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[英]Python: Find the two prime factors of a number. how to make this code more efficiency?
[英]python code trying to find prime number. Code is counting not prime number. Can't find why
我試圖在用戶選擇的范圍內找到每個素數,列出它們並計算它們。 我的代碼計數和列表編號不是素數。 我真的找不到原因? 有人可以幫我嗎。
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)
你的素數部分是錯誤的。 我將把它抽象為一種不同的方法,以使其更容易。 要成為素數, n%x
必須支持任何x
,使得x
在數字集合[2,ceil(sqrt(n))]
。
確保您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,只需更換
if num > 1:
for i in range(2,num):
if (num % i) == 0:
break
else:
prime.append(num)
break
和,
if num>1:
if is_prime(num):
prime.append(num)
您的問題是您不等待所有除法檢查將數字附加到素數列表:
if (num % i) == 0:
break
else:
prime.append(num)
break
您的代碼的修復:
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)
問題就在這里,
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
您應該像這樣檢查條件: If num is not divisible by ANY i, it is prime
。
使用 else 和 for 進行最少的更改,
for i in range(2,num):
if (num % i) == 0:
break
else:
prime.append(num)
這是從列表中獲取素數的解決方案。 詢問用戶應創建質數之間的范圍並將其存儲在list
:
list4=list(range(3,10,1))
prime_num=[x for x in list4 if x%2!=0 and x%3!=0]
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