以下是一个可能的解决方案：

def is_prime(n: int) -> bool:
    if n <= 1:
        return False
    if n == 2:
        return True
    if n % 2 == 0:
        return False
    i = 3
    while i**2 <= n:
        if n % i == 0:
            return False
        i += 2
    return True


cd = int(input("Enter the number of elements: "))
A = []
for i in range(cd):
    el = int(input(""))
    A.append(el)

# Dictionary `prime_counts` will contain the number of times that each prime
# number appears in list `A`. The dictionary will have the following structure:
# - Keys: prime number
# - Values: number of times the prime number appears on list `A`
prime_counts = {}

for el in A:
    if is_prime(el):  # Check if number is prime
        # Add new key to the `prime_counts` dictionary if key doesn’t already
        # exist..
        if el not in prime_counts.keys():
            prime_counts[el] = 0
        # Add 1 to the prime counting
        prime_counts[el] += 1

# Finding which prime number repeats the most
max_count = max(prime_counts.items(), key=lambda kv: kv[1]) # Returns a tuple, like (139, 121)
# Where:
# - 139: is the prime number that repeats the most
# - 121: frequency that the number repeats

print(f"The prime number {max_count[0]:,} repeats most frequently: {max_count[1]:,}")
# Prints something like:
# "The prime number 139 repeats most frequently: 121"

Testing

为了测试该解决方案，我们将使用random包生成20000个值在0到200之间的随机数。

import random

cd = 20_000
A = [random.randint(0, 200) for _ in range(cd)]

prime_counts = {}
for el in A:
    if is_prime(el):
        if el not in prime_counts.keys():
            prime_counts[el] = 0
        prime_counts[el] += 1

max_count = max(prime_counts.items(), key=lambda kv: kv[1])
print(f"The prime number {max_count[0]:,} repeats most frequently: {max_count[1]:,}")
# Prints:
# "The prime number 173 repeats most frequently: 127"

Detailed Overview

我们可以将问题分解为几个步骤：

1. Extract all prime numbers from the array.
2. Count the occurrences of each prime number.
3. Identify the prime number with the highest count.

您开始使用的代码并不完全正确，因为它试图检查列表中每个数字的素性，但它没有跟踪计数。

定义函数is_prime有助于我们检查一个数字是否为素数：

def is_prime(n):
    if n <= 1:
        return False
    if n == 2:
        return True
    if n % 2 == 0:
        return False
    i = 3
    while i * i <= n:
        if n % i == 0:
            return False
        i += 2
    return True

Next, we can modify your original code to create a list of prime numbers and count their occurrences:

cd = int(input("Enter the number of elements: "))
A = []
for i in range(cd):
    el = int(input(""))
    A.append(el)

prime_counts = {}
for el in A:
    if is_prime(el):
        if el not in prime_counts.keys():
            prime_counts[el] = 0
        prime_counts[el] += 1

最后，使用max函数，我们可以找到最常出现的素数：

max_count = max(prime_counts.items(), key=lambda kv: kv[1]) # Returns a tuple, like (139, 121)
# Note: kv stands for "key-value"
print(f"The prime number {max_count[0]:,} repeats most frequently: {max_count[1]:,}")