[英]How do you find the first N prime numbers in python?
I am pretty new to python, so I don't fully understand how to use loops.我对 python 很陌生,所以我不完全理解如何使用循环。 I am currently working on a piece of code that I have to find the first N prime numbers.
我目前正在编写一段代码,我必须找到前 N 个素数。 The result that is desired is if you input 5, it outputs 2, 3, 5, 7, and 11, but no matter what I input for 'max', the output always ends up being 2 and 3. Is there a way to improve this?
所需的结果是,如果您输入 5,它会输出 2、3、5、7 和 11,但无论我为“max”输入什么,输出总是最终为 2 和 3。有没有办法改善这个?
max=int(input("How many prime numbers do you want: "))
min=2
while(min<=(max)):
for c in range(2, min):
if min%c==0:
break
else:
print min
min=min+1
You only increment min
in the else
block, ie, if min % c
is nonzero for all c
, ie, if min
is prime.您只在
else
块中增加min
,即,如果min % c
对于所有c
不为零,即,如果min
是素数。 This means that the code won't be able to move past any composite numbers.这意味着代码将无法越过任何合数。 You can fix this by unindenting
min=min+1
one level so that it lines up with the for
and else
.您可以通过将
min=min+1
取消缩进一级来解决此问题,使其与for
和else
。
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)
I guess this will help, let me know. 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
Solution: Get the nth prime number entry.解决方案:获取第 n 个质数条目。 Iterate through each natural numbers for prime number and append the prime number to a list.
遍历每个自然数以获得质数并将质数附加到列表中。 Terminate the program when length of a list satisfies the user nth prime number entry.
当列表的长度满足用户第 n 个素数条目时终止程序。
# 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
The following code will give you prime numbers between 3 to N, where N is the input from user:以下代码将为您提供 3 到 N 之间的素数,其中 N 是用户的输入:
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)
You can see to check a number i to be prime you only have to check its divisibility with numbers till n/2.你可以看到检查一个数字 i 是素数,你只需要检查它与数字的整除性,直到 n/2。
Try that :试试看:
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)
Its simple.这很简单。 Check the below code, am sure it works!
检查下面的代码,确保它有效!
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
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