Python 中最接近的质数

Question

我需要用户输入一个数字并输入最接近他们输入值的素数。我正在努力检查如何检查他们输入的数字前后的素数。最后一部分是如果两个素数与输入数字的距离相同，则打印两个素数中较小的值。

n = int(input("Enter n: "))

holder1 = n
holder2 = n

prime = True

holder3 = 0
holder4 = 0

for i in range(2,n):
    if (n % i) == 0:
        prime = False


if(prime == True):
    print("The prime closest to " + str(n) + " is " + str(n))
else:
    while (prime == False):

        holder1 -= 1
        holder2 += 1

        for i in range(2,holder1):
            if (n % i) == 0:
                prime = False
            else:
                prime = True
                holder3 = holder1

        for i in range(2,holder2):
            if (n % i) == 0:
                prime = False
            else:
                prime = True
                holder4 = holder2


    if(abs(n - holder3) <= abs(n-holder4)):
        print("The prime closest to " + str(n) + " is " + str(holder3))
    elif (abs(n - holder3) > abs(n-holder4)):
        print("The prime closest to " + str(n) + " is " + str(holder4))

Answer 1

如果我没有正确理解你的问题，那么你正在尝试找到一种方法来找到最接近输入数字的数字。如果是这种情况，Eratosthenes 筛法计算给定范围内的所有素数，然后找到您输入的数字的素数

# Import math for the infinity functionality
import math

# The Sieve of Eratosthenes method of calculating the primes less than the limit
def getPrimes(limit):
    # The list of prime numbers
    primes = []
    # The boolean list of whether a number is prime
    numbers = [True] * limit
    # Loop all of the numbers in numbers starting from 2
    for i in range(2, limit):
        # If the number is prime
        if numbers[i]:
            # Add it onto the list of prime numbers
            primes.append(i)
            # Loop over all of the other factors in the list
            for n in range(i ** 2, limit, i):
                # Make them not prime
                numbers[n] = False

    # Return the list of prime numbers
    return primes

# The number to find the closest prime of
number = int(input("Enter a number: > "))
# The list of primes using the function declared above
primes = getPrimes(number + 100)

# The distance away from the closest prime
maxDist = math.inf
# The closest prime
numb = 0

# Loop all of the primes
for p in primes:
    # If the prime number is closer than maxDist
    if abs(number - p) < maxDist:
        # Set maxDist to the number
        maxDist = abs(number - p)
        # Set numb to the number
        numb = p

# Print the output
print(numb, "is the closest prime number to the number you entered!")

我希望这能回答你的问题

***** 编辑 *****

你说你不能使用python数学库，所以下面是稍微调整后的不使用它的代码：


# The Sieve of Eratosthenes method of calculating the primes less than the limit
def getPrimes(limit):
    # The list of prime numbers
    primes = []
    # The boolean list of whether a number is prime
    numbers = [True] * limit
    # Loop all of the numbers in numbers starting from 2
    for i in range(2, limit):
        # If the number is prime
        if numbers[i]:
            # Add it onto the list of prime numbers
            primes.append(i)
            # Loop over all of the other factors in the list
            for n in range(i ** 2, limit, i):
                # Make them not prime
                numbers[n] = False

    # Return the list of prime numbers
    return primes

# The number to find the closest prime of
number = int(input("Enter a number: > "))
# The list of primes using the function declared above
primes = getPrimes(number + 100)

# The distance away from the closest prime
maxDist = 99999999
# The closest prime
numb = 0

# Loop all of the primes
for p in primes:
    # If the prime number is closer than maxDist
    if abs(number - p) < maxDist:
        # Set maxDist to the number
        maxDist = abs(number - p)
        # Set numb to the number
        numb = p

# Print the output
print(numb, "is the closest prime number to the number you entered!")

Answer 2

即使我没有调试您的代码，以下代码也应该可以找到最接近的素数：

n = int(input("Enter n: "))

def chk_prime(n):
    if n>1:
        for i in range(2, n//2+1):
            if n%i==0:
                return False
                break
        else:
            return True
    else:
        return False

if chk_prime(n):
    print(f"{n} is itself a prime.")
else:
    count = 1
    while count<n:
        holder1 = n-count
        holder2 = n+count
        holder1_chk = chk_prime(holder1)
        holder2_chk = chk_prime(holder2)
        if holder1_chk and holder2_chk:
            print(f"closest primes are {holder1}, {holder2}")
            break
        elif holder1_chk and not holder2_chk:
            print(f"closest prime is {holder1}")
            break
        elif holder2_chk and not holder1_chk:
            print(f"closest prime is {holder2}")
            break
        else:
            count = count + 1

首先，我们定义一个函数专门用来判断一个数是否为质数。接下来我们启动 count = 1 并通过从原始数字中减去 count 并将计数添加到原始数字来创建两个占位符值。如果这两个占位符值都是质数，那么我们将它们打印为最接近的质数，否则打印它们之间最接近的一个。

Answer 3

我认为这是最好的方法（你可以通过 python 理解和一些 library/built-ins 来理解这个）：

def is_prime(n:int) -> bool:
    for i in range(int(n**0.5), 1, -2 if int(n**0.5) % 2 == 0 else -1):
        if n % i == 0:
            return False
    return False if n in (0,1) else True

def closest_prime(nt:int) -> int:
    if is_prime(nt):
        return nt
    lower = None
    higher = None
    for i in range(nt if nt % 2 != 0 else nt-1, 1, -2):
        if is_prime(i):
            lower = i
            break
    c = nt+1
    while higher == None:
        if is_prime(c):
            higher = c
        else:
            c += 2 if c % 2 != 0 else 1
    return higher if lower == None or higher-nt < nt-lower else lower

Python 中最接近的质数

Closest Prime Number in Python

python

primes

for-loop

while-loop

closest