使用循环和 if/else 的现金找零程序

Question

编写一个程序，允许用户输入最多 200.00 美元的现金金额，然后计算并打印以下面额(20、10、5、1、.25、.10、.05、.01)的金额。我想我已经弄清楚了如何获得面额(除法/模数)的基础知识，但 do/while 和 if/else 的结构给我带来了麻烦。我不断收到一个错误，即我需要一个 while 语句，即使我已经输入了它及其条件(如下所示)，但我也对在哪里放置范围提示感到有点困惑(如果用户输入了一些内容)负数或高于 200)。任何建议/指导将不胜感激!

double amt_ent;
    int twenty, ten, five, one, quarter, dime, nickel, penny, remainder;

    printf ("Enter a dollar amount up to $200.00:");
    scanf ("%lf", &amt_ent);

  do
  { printf ("Name - Assignment 2 - Change-O-Matic\n");
      printf ("Amount entered: $%.2lf\n", ((amt_ent*100)/100));
      printf ("Change breakdown:\n");

      { /*Change in twenties*/
      twenty= (int) amt_ent/20;
      if (twenty >= 2)
      printf("%i\t$20.00s\n", twenty);
      if (twenty == 1)
      printf ("%i\t$20.00\n", twenty);
      else 

     /*Change in tens*/
      remainder = twenty % 20;
      ten = remainder/10;

      if (ten >=2)
      printf ("%i\t$10.00s\n", ten);
      if (ten == 1)
      printf ("%i\t$10.00\n", ten);
      else 

    /*Change in fives*/
     remainder = ten % 10;
     five = remainder/10;

     if (five >= 2)
     printf ("%i\t$5.00s\n", five);
     if (five == 1)
     printf ("%i\t$5.00\n", five);
     else 

    /*Change in ones*/
     remainder = five % 5;
     one = remainder/1;

     if (one >= 2)
     printf ("%i\t$1.00s\n", one);
     if (one == 1)
     printf ("%i\t$1.00\n", one);
     else 

    /*Change in quarters*/
     remainder = one % 1;
     quarter = remainder/.25;

     if (quarter >= 2)
     printf ("%i\t$.25s\n", quarter);
     if (quarter == 1)
         printf ("%i\t$.25\n", quarter);
     else

    /*Change in dimes*/
     remainder = quarter % 4;
     dime = remainder/.10;

     if (dime >= 2)
     printf ("%i\t$.10s\n", dime);
     if (dime == 1)
     printf ("%i\t$.10\n", dime);
     else

    /*Change in nickels*/
    remainder = dime % 10;
          nickel = remainder/.05;

          if (nickel >= 2)
              printf ("%i\t$.05s\n", nickel);
          if (nickel == 1)
              printf ("%i\t$.05\n", nickel);
          else 

    /*Change in pennies*/
              remainder = nickel % 20;
          penny = remainder/100;

          if (penny >= 2)
              printf ("%i\t$.01s\n", penny);
          if (penny == 1)
              printf ("%i\t$.01\n", penny);
      }

    while ((amt_ent <= 200.00) && (amt_ent >= 00.00));}

return 0;

Answer 1

关于您的错误， printf ("Change breakdown:\n"); 之后有额外的大括号。并且您不需要在 while 语句之后放置右大括号。 while ((amt_ent <= 200.00) && (amt_ent >= 00.00));}也删除它。

对于您的有效金额处理问题，有一个 continue 命令跳过剩余的循环并在遇到时从头开始重复。您可以使用它。

do
  { 
      printf ("Enter a dollar amount up to $200.00:");
      scanf ("%lf", &amt_ent);
      if(amount_ent < 00.00 || amount_ent>200.00)
           continue;
      printf ("Name - Assignment 2 - Change-O-Matic\n");
      printf ("Amount entered: $%.2lf\n", ((amt_ent*100)/100));
      printf ("Change breakdown:\n");

如果输入的金额无效，则 continue语句将执行并保持循环被跳过。然后printf ("Enter a dollar amount up to $200.00:");将被执行。因此您可以看到，除非输入正确的金额值，否则用户将无法继续继续。