C 编程 - 分解金钱

Question

我需要在他们各自的钞票/硬币上 segmentation 金额。

输出有点像这样:

到目前为止，这是我的代码:(我将最后几段代码作为注释，因为错误来自那里)

  { 
  int x,y;

  printf("Enter input: ");
  scanf("%d",&x);

  y=x/1000;
  printf("\n# of $1000 bill: %d",y);
  x = x%1000;

  y=x/500;      
  printf("\n# of 4500 bill: %d",y);    
  x = (x%500);

  y=x/200;
  printf("\n#. of $200 bill: %d",y);    
  x = (x%200);

  y=x/100;
  printf("\n# of $100 bill: %d",y);    
  x = (x%100);

  y=x/50;
  printf("\n# of $50 bill: %d",y);    
  x = (x%50);

  y=x/20;
  printf("\n# of $20 bill: %d",y);    
  x = (x%20);

  y=x/10;
  printf("\n#. of $10 coin: %d",y);    
  x = (x%10);

  y=x/5;
  printf("\n#. of $5 coin: %d",y);    
  x = (x%5);

  y=x/1;
  printf("\n# of $1 coin: %d",y);    
  x = (x%1);


  getch();
  return 0;
  }

我希望你能帮我解决这个问题。 :/

谢谢!

Answer 1

这是一个简单的启发式问题。你已经有了正确的想法。首先，如果您想使用 int(这可能是避免 fp 精度问题和使用单独 fmod 的需要的好主意)，您必须缩放浮点输入和模数/除数除以 100。还要减少一些冗余，请考虑使用如下函数:

// returns number of units and subtracts unit_size * result
// from val
int units(int* val, int unit_size)
{
    int num = *val / unit_size;
    *val %= unit_size;
    return num;
}

printf("No. of P1000 bill: %d\n",units(&x, 1000 * 100) );
printf("No. of P500 bill: %d\n",units(&x, 500 * 100) );
printf("No. of P200 bill: %d\n",units(&x, 200 * 100) );
etc.

这应该会减少一些冗余代码。也许不值得太花哨，我怀疑这是作业。完整的解决方案:

// returns number of units and subtracts unit_size * result
// from val
int units(int* val, int unit_size)
{
    int num = *val / unit_size;
    *val %= unit_size;
    return num;
}

int main()
{
    printf("Enter input: ");

    float amount;
    scanf("%f",&amount);
    int x = (int)(amount * 100.0 + 0.5);

    printf("No. of P1000 bill: %d\n", units(&x, 1000 * 100) );
    printf("No. of P500 bill: %d\n", units(&x, 500 * 100) );
    printf("No. of P200 bill: %d\n", units(&x, 200 * 100) );
    printf("No. of P100 bill: %d\n", units(&x, 100 * 100) );
    printf("No. of P50 bill: %d\n", units(&x, 50 * 100) );
    printf("No. of P20 bill: %d\n", units(&x, 20 * 100) );
    printf("No. of P10 coin: %d\n", units(&x, 10 * 100) );
    printf("No. of P5 coin: %d\n", units(&x, 10 * 100) );
    printf("No. of P1 coin: %d\n", units(&x, 1 * 100) );
    printf("No. of 25 cents: %d\n", units(&x, 25) );
    printf("No. of 1 cent: %d\n", units(&x, 1) );

    return 0;
}

[编辑] 如果您在理解指针方面有困难，那么就按照您编写的方式进行操作，而不使用 units 函数，但相应地修改它以读入 float 并乘以 100，如上例所示。

[编辑]要求:

int main()
{
    printf("Enter input: ");

    float amount;
    scanf("%f",&amount);
    int x = (int)(amount * 100.0 + 0.5); // x stores the user input in cents

    int y = x / 100000; // 1000 dollars is 100,000 cents
    printf("\nNo. of P1000 bill: %d",y);
    x = x % 100000;

    ...

    y=x / 25; // we're working with cents, so 25 = 25 cents
    printf("\nNo. of 25 cents: %d",y);    
    x = (x % 25);

    ...
}