需要 2.1 秒对 67 秒。

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附录

我自己是 Julia 的新手，并且在玩这个的过程中一直在学习。这导致了下面的版本，我用 && 和 foreach 替换了条件循环。我删除了注释以强调需要的代码有多么少:

UPPERLIMIT = 100

is_prime = [(i % 2) > 0 for i in 1:UPPERLIMIT]
is_prime[2] = true  # 2 is the exception, an even prime

for num in 3:2:floor(Int,sqrt(UPPERLIMIT))
    is_prime[num] && foreach(i -> is_prime[i] = false, num*num:2*num:UPPERLIMIT)
end

foreach(i -> is_prime[i] && println(i), 2:UPPERLIMIT)

这个变体不仅涉及更少的代码，而且运行速度更快——1.41 秒对比 2.1 秒生成高达 10M 的素数。