for num in range(100, 1000): if sum(map(lambda x:int(x)**3, str(num))) == num: print(num) 详细解释这段代码

这段代码是用来找出三位数中的所有阿姆斯特朗数的。阿姆斯特朗数（又称水仙花数）是指一个 n 位正整数（n≥3）的每个位上的数字的 n 次幂之和等于它本身。例如，153 是一个阿姆斯特朗数，因为 1^3 + 5^3 + 3^3 = 153。

现在来详细解释这段代码：

for num in range(100, 1000):  # 遍历所有三位数
    if sum(map(lambda x:int(x)**3, str(num))) == num: # 判断是否为阿姆斯特朗数
        print(num)  # 输出阿姆斯特朗数

代码中的 `range(100, 1000)` 表示遍历所有从 100 到 999 之间的整数，即所有的三位数。遍历过程中，对于每一个数 `num`，都会执行 `if` 语句中的判断。

`map(lambda x:int(x)**3, str(num))` 这一部分是将 `num` 转换为字符串后，对每个字符执行 `int(x)**3` 操作，然后将结果组成一个列表。这样可以得到所有位数的立方和。例如，对于数字 153，`map` 函数会生成一个列表 `[1**3, 5**3, 3**3]`。

接着，`sum` 函数对上面生成的列表求和，得到这个数的立方和。如果这个和等于 `num`，那么这个数就是阿姆斯特朗数，满足条件，就会被输出。

相关问题

for num in range(100,1000): r=map(lambda x:int(x)**3,str(num))

if sum(r)==num: print(num)

This code finds all the Armstrong numbers between 100 and 999. An Armstrong number is a number that is equal to the sum of its own digits raised to the power of the number of digits. For example, 153 is an Armstrong number because 1^3 + 5^3 + 3^3 = 153.

The code uses a for loop to iterate through all the numbers between 100 and 999. It then converts each number to a string and applies a lambda function to each digit to raise it to the power of 3. The resulting list of cubed digits is then summed up and checked against the original number. If they are equal, the number is printed out as an Armstrong number.

for num in range(100,1000): r=map(lambda x:int(x)**3,str(num)) if sum(r)==num: print(num)

As an AI language model, I cannot execute code, but I can explain what this code does.

This code is checking for all three-digit numbers that are equal to the sum of the cubes of their digits.

The code first iterates through all numbers from 100 to 999 using the `range()` function.

For each number, it converts it to a string and then maps the `int()` function to each digit in the string to convert it to an integer. It then cubes each integer using the `**` operator and returns the list of cubes.

The `sum()` function is then used to add up all the cubes, and if the sum equals the original number, it is printed to the console.

This code is an example of a brute-force approach to solving a problem, as it simply checks every possible three-digit number until it finds the ones that meet the criteria.