Question

程序从用户那里收到一个正数k，并应检查方程有多少解

3*x+5*y=k

在有许多解决方案的情况下，该函数采用所有解决方案中|x-y|的更大的绝对值。如果只有一种解决方案，则将其打印出来。例如：

如果用户键入k=34，则(x,y)有两种解决方案：(3,5)和(8,2)。因此，程序有两种解决方案，分别计算|8-2|和|3-5|并采用较大的6。
如果示例的用户输入k=8，则只有一个解决方案(1,1)。

可悲的是，我的老师要求我只使用循环，if和else语句。没有递归，没有辅助函数，也没有数组。她还希望程序高效，这样我就不能在循环内部使用循环。

我试图编写代码，但是程序没有响应。我定义了counter来计算方程的可能解数，并且distance定义了更大的绝对值：

void check_how_many_solutions(int n) {
    int  y = 0, counter = 0, distance = 0, equation1 = 0, equation2 = 0, equation3 = 0; 
    while (equation3 <= n) {
        equation1 = (n - (5 * y)) / 3;
        equation2 = (n - (3 * equation1)) / 5; 
        equation3 = (3 * equation1) + (5 * equation2);
        if (equation3 == n) { 
            counter++;  
            if (fabs(equation1 - equation2) > distance)
                distance = fabs(equation1 - equation2);
        }
        y++;
    }   
    if (counter > 1)
        printf("The values of x and y are (%d,%d)\n", equation1, equation2); 
    else
        printf("The greater absolute value of |x-y| is %d\n", distance);
}

代码可以运行，但是没有结果。

Answer 1

这是我制作的一个程序，我相信它可以回答您所提出的问题。您只想遍历x的整数值，并检查计算出的y是否也是整数。我还跟踪找到的解决方案总数以及x和y之间距离最大的解决方案，并根据这些信息给出最终答案。

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

void count_solns(double);

int main(int argc, char *argv[])
{
    if (argc == 2 && argv[1]) {
        int k = atoi(argv[1]);
        count_solns(k);
    }
    else {
        puts("Usage: count_solns <positive_integer>");
        return 1;
    }
    return 0;
}

void count_solns(double k)
{
    printf("Print positive integer solutions to 3x + 5y = %.3f\n", k);
    int solns_found = 0;
    int distance = -1;
    int final_x, final_y;
    double x, y;
    for (x = 0; x <= (k/3); x++) {
        y = (k-3*x)/5;
        if (y == floor(y)) {
            printf("Solution found at (x, y) == (%d, %d)\n", (int) x, (int) y);
            solns_found++;
            if (fabs(x-y) > distance) {
                final_x = x;
                final_y = y;
            }
        }
    }
    if (solns_found == 0)
        printf("No whole number solutions found for 3x + 5y = %.3f", k);
    else if (solns_found == 1)
        puts("This is the only solution where x and y are both whole numbers");
    else
        printf("The whole number solution with the highest distance between x and y is (x, y) == (%d, %d)", final_x, final_y);
    return;
}

用法如下：

$ ./count_solns 70
Print positive integer solutions to 3x + 5y = 70.000
Solution found at (x, y) == (0, 14)
Solution found at (x, y) == (5, 11)
Solution found at (x, y) == (10, 8)
Solution found at (x, y) == (15, 5)
Solution found at (x, y) == (20, 2)
The solution with the highest distance between x and y is (x, y) == (20, 2)

Answer 2

在查看max1000001的解决方案之前，我建议您先找到拧紧的位置，修复它，然后进行比较。

您的代码当前无限循环。尝试仅遍历代码的几次迭代并打印变量，然后查看它们对于每次迭代是否对您有意义。这是简化的相同代码，循环限制为20次迭代，并在每次迭代中打印变量：

void check_how_many_solutions(int n) {
    int  y = 0, counter = 0, distance = 0, equation1 = 0, equation2 = 0, equation3 = 0; 
    while( y < 20){
    //while (equation3 <= n) {
        equation1 = (n - (5 * y)) / 3;
        equation2 = (n - (3 * equation1)) / 5; 
        equation3 = (3 * equation1) + (5 * equation2);
        printf("y: %i, equation1: %i, equation2: %i, equation3: %i\n",y,equation1,equation2,equation3);
        y++;
    }   
}

查找方程的（x，y）解

2 个答案: