编写一个python脚本来打印两个给定数字(包括两个值(之间的所有素数有人能告诉我我在这里做错了什么吗？
a = int(input("Enter the value of a : "))
b = int(input("Enter the value of b : "))
for k in range(a,b):
for i in range(2,k):
if k%i!=0:
if k!=i:
continue
elif k==i:
print(k)
break
elif k!=i:
break
您正在检查一个数字是否以错误的方式素数——有很多代码中未处理的情况，例如如果a大于b您将从"0"开始循环；a"；无论如何即使数值设置正确，算法也不正确这是我的最佳解决方案，希望它能帮助
a = int(input("Enter the value of a : "))
b = int(input("Enter the value of b : "))
def is_prime(n):
# negative numbers cannot be primes 1 and 0 are also not primes
if n <= 1:
return False
# since 2 is the first prime we will start looping from it
# until n since you mentioned that n is included
for i in range(2, n + 1):
# if n is cleanly divisible by any number less than n
# that means that n is not prime
if n % i == 0 and n != i:
return False
return True
for k in range(a,b):
if a > b:
print ("a cannot be bigger than b")
if is_prime(k):
print(k)
以下是解决方案。我添加了一些注释来解释代码的作用。
a = int(input("Enter the value of a : "))
b = int(input("Enter the value of b : "))
# Include b in the range by adding 1
for num in range(a, b + 1):
# Prime numbers are greater than 1
if num > 1:
for i in range(2, num):
# If the number is divisible, it is not prime
if num % i == 0:
# Check if the number is equal to the number itself
if num != i:
break
else:
# If the loop was not broken, the number isn't divisible
# by other numbers except itself and 1, so it's prime
print(num)
好吧，素数是"一个只能被自身和1〃整除的数；，所以实际上，首先我只去CCD_ 1。这种方法是最直接的方法之一，并且在计算上不友好。有一些算法专门研究这种素数的寻找。
我已经修复了代码，简化了表达式，并添加了前面提到的+1以实现包容性。
a = int(input("Enter the value of a : "))
b = int(input("Enter the value of b : "))
for k in range(a,b+1):
for i in range(2,k+1):
if k==i:
print(k)
elif k%i!=0:
continue
else: # is divisible and isn't 1 or the number
break
我鼓励你看看这篇文章，他们是如何做到的。
另一种方法是：
#Check only 1 number at time:
def check_prime(check):
if check > 1:
for i in range(2, check):
if check % i == 0:
return False
return True
return False
#then check your numbers in a loop
list_prime = []
for i in range(50, 250):
if check_prime(i):
list_prime.append(i)
print(list_prime)
这样你就可以一次检查1个可能的素数。如果你需要介于两者之间的数字，就把它放在循环中。
通常，当使用某种形式的"Erotothenes筛"；算法，只需要检查分母值，直到被评估数字的平方根。考虑到这一点，这里还有一个可能的素数循环测试。
import math
a = int(input("Enter the value of a : "))
b = int(input("Enter the value of b : "))
for k in range(a,b):
if k == 2 or k == 3: # Pick up the prime numbers whose square root value is less than 2
print(k)
x = int(math.sqrt(k) + 1) # Need only check up to the square root of your test number
for i in range(2,x):
if k%i!=0:
if x-1 > i:
continue
else:
print(k)
else: # If the remainder is zero, this is not a prime number
break
要尝试的另一个版本。