Question

我试图在用户选择的范围内找到每个素数，列出并计算它们。我的代码计数和列表编号不是素数。我真的找不到原因？有人可以帮助我。

print("This code will count how many prime number exist in a certain range")
count = 0
lower = int(input("Enter lower range: "))
upper = int(input("Enter upper range: "))
prime = []
for num in range(lower, upper + 1):
    if num > 1:
        for i in range(2,num):
            if (num % i) == 0:
                break
            else:
                prime.append(num)
                break
print(prime)
print("There are", len(prime), "prime number between", lower, "and", upper)

Answer 1

你的素数部分是错误的。我将把它抽象为另一种方法以使其更容易。要成为素数，n%x必须支持任何x，x位于数字[2,ceil(sqrt(n))]中。

确保import math

def is_prime(num):
    for i in range(2, math.ceil(num**(1/2))):
        if num%i==0:
            return False
    return True

然后，只需替换

if num > 1:
    for i in range(2,num):
        if (num % i) == 0:
            break
        else:
            prime.append(num)
            break

使用，

if num>1:
   if is_prime(num):
      prime.append(num)

Answer 2

你的问题是你不要等所有分区检查将数字附加到素数列表：

            if (num % i) == 0:
                break
            else:
                prime.append(num)
                break

代码修复：

        is_prime =True
        for i in range(2,num):
            if num % i == 0:
                is_prime = False
                break

        if is_prime:
            prime.append(num)

$ python prime_test.py 
This code will count how many prime number exist in a certain range
Enter lower range: 12
Enter upper range: 20
[13, 17, 19]
('There are', 3, 'prime number between', 12, 'and', 20)

Answer 3

问题出在这里，

    for i in range(2,num):
        if (num % i) == 0:    #If num is divisible by SOME i, it is not prime. Correct.
            break
        else:                 #If num is not divisible by SOME i, it is not prime. WRONG.
            prime.append(num)
            break

您应该检查这样的条件：If num is not divisible by ANY i, it is prime。

使用for，

进行最少的更改

    for i in range(2,num):
        if (num % i) == 0:
            break
    else:
        prime.append(num)

了解详情：Why does python use 'else' after for and while loops?

Answer 4

这是从列表中获取素数的解决方案。向用户询问应创建素数之间的范围，并将其存储在list中：

list4=list(range(3,10,1))  
prime_num=[x for x in list4 if x%2!=0 and x%3!=0]

python代码试图找到素数。代码不是素数。无法找到原因

4 个答案: