[英]Why doesn't this prime number algorithm say that 2 is not prime?
為什么當您在該程序中輸入2時卻返回“ 2 is prime”? 根據該代碼,如果除以i
的余數等於0,則i
不是素數, i
是從2(包括2)到該數字的任何數字。 但是2的余數除以2就是0,那么為什么程序會說2是質數呢?
# Python program to check if the input number is prime or not
# take input from the user
num = int(input("Enter a number: "))
# prime numbers are greater than 1
if num > 1:
# check for factors
for i in range(2,num):
if (num % i) == 0:
print(num,"is not a prime number")
print(i,"times",num//i,"is",num)
break
else:
print(num,"is a prime number")
# if input number is less than
# or equal to 1, it is not prime
else:
print(num,"is not a prime number")
因為for i in range(2,2):
永遠不會為true /將不會執行。
考慮一下range(start, stop)
...開始和停止相同,因此不會進入for循環。
2是質數,但在這種情況下,無需計算if語句即可確定其質數
range(2, 2)
實際上是一個空列表,因此for循環根本不會迭代。 它只是直接跳轉到else
子句:
>>> num = 2
>>> range(2, num)
[]
>>> for i in range(2, num):
... print "executing loop"
... else:
... print "done!"
...
done!
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