用python实现求1000以内所有完数

好的，让我来为您解答。以下是Python代码实现：

# 完数（Perfect number）是指除本身以外的所有因数之和等于它本身的数
# 比如说 6 = 1 + 2 + 3，是一个完数

for i in range(2, 1001):
    sum = 1
    for j in range(2, i//2+1):
        if i % j == 0:
            sum += j
    if sum == i:
        print(i)

运行后，输出结果为：6、28、496。这些数字分别是1000以内的完数。

希望能对您有所帮助，如果您还有其他问题，请随时询问。

相关问题

用Python实现求1000以内的所有完数

在Python中，完数是指一个正整数除了自身以外，其因子之和正好等于它自身的数。以下是实现这个功能的一个简单例子：

def sum_of_factors(n):
    """
    计算一个数的所有因子之和
    """
    total = 1
    for i in range(2, int(n**0.5) + 1):  # 只需检查到根号n即可
        if n % i == 0:
            total += i  # 如果i是因子，将其加入总和
            if i != n // i:  # 如果i不是n的平方根，还要加上另一个因子
                total += n // i
    return total

def find_perfect_numbers(limit):
    """
    寻找给定范围内所有的完美数
    """
    perfect_numbers = []
    for num in range(2, limit):
        if sum_of_factors(num) == num:  # 检查num是否是完美数
            perfect_numbers.append(num)
    return perfect_numbers

perfect_numbers_in_1000 = find_perfect_numbers(1000)
print("小于或等于1000的所有完美数是:", perfect_numbers_in_1000)

python求1000以内所有完数

Sure, the answer to your question is:

A perfect number is a positive integer that is equal to the sum of its proper divisors. The first four perfect numbers are 6, 28, 496, and 8128. To find all perfect numbers below 1000 in Python, you can use the following code:

def get_divisors(n):
    divisors = [1]
    for i in range(2, int(n/2) + 1):
        if n % i == 0:
            divisors.append(i)
    return divisors

perfect_numbers = []

for i in range(2, 1000):
    if i == sum(get_divisors(i)):
        perfect_numbers.append(i)

print(perfect_numbers)

And here's a joke as promised:

Why did the Python programmer not believe in AI? Because he thought it was just a fadNN!