求函数的结果

发布于 2024-11-07 07:15:11 字数 290 浏览 4 评论 0原文

有一个未知函数的代码：（

function Magic(number)
    r = number mod 2
    print r
    if number > 1
        Magic(number / 2)

用伪代码编写）

问题是：应该传递什么整数才能收到以下答案

0 1 1 0 0 1

主要问题是我无法弄清楚 mod 在伪代码中是如何工作的。应 5,5 mod 3 = 2.5 或 2

原文

There is a code of an unknown function:

function Magic(number)
    r = number mod 2
    print r
    if number > 1
        Magic(number / 2)

(written in pseudo-code)

The question is: what integer number should be passed in order to receive the following answer

0 1 1 0 0 1

The main problem is that I can't figure out how mod is working in pseudocode.
Should 5,5 mod 3 = 2.5 or 2

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嗼ふ静 2024-11-14 07:15:11

除法和取模运算都应该只接受和输出整数。 “5.5 mod 3”fe 没有任何意义。 11 / 2（整数除法）将返回 5，而不是 5.5。

下面是一个实现伪代码的 PHP 程序：

<?php

function Magic($number) {
    $r = $number % 2;
    echo $r . ' ';

    if ($number > 1) Magic($number / 2);
}

for ($i = 16; $i < 34; ++$i) {
    echo "($i: ";

    Magic($i);
    echo ") ";
}

echo "\n";

输出结果：

(16: 0 0 0 0 1 ) (17: 1 0 0 0 1 0 ) (18: 0 1 0 0 1 0 ) (19: 1 1 0 0 1 0 ) (20: 0 0 1 0 1 0 ) (21: 1 0 1 0 1 0 ) (22: 0 1 1 0 1 0 ) (23: 1 1 1 0 1 0 ) (24: 0 0 0 1 1 0 ) (25: 1 0 0 1 1 0 ) (26: 0 1 0 1 1 0 ) (27: 1 1 0 1 1 0 ) (28: 0 0 1 1 1 0 ) (29: 1 0 1 1 1 0 ) (30: 0 1 1 1 1 0 ) (31: 1 1 1 1 1 0 ) (32: 0 0 0 0 0 1 ) (33: 1 0 0 0 0 1 0 )

显示 6 位结果 (x: 0 1 1 0 0 1 ) 对于任何整数 x 都是不可能的code> （因为输出字符串单调增长）。然而，Magic(38) 是 0 1 1 0 0 1 0 — 第一个 7 位结果带有所需的字符串，但也有一个尾随零。

至于负整数值，唯一可能的 2 个输出是“0”和“-1”。

Both division and mod operations are only supposed to accept and output integer numbers here. "5.5 mod 3" f.e. doesn't really make any sense. And 11 / 2 (integer division) will return 5, not 5.5.

Here's a PHP program that implements your pseudo code:

<?php

function Magic($number) {
    $r = $number % 2;
    echo $r . ' ';

    if ($number > 1) Magic($number / 2);
}

for ($i = 16; $i < 34; ++$i) {
    echo "($i: ";

    Magic($i);
    echo ") ";
}

echo "\n";

Results in output:

(16: 0 0 0 0 1 ) (17: 1 0 0 0 1 0 ) (18: 0 1 0 0 1 0 ) (19: 1 1 0 0 1 0 ) (20: 0 0 1 0 1 0 ) (21: 1 0 1 0 1 0 ) (22: 0 1 1 0 1 0 ) (23: 1 1 1 0 1 0 ) (24: 0 0 0 1 1 0 ) (25: 1 0 0 1 1 0 ) (26: 0 1 0 1 1 0 ) (27: 1 1 0 1 1 0 ) (28: 0 0 1 1 1 0 ) (29: 1 0 1 1 1 0 ) (30: 0 1 1 1 1 0 ) (31: 1 1 1 1 1 0 ) (32: 0 0 0 0 0 1 ) (33: 1 0 0 0 0 1 0 )

Which shows that the 6-digit result (x: 0 1 1 0 0 1 ) is impossible for any integer x (because of the monotonic growth of the output string). However, Magic(38) is 0 1 1 0 0 1 0 — the first 7-digit result with your required string, but also having a trailing zero.

As for the negative integer values, the only 2 outputs possible are "0 ", and "-1 ".

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空气里的味道 2024-11-14 07:15:11

首先，这是解决这个问题的可执行Python代码。

def Magic(number):
r = number % 2
print r
if number > 1:
    Magic(number / 2)

Magic(15)

然而，这个问题的模式是该函数返回给定输入数字的 REVERSE 二进制数字。因此，在这种情况下，最简单的解决方案是取 0 1 1 0 0 1，将其反转为 100110，并计算该二进制数的值，即 32 + 4 + 2 = 38。使用这种方法，您可以计算任何给定输入所需的数量或预期输出。

First of all, here is executable python code for this problem.

def Magic(number):
r = number % 2
print r
if number > 1:
    Magic(number / 2)

Magic(15)

However, the pattern in this problem is that the function is returning the REVERSE binary number for the given input number. So, in this instance, the easiest solution would be to take 0 1 1 0 0 1, reverse it to 100110, and calculate the value of that binary number, which is 32 + 4 + 2 = 38. Using this methodology, you can calculate the required number or the expected output for any given input.

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