前 100 个素数 [英] First 100 prime numbers

问题描述

我知道有多种方法可以找到前 100 个素数，但请帮助我解决问题.我发现 count 的值正在增加，但由于某种原因 while 循环条件不适用:

count = 0而(计数 <= 20):对于范围内的 i (2, 20):对于范围内的 j (2, i):如果我＜j:打印(数字"，i，是素数")elif i % j == 0:休息别的:打印(数字"，i，是素数")计数 = 计数 + 1打印(计数)

解决方案

您可以使用 Eratosthenes 筛网 找到前 n 个素数:

def primes_upto(limit):素数 = [真] * 限制对于范围内的 n(2，限制):如果素数[n]:yield n # n 是素数对于范围内的 c (n*n, limit, n):prime[c] = False # 标记复合

要获得前 100 个质数:

<预><代码>>>>列表(primes_upto(542))[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, ... ,499、503、509、521、523、541]

要找到前 n 个素数，您可以估计第 n 个素数(通过上限) 或使用 无限质数生成器 并根据需要获取尽可能多的数字，例如，使用 list(itertools.islice(gen, 100)).

I know there are a number of ways to find the first 100 prime numbers but please help me in my approach. I find the value of count to be increasing but for some reason the while loop condition doesn't apply:

count = 0

while(count <= 20):
    for i in range(2, 20):
        for j in range(2, i):
            if i < j:
                print("The number",i,"is prime")
            elif i % j == 0:
                break
        else:
            print("The number",i,"is prime")
            count = count + 1
            print(count)

解决方案

You could use Sieve of Eratosthenes to find the first n prime numbers:

def primes_upto(limit):
    prime = [True] * limit
    for n in range(2, limit):
        if prime[n]:
            yield n # n is a prime
            for c in range(n*n, limit, n):
                prime[c] = False # mark composites

To get the first 100 primes:

>>> list(primes_upto(542))
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, ... ,
 499, 503, 509, 521, 523, 541]

To find the first n primes, you could estimate n-th prime (to pass the upper bound as the limit) or use an infinite prime number generator and get as many numbers as you need e.g., using list(itertools.islice(gen, 100)).

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