c - C三个堆栈并尝试所有选项

Question

我有三个筹码和数字，我需要的是尝试分配的所有选项。我将计算每个选项并检查其是否符合我的要求。

示例：3个数字：1 2 3
123 ||
12 | 3 |
12 || 3
1 | 23 |
1 || 23 ...

编辑：添加代码：

int array[] = { 1,2,3 };
int stackA[10];
int stackB[10];
int stackC[10];
...
printf("stackA contains 123, stackB contains, stackC contains");
printf("stackA contains 12, stackB contains 3, stackC contains");
printf("stackA contains 12, stackB contains, stackC contains 3");

Answer 1

编辑：这是用于计算所有可能性的有效C代码：

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

/* exponentiation: base b, exponent n */
int power(int b, unsigned int n)
{
    int result = 1;

    while (n != 0)
    {
        if ((n & 1) == 1)
        {
            result *= b;
        }

        n = n >> 1;
        b *= b;
    }

    return result;
}

void print_instance_of_solution_set(int* a, char** stack, int number_of_stacks, unsigned int number_of_elements)
{
    int k, j;

    for (k = 0; k < number_of_stacks; k++)
    {
        printf("Stack %i = { ", k);
        for (j = 0; j < number_of_elements; j++)
        {
            if (stack[k][j] == 1)
            {
                printf("%i ", a[j]);
            }
        }
        printf("}; ");
    }
    printf("\n");
}

int main()
{
    /* configuration parameters */
    const int number_of_stacks = 3;
    const unsigned int number_of_elements = 5;
    int a[] = { 0, 2, 4, 6, 8 }; /* has to contain number_of_elements elements */

    /* other variables */
    int i, j, k;
    int total_count;
    char** stack;

    /* allocate memory for stacks */
    stack = (char**)malloc(number_of_stacks*sizeof(char*) + number_of_stacks*number_of_elements*sizeof(char));
    if (stack == 0)
    {
        fprintf(stderr, "Memory allocation error!\n");
        return -1;
    }
    for (i = 0; i < number_of_stacks; i++)
    {
        stack[i] = (char*)stack + number_of_stacks*sizeof(char*) + i*number_of_elements*sizeof(char);
    }

    /* calculate cardinality of solution set */
    /* the cardinality of solution set is number_of_stacks raised to the power number_of_elements */
    total_count = power(number_of_stacks, number_of_elements);

    /* iterate over all instances */
    for (i = 0; i < total_count; i++)
    {
        int tmp = i;
        for (j = 0; j < number_of_elements; j++)
        {
            for (k = 0; k < number_of_stacks; k++)
            {
                stack[k][j] = (k == tmp % number_of_stacks) ? 1 : 0;
            }
            tmp /= number_of_stacks;
        }

        /* output of stack elements */
        print_instance_of_solution_set(a, stack, number_of_stacks, number_of_elements);
    }

    return 0;
}