我isPrime的範圍不工作，就只給出第一個數字

Question

#!/usr/local/bin/python3.6 

def isPrime(n): 


if n == 1: 
    print("1 is a special") 
    return False 


for x in range(2, n): 
    if n % x == 0: 
     print("{} is equal to {} * {}".format(n, x, n // x)) 
     return False 
    else: 
     print(n, " is a prime number") 
     return True 


for n in range(1, 21): 
    isPrime(n) 

isPrime(2) 
isPrime(21) 
isPrime(25) 

，結果它給我的是：

1 is a special 
3 is a prime number 
4 is equal to 2 * 2 
5 is a prime number 
6 is equal to 2 * 3 
7 is a prime number 
8 is equal to 2 * 4 
9 is a prime number 
10 is equal to 2 * 5 
11 is a prime number 
12 is equal to 2 * 6 
13 is a prime number 
14 is equal to 2 * 7 
15 is a prime number 
16 is equal to 2 * 8 
17 is a prime number 
18 is equal to 2 * 9 
19 is a prime number 
20 is equal to 2 * 10 

21 is a prime number 
25 is a prime number 

存在的（2）， 沒有結果，也這是結果「奇偶」號碼，而不是「isPrime」，因爲在代碼「在範圍X（2，n）的」，它已計算出僅與號碼2對於x

什麼錯誤我的代碼？

Answer 1

不要返回True形成第一次迭代。不是被2整除並不能使它總理：

for x in range(2, n): # it would be enough to loop to sqrt(n) 
    if n % x == 0: 
     # you know it is NOT prime after first divisor found 
     return False 
# you only know it IS prime after you tried all possible divisors 
return True 

Answer 2

這樣做：

def isPrime(n): 
    if n == 1: 
     print("1 is a special") 
     return False 
    for x in range(2, n): 
     if n % x == 0: 
      print("{} is equal to {} * {}".format(n, x, n // x)) 
      return False 
    print(n, " is a prime number") 
    return True 

for n in range(1, 21): 
    isPrime(n) 

Answer 3

for x in range(2, n): 
    if n % x == 0: 
     print("{} is equal to {} * {}".format(n, x, n // x)) 
     return False 
else: 
    print(n, " is a prime number") 
    return True 

觀看縮進：別人現在是在與（不與IF）對齊，它應該解決你的問題。否則意味着x在n上重複，所以n％x從不0 =>你有一個素數