（原问题见算法导论第三版2.1-4）
问题：考虑把两个n位二进制数加起来的问题。这两个整数分别存储在两个n元数组A和B中。这两个整数的和应按二进制形式存储在一个（n + 1）元数组C中。
解答：
代码如下。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>


void binary_add(int *, int *, int *, int);

int main(int argc, char const *argv[])
{
    /* code */
    if (argc < 3) {
        fprintf(stderr, "please input enough binary number\n");
        return -1;
    }

    const char* num1 = argv[1];
    const char* num2 = argv[2];

    int n1 = strlen(num1);
    int n2 = strlen(num2);
    int n = (n1 > n2) ? n1 : n2;

    int * left = calloc(n1, sizeof(int));
    int * right = calloc(n2, sizeof(int));
    int * result = calloc(n + 1, sizeof(int));

    for (int i = 0; i < n1; ++i) {
        if (num1[i] == '1') {
            left[n1 - i - 1] = 1;
        } else if (num1[i] == '0') {
            left[n1 - i - 1] = 0;
        } else {
            fprintf(stderr, "error input left number, please input binary number\n");
            return -1;
        }
    }
    for (int i = 0; i < n2; ++i) {
        if (num2[i] == '1') {
            right[n2 - i - 1] = 1;
        } else if (num2[i] == '0') {
            right[n2 - i - 1] = 0;
        } else {
            fprintf(stderr, "error input right number, please input binary number\n");
            return -1;
        }
    }
    for (int i = 0; i < n + 1; ++i)
    {
        if (i < n + 1 - n1) {
            printf(" ");
        } else {
            printf("%d", left[n1 - (i - n - 1 + n1) - 1]);
        }
    }
    printf("\n");
    for (int i = 0; i < n + 1; ++i)
    {
        if (i < n + 1 - n2) {
            printf(" ");
        } else {
            printf("%d", right[n2 - (i - n - 1 + n2) - 1]);
        }
    }
    printf("\n");

    binary_add(left, right, result, n);

    for (int i = 0; i < n + 1; ++i)
    {
        printf("%d", result[n - i]);
    }
    printf("\n");
    return 0;
}

void binary_add(int *left, int *right, int *result, int n) {
    int c = 0;
    int temp;
    for (int i = 0; i < n; i++) {
        temp = left[i] + right[i] + c;
        result[i] = temp % 2;
        c = temp / 2;
    }
    if (c != 0) {
        result[n] = c;
    }
}