如何修复一个质数三角形

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英文:

How to fix a triangle of prime numbers

问题

在上面提到的代码中,我尝试打印出一个类似以下的质数三角形:

2

2 3

2 3 5

2 3 5 7

2 3 5 7 11

但是,它打印出了10而不是11。在我尝试创建质数三角形时,我还得到了一些复合数。

当我在第2行中使用了range(15)时,我得到了以下结果:

2

2 3

2 3 5

2 3 5 7

2 3 5 7 10

2 3 5 7 10 11

2 3 5 7 10 11 13

2 3 5 7 10 11 13 17

2 3 5 7 10 11 13 17 19

2 3 5 7 10 11 13 17 19 22

2 3 5 7 10 11 13 17 19 22 23

2 3 5 7 10 11 13 17 19 22 23 25

2 3 5 7 10 11 13 17 19 22 23 25 29

2 3 5 7 10 11 13 17 19 22 23 25 29 31

2 3 5 7 10 11 13 17 19 22 23 25 29 31 34

我无法理解为什么在代码本应只打印质数的情况下,也会出现复合数。

英文:
n=3

for i in range(5):

    print(2, end=" ")

    for j in range(i):

        for k in range(2,n):

            if (n%k==0):

                n+=1

            else:

                print (n, end=" ")

                n+=1

                break

    n=3

 print()


In the code mentioned above , I'm trying to print out prime numbers in a triangle , like :

2

2 3

2 3 5

2 3 5 7

2 3 5 7 11

But , it prints out 10 instead of 11 .

Please help .

In the prime numbers triangle that I was trying to create , I got some composite numbers too in the way .

Say when I put range(15) in 2nd line , I got :

2

2 3

2 3 5

2 3 5 7

2 3 5 7 10

2 3 5 7 10 11

2 3 5 7 10 11 13

2 3 5 7 10 11 13 17

2 3 5 7 10 11 13 17 19

2 3 5 7 10 11 13 17 19 22

2 3 5 7 10 11 13 17 19 22 23

2 3 5 7 10 11 13 17 19 22 23 25

2 3 5 7 10 11 13 17 19 22 23 25 29

2 3 5 7 10 11 13 17 19 22 23 25 29 31

2 3 5 7 10 11 13 17 19 22 23 25 29 31 34

I am not able to understand that how do composite numbers , too , come in the triangle , when the code was supposed to print only prime numbers there .

答案1

得分: 0

你的代码逻辑有问题。这两行:

if (n%k==0):
    n+=1

在检测到n不是质数后,没有将k重置为2。因此,在增加n后,k会继续从之前的值开始。此外,这个语句:

else:
    print (n, end=" ")

意味着如果检查的第一个k值不是n的因子,就会自动假设n是质数。这是不正确的,因为如果n是合数,可能会有其他值的k是它的因子。

可以尝试像这样修改代码:

n=3

for i in range(5):

    print(2, end=" ")

    for j in range(i):

        ok = False
        
        while not ok:

            ok = True

            for k in range(2,n):

                if (n%k==0):

                    ok = False
                    
                    break
                
            if not ok:
                
                n += 1
                
        print (n, end=" ")
        
        n += 1

    n=3

    print()

在这里,ok 跟踪当前的 n 值是否为质数。

英文:

The logic of your code is flawed. These two lines:

if (n%k==0):
    n+=1

do not reset k to 2 after n has been detected as not being prime. Therefore, after you increment n, k continues where it has left off. Furthermore, this statement

else:
    print (n, end=" ")

means that n is automatically assumed as prime if the first value of k checked is not a factor of n. This is incorrect as there may be other values of k that are factors of n if n is composite.

Try something like this instead:

n=3

for i in range(5):

    print(2, end=" ")

    for j in range(i):

        ok = False
        
        while not ok:

            ok = True

            for k in range(2,n):

                if (n%k==0):

                    ok = False
                    
                    break
                
            if not ok:
                
                n += 1
                
        print (n, end=" ")
        
        n += 1

    n=3

    print()

Here, ok tracks if the current value of n is prime or not.

huangapple
  • 本文由 发表于 2023年6月6日 10:01:40
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