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=3for i in range(5):print(2, end=" ")for j in range(i):for k in range(2,n):if (n%k==0):n+=1else:print (n, end=" ")n+=1breakn=3print()
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=3for i in range(5):print(2, end=" ")for j in range(i):ok = Falsewhile not ok:ok = Truefor k in range(2,n):if (n%k==0):ok = Falsebreakif not ok:n += 1print (n, end=" ")n += 1n=3print()
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=3for i in range(5):print(2, end=" ")for j in range(i):ok = Falsewhile not ok:ok = Truefor k in range(2,n):if (n%k==0):ok = Falsebreakif not ok:n += 1print (n, end=" ")n += 1n=3print()
Here, ok tracks if the current value of n is prime or not.
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论