重新排列数字块 [英] Rearranging blocks of digits

问题描述

我遇到了一个棘手的问题，我不知道该答案的答案:用递归函数从两个整数中重新排列整数"这是一个例子:

I encountered a hard question I don't know the answer to: "Rearrange the digits from an integer in blocks of two with a recursive function" here's an example:

输入:123456

unsigned long pairinvPrint(unsigned long number) {
    printf("%d", number % 100);

    if ((number / 100) <= 99) {
        printf("%d", number / 100);
    }
    else {
        pairinv(number / 100);
    }
}

输出:563412
更多I/O示例:42->42;1234->3412

Output: 563412
More I/O Examples: 42 -> 42; 1234 -> 3412

但是，执行此操作的设置环境很困难(没有循环，数组，指针，全局或静态变量，没有库)，并且不应直接打印解决方案，而应在这样的调用时将其返回:

However, the set circumstances to do this are hard (no loops, arrays, pointers, global- or static variables, no libraries) and it should not print the solution directly, rather return it upon a call like this:

printf("Rearrange int (%lu) = %lu", input, pairinvert(input));

幸运的是，有一种情况使输入变得更容易，输入数字的数量始终为偶数.

Luckily there's one circumstance to make it easier, the number of the input digits is always even.

现在我尝试了一段时间，但无法提出一个可行的解决方案，除了使用 printf 的无效解决方案.

Now I experimented for a while, but cant come up with a working solution, except the invalid one using printf.

有人对我有启发性或想法如何解决这个问题吗?

Does anyone have some inspiration for me or idea how to tackle this?

推荐答案

我会咬:-)

unsigned long p(unsigned long p1, unsigned long p2) {
    // no loops, no arrays, no pointers, no global, no static, no variables, no libraries
    if (p1 < 100) return p2*100 + p1;
    return p(p1/100, p2*100 + p1%100);
}

unsigned long pairinvert(unsigned long n) {
    // no loops, no arrays, no pointers, no global, no static, no variables, no libraries
    if (n < 100) return n;
    return p(n/100, n%100);
}

// need <stdio.h> for printf()
#include <stdio.h>

int main(void) {
    unsigned long input;
    input = 123456;
    printf("Rearrange int (%lu) = %lu\n", input, pairinvert(input));
    input = 42;
    printf("Rearrange int (%lu) = %lu\n", input, pairinvert(input));
    input = 1234;
    printf("Rearrange int (%lu) = %lu\n", input, pairinvert(input));
}

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