c 8个硬币组合码[关闭]

Question

我的代码应该计算总组合。使用这些硬币支付 1-500 美元之间的现金：1、2、5、10、20、50、100、200。但它计算不正确。我做错了什么？ 您可以假设输入是正确的，并且您只能使用循环和 if 语句。你不能使用递归。

int pr, a, b, c, d, e, f, g,h, poss = 0;
printf_s("What is the amount that you like to check? (or press '0' to exit)\n");
scanf_s("%d", &pr);

for (a = 0; a <= pr; a++)
{
    for (b = 0; b <= (pr/2); b++)
    {
        for (c = 0; c <= (pr /5); c++)
        {
            for (d = 0; d <= (pr /10); d++)
            {
                for (e = 0; e <= (pr /20); e++)
                {
                    for (f = 0; f <= (pr / 50); f++)
                    {
                        for (g = 0; g <= (pr / 100); g++)
                        {

                            for (h = 0; h <= (pr/200); h++)
                            {
                                if (1 * a + 2 * b + 4 * c + 10 * d + 20 * e + 40 * f + 100 * g + h * 200 == pr)

                                    poss += 1;

                            }
                        }
                    }
                }
            }
        }
   }printf_s("The number of possibilities is: %d.\n", poss);
}

Answer 1

当输入5 时，会报告正确的权限数，但扩展代码会打印错误的值。因为这一行

if (1 * a + 2 * b + 4 * c + 10 * d + 20 * e + 40 * f + 100 * g + h * 200 == pr)

有错误的面额。应该是

if (1 * a + 2 * b + 5 * c + 10 * d + 20 * e + 50 * f + 100 * g + h * 200 == pr)

你也应该移动这条线

printf_s("The number of possibilities is: %d.\n", poss);

在循环之外。

Answer 2

您的最终printf 需要脱离循环

    for (a = 0; a <= pr; a++)
    {
        for (b = 0; b <= (pr/2); b++)
        {
            for (c = 0; c <= (pr /5); c++)
            {
                for (d = 0; d <= (pr /10); d++)
                {
                    for (e = 0; e <= (pr /20); e++)
                    {
                        for (f = 0; f <= (pr / 50); f++)
                        {
                            for (g = 0; g <= (pr / 100); g++)
                            {

                                for (h = 0; h <= (pr/200); h++)
                                {
                                    if (1 * a + 2 * b + 4 * c + 10 * d + 20 * e + 50 * f + 100 * g + h * 200 == pr)

                                        poss += 1;

                                }
                            }
                        }
                    }
                }
            }
       }
    }
    printf("The number of possibilities is: %d.\n", poss);

Answer 3

正如Weather Vane 回答中提到的，OP 在内部条件中使用了一些错误的乘数（4 和 40 而不是 5 和 50）。

值得注意的是，即使使用这种蛮力方法，我们也可以通过避免在（过于嵌套的）循环内进行不必要的计算并将这些循环的范围限制在更小的范围内，从而节省一些 CPU 时间。 考虑以下重构：

#include <stdio.h>

int number_of_possibilities(int price)
{
    int poss = 0;
// It takes less time to consume the bigger pieces earlier    
    for ( 
// 'a' represent the sum of the values of 200$ pieces, not the number of 200$
// pieces, which is 'a / 200'           
          int a = 0; a <= price;
// add the value of a single 200$ piece to move forward
          a += 200 )
    {
      for ( int b = 0,
// 'dif_b' is what is left from the price, once the 200$ pieces are counted
                dif_b = price - a;
// we don't need to iterate from 0 to 'price', but only to what is left
            b <= dif_b; b += 100 )                
      {
// 'dif_c' is what is left once the 100$ and 200$ pieces counted so far are
// subctracted from the original price. The same holds for the inner loops     
        for ( int c = 0, dif_c = dif_b - b; c <= dif_c; c += 50 )
        {
          for ( int d = 0, dif_d = dif_c - c; d <= dif_d; d += 20 )
          {
            for ( int e = 0, dif_e = dif_d - d; e <= dif_e; e += 10 )
            {
              for ( int f = 0, dif_f = dif_e - e; f <= dif_f; f += 5 )
              {
                for ( int g = 0, dif_g = dif_f - f; g <= dif_g; g += 2 )
                {
// now that only the 1$ coins are left to consider, we can avoid another inner
// loop and just realize that we need exactly 'dif_g - g' 1$ coins to pay the 
// full price, so there is one and only one possible combination.
                  ++poss;
                }
              }
            }
          }
        }      
      }
    }
    return poss;
}


int main(void)
{
    printf("  i   number of possibilities\n\n");
    for ( int i = 0; i < 501; ++i )
    {
        printf("%4d %16d\n", i, number_of_possibilities(i));
    }
    return 0;
}

这给出了以下内容：

  i   number of possibilities

   0                1
   1                1
   2                2
   3                2
   4                3
   5                4
   6                5
   7                6
   8                7
   9                8
  10               11

 ...

  99             4366
 100             4563
 101             4710    

 ...

 498          6159618
 499          6224452
 500          6295434

在 1.6 GHz 的旧 Atom N270 上执行不到 2 秒...