二进制的二三事 | Parallel Labs


二进制的二三事 | Parallel Labs
Task1:求一个固定长度集合所有子集
例如集合{a, b, c, d, e}中的子集{a, b, c}可以用“11100”来表示,子集{c, d, e}可以用“00111”来表示,空子集{}可以用“00000”来表示。
那么对任意固定长度的集合a[n],我们可以从a[0]开始穷举a[0]在子集中和a[0]不在子集中两种情况,再依次类推从a[1]开始一直穷举到a[n-1]为止。于是我们可以得到一个很直观的递归程序:
void sub_sets_recu(int index, int length, char *array, char *bits)
{
    int j = 0;
 
    if (index >= length) {
        for (j = 0; j < length; j++) {
            if (bits[j] == '1') {
                printf("%c", array[j]);
            }
        }
        printf("\n");
    }
    else {
        bits[index] = '1';
        sub_sets_recu(index+1, length, array, bits);
        bits[index] = '0';
        sub_sets_recu(index+1, length, array, bits);
    }
}
二进制版本
void sub_sets_iter(int length, char *array)
{
    int i = 0;
    int j = 0;
    for (i = 0; i < (1<<length); i++) {
        for (j = 0; j < length; j++) {
            if ((i & (1<<j)) != 0) {
                printf("%c", array[j]);
            }
        }
        printf("\n");
    }
}

Task2:小白鼠试毒药问题
实验室里有1000个一模一样的瓶子,但是其中的一瓶有毒。可以用实验室的小白鼠来测试哪一瓶是毒药。如果小白鼠喝掉毒药的话,会在一个星期的时候死去,其他瓶子里的药水没有任何副作用。请问最少用多少只小白鼠可以在一个星期以内查出哪瓶是毒药?
这个问题关键是跳出思维,并不是说瓶子和小白鼠必须一一对应,然后我们用1000只小白鼠,每只各试一瓶药水,等一个礼拜然后出结果。其实我们仔细想想,小白鼠试毒只有两种结果,非“死”即“生”,又是一个0/1状态。很显然我们可以再次构建一个0/1组成的二叉树,由此即可表示不同的试毒结果。假设我们需要n个小白鼠,那么2^n-1应大于等于1000(因为有1000种可能的结果,减一是指要除去小白鼠全部不死的情况 — 多谢danqi指正),那么最小的n就是10了
解决此类问题的关键在于找到能用0/1表示的状态,然后想办法用0/1组成的二进制数来表示不同的结果,从而达到节省存储空间,提高解题效率的目的。

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