C 三个堆栈并尝试所有选项答案

【问题标题】：C three stacks and try all optionsC 三个堆栈并尝试所有选项
【发布时间】：2016-04-01 16:52:44
【问题描述】：

我有三个堆栈和数字，我只需要尝试所有分发选项。我会计算每个选项并检查它是否符合我的要求。

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");

【问题讨论】：

到目前为止你的代码在哪里？
它只是作为输入和输出的数字数组都是组合
请发布您的代码
它没有展示您尝试做的任何事情。没有循环或任何我或其他人可以告诉你你做错了什么或你可以改进的地方。请更好地解释你的问题和你的代码，其中包含数组但没有对它们做任何事情......

标签： c options cycle

【解决方案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;
}

【讨论】：