for i, j in enumerate(prime_gen()): # if i < n: if i < (n+1): primes.append(j) else: break # return primes return primes[1:] print("Find the first M prime numbers") # python 2.x输入数据的话,要作相应修改 M = input("M? ")
# Python Program to find prime numbers in a range import time def is_prime(n): if n <= 1: return False for i in range(2,n): if n % i == 0: return False return True # Driver function t0 = time.time() c = 0 #for counting for n in range(1,100000): x = is_prime(n)...
numbers = list() num = n while num <= m: i = 2 while i < num: if (num % i == 0) and (num != i): break else: i += 1 if num == i: numbers.append(num) num += 1 return numbers else: return "error input" print(find_prime_number(1, 100)) 取随机数扩展。取随机数...
import math def find_num(n): if (math.sqrt(n + 100)).is_integer() and (math.sqrt(n + 100 + 168)).is_integer(): return True return False if __name__ == '__main__': print('这样的数有:{}'.format(list(filter(find_num, range(7000000000))) 1. 2. 3. 4. 5. 6. 7. ...
题目:一个整数,它加上100后是一个完全平方数,再加上168又是一个完全平方数,请问该数是多少? 完全平方数:可以拆分成一个数的方,如121 = 11^2 importmathdeffind_num(n):if(math.sqrt(n +100)).is_integer()and(math.sqrt(n +100+168)).is_integer():returnTruereturnFalseif__name__ =='__main...
while count < NUMBER_OF_PRIMES: # Repeatedly find prime numbers...isPrime = True #Is the current number prime?...divisor = 2...while divisor <= number / 2:...if number % divisor == 0:# If true, the number is not prime...isPrime = False # Set isPrime to false...break # Exi...
我在Youtube上看到了关于如何解决Euler`s问题7的教程,并找到了以下代码: x = 2 while (len(list_of_primes) < numbers_prime_to_find): if all(x % prime!=0 for prime in list_of_primes): # for prime in li 浏览2提问于2020-07-30得票数 1 回答已采纳 6回答 Eratosthenes的python素数筛 、、...
/usr/bin/python # -*- coding: UTF-8 -*- sStr1 = 'abcdefg' sStr2 = 'cde' print sStr1.find(sStr2) 以上实例输出结果为: 2 题目63:画椭圆。 程序分析:使用 Tkinter。 程序源代码: 实例 #!/usr/bin/python # -*- coding: UTF-8 -*- if __name__ == '__main__': from Tkinter ...
In this program, we have checked if num is prime or not. Numbers less than or equal to 1 are not prime numbers. Hence, we only proceed if the num is greater than 1. We check if num is exactly divisible by any number from 2 to num - 1. If we find a factor in that range, th...
def test_02_v1(numbers): my_list_length = len(numbers) output_list = [] foriinrange(my_list_length): output_list.append(i * 2) returnoutput_list 通过将列表长度计算移出for循环,加速1.6倍,这个方法可能很少有人知道吧。 # Summary Of ...