Question

我试图编写一个遵循这一行的程序（示例其工作方式）；用户输入：

Give lower limit: 1
Give upper limit: 10

程序应从1-10开始打印：

1 is not a valid start number.
2 is a prime number.
3 is a prime number.
4 is not a prime number, because 2 * 2 = 4
5 is a prime number.
...
10 is not a prime number, because 2 * 5 = 10

检查给定范围后，应打印：

10 numbers was examined, 4 of which were prime numbers
The last found prime number is 7

我试图解决这个问题已经很长时间了，我太沮丧了，失去了一切。如果有人对此有解决方案，我很乐意看到它并进行分析，同时还要从中学习。

感谢，这将是巨大的帮助！

Answer 1

lower = int(input("Give lower limit: "))               # Get lower bound, must be an int
upper = int(input("Give upper limit: "))               # Get upper bound, must be an int
print()                                                # Print an empty line
primes = []                                            # An array to store the primes in
for i in range(lower, upper+1):                        # Go through the range
    if i == 1:
        print("1 is not a valid start number")
    else:
        primeFlag = True                               # Will only remain as true if the number is prime
        for div in range(2, i):                        # Go from 2 to the current number
            if i % div == 0:
                primeFlag = False                      # If this happens, it isnt prime
        if(primeFlag):                                 # If it is a prime
            primes.append(i)                           # Add to primes array
            print(i," is a prime number")
        else:
            print(i," is not a valid prime")

print((upper-lower+1), " numbers were examined, ", len(primes), " were prime numbers")
print("The last found prime number is ", primes[-1:])

会产生：

Give lower limit: 1
Give upper limit: 10

1 is not a valid start number
2  is a prime number
3  is a prime number
4  is not a valid prime
5  is a prime number
6  is not a valid prime
7  is a prime number
8  is not a valid prime
9  is not a valid prime
10  is not a valid prime
10  numbers were examined,  4  were prime numbers
The last found prime number is  [7]

我希望这会有所帮助，它可以使您几乎获得所需的输出，请花一些时间来理解和学习。

Answer 2

我首先创建了一个范围为low and high的列表。从那里使用1消除了is not a valid start，然后我创建了一个简单的检查表，运行i % n，其中n是范围2, i中的值。现在，如果all(check)表示所有结果均返回非零值，则该数字为质数。如果不是，我声明数字不是素数，如果enumerate, start = 2 0 item中的check is不是项目,真will return idx { {1}} 0`的值来解释为什么它不是素数。

and you can use the

of that

给定范围内的素数Python

2 个答案: