Question

我对python很陌生，所以我不完全了解如何使用循环。我目前正在编写一段代码，我必须找到前N个素数。理想的结果是，如果您输入5，则输出2、3、5、7和11，但是无论我为“ max”输入什么，输出总是以2和3结尾。改善这个？

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    server = smtplib.SMTP_SSL('smtp.yandex.com.tr', 465)
    ...

Answer 1

您只能在min块中递增else，也就是说，如果min % c对于所有c都不为零，即min是质数。这意味着代码将无法越过任何复合数字。您可以通过将min=min+1取消缩进一级来使其与for和else对齐来解决此问题。

Answer 2

number = int(input("Prime numbers between 2 and "))
for num in range(2,number + 1):
       if num > 1:
            for i in range(2,num):
                if (num % i) == 0:
                     break
            else:
                print(num)

Answer 3

我想这会有所帮助，让我知道。

number = int(input("How many prime numbers do you want ? "))

cnt = 0
for i in range (1, number+1):
    for j in range(1, i+1):
        if(i%j == 0):
            cnt += 1
    if(cnt == 2):
        print(i)
    cnt = 0

Answer 4

解决方案：获取第n个质数项。遍历每个自然数以获取素数，并将素数附加到列表中。当列表的长度满足用户第n个素数项时，终止程序。

# Get the number of prime numbers entry.
try:
    enterNumber = int(input("List of nth prime numbers: "))
except:
    print("The entry MUST be an integer.")
    exit()
startNumber = 1
primeList = []
while True:
    # Check for the entry to greater than zero.
    if enterNumber <= 0:
        print("The entry MUST be greater than zero.")
        break
    # Check each number from 1 for prime unless prime number entry is satisfied.
    if startNumber > 1:
        for i in range(2,startNumber):
            if (startNumber % i) == 0:
                break
        else:
            primeList.append(startNumber)
    if (len(primeList) == enterNumber):
        print(primeList)
        break
    else:
        startNumber = startNumber + 1
        continue

Answer 5

以下代码将为您提供3到N之间的质数，其中N是来自用户的输入：

number = int(input("Prime numbers between 2, 3 and ")) 
for i in range(2,number):
    for j in range(2,int(i/2)+1):
        if i%j==0:
    break
        else:
            if j==int(i/2):
                print(i)

您可以看到检查数字i为素数，只需检查数字的可除性，直到n / 2。

Answer 6

尝试：

n = int(input("First N prime number, N ? "))
p = [2]
c = 2

while len(p) < n:
    j = 0
    c += 1
    while j < len(p):
        if c % p[j] == 0:
            break
        elif j == len(p) - 1:
            p.append(c)
        j += 1
print(p)

Answer 7

很简单。检查下面的代码，确定它可以正常工作！

N = int(input('Enter the number: ')
i=1
count=0
while(count<N):
      for x in range(i,i+1):
          c=0
          for y in range(1,x+1):
              if(x%y==0):
                 c=c+1
          if(c==2):
             print(x)
             count=count+1
      i=i+1

如何在python中找到前N个素数？

7 个答案: