c - 我如何获得最大数量的奖学金？

Question

我无法找到一种方法来“切断”该计划提供超过 5 个 1,000 美元奖学金和 10 个 500 美元奖学金的机会。在 5 和 8 之后我如何将它从我已经拥有的东西中删除？我的代码位于问题下方

<小时/>

“程序员共创美好明天”的一个部门是他们的奖学金捐赠基金。他们 每年为需要帮助的学生提供金额为 1000、500 和 250 的奖学金 美元。 这些奖学金的资金来自以前捐款的利息和 投资。

您将创建一个程序来计算基金的年利息并 确定可以授予多少 1000 美元、500 美元和 250 美元的奖学金。 例如，如果基金在 2016 年 9 月 30 日有 500,000 美元，并且每年 利率为 3%，则本期结束时基金内有 515,000 美元 九月。这为他们提供了 15,000 美元作为奖学金。

如果可能，基金倾向于颁发 5 名 1000 美元奖学金、10 名 500 美元奖学金，以及尽可能多的奖学金 250 美元，因为他们还有钱。凭借 15,000 美元，该基金可颁发 5 名 1000 美元的奖学金、10 名 500 美元奖学金和 20 名 250 美元奖学金。您的程序应该打印此信息 用户。

如果不可能，该基金将尽可能提供 1000 美元和 500 美元的奖学金。

例如，如果他们有 4,750 美元，他们将颁发 4 份 1000 美元的奖学金、1 份 500 美元的奖学金，以及 1 美元 250 美元的奖学金。

输入规范

1. 截至一年前的基金金额 n，其中 n 大于或等于 到 0。(n 可以包含小数位)
2. 年百分比 p，为整数，其中 p 大于零。

输出规范
使用以下格式输出结果:

X $1000 scholarships will be awarded.
Y $500 scholarships will be awarded.
Z $250 scholarships will be awarded

到目前为止我的代码:

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


//main function
int main() {

        int ten, five, twofive, leftovers_ten, leftovers_five, scholarship_money;
        float fund, interest;

printf("How much was in the fund last year?\n");
scanf("%f", &fund);

printf("What is the yearly percentage rate?\n");
scanf("%f", &interest);

    scholarship_money = fund * (interest / 100);

    {
    if(ten < 5) {
    ten = scholarship_money / 1000;
    printf("%d $1000 scholarships will be awarded.\n", ten);

    }
    else {
        ten = 5;
        printf("5 $1000 scholarships will be awarded.\n");
    }
    }

        leftovers_ten = scholarship_money - (ten * 1000);
    {

    if(five < 10) {
    five = leftovers_ten / 500;
    printf("%d $500 scholarships will be awarded.\n", five);

    }

    else {
            five = 10;
        printf("10 $500 scholarships will be awarded.\n");
    }
    }

    leftovers_five = leftovers_ten - (five * 500);

    twofive = leftovers_five / 250;
    printf("%d $250 scholarships will be awarded.\n", twofive);

    return 0;
}

Answer 1

最简单的调试工具就是打印中间结果。这样你就会发现评论者试图告诉你什么。如果我可以冒昧地给你一个简短的草图:

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

// ALL CHECKS OMMITTED!

/*
If possible, the Fund prefers to award 5 $1000 scholarships, 10 $500 scholarships, and as many $250 as they have money left for. With $15,000 the Fund can award 5 $1000 scholarships, 10 $500 scholarships, and 20 $250 scholarships.

If that is not possible, the Fund will award as many $1000 and $500 scholarships as they can.

Input Specification 
  1. The amount of money in the fund, n, as of one year ago where n is greater than
     or equal to 0. (n may include decimal places)
  2. The yearly percent rate, p, as an integer where p is greater than zero.

Output Specification

  Output the result using the format below:
      X $1000 scholarships will be awarded.
      Y $500 scholarships will be awarded.
      Z $250 scholarships will be awarded
*/

int main()
{

  int ten, five, twofive, interest;
  int res;
  float fund, leftovers_ten, leftovers_five, scholarship_money;

  printf("How much was in the fund last year?\n");
  // scanf returns the number of elements it had read
  res = scanf("%f", &fund);
  if (res != 1) {
    // just bail out here for simplicity
    fprintf(stderr, "Input for fund incorrect\n");
    exit(EXIT_FAILURE);
  }
  // 1. The amount of money in the fund, n, as of one year ago where n is greater than
  //    or equal to 0. (n may include decimal places)
  if (fund < 0.0) {
    fprintf(stderr, "Fund must be bigger than or equal to zero but is %f\n",
            fund);
    exit(EXIT_FAILURE);
  }

  printf("What is the yearly percentage rate?\n");
  res = scanf("%d", &interest);
  if (res != 1) {
    // just bail out here for simplicity
    fprintf(stderr, "Input for interrest incorrect\n");
    exit(EXIT_FAILURE);
  }
  // 2. The yearly percent rate, p, as an integer where p is greater than zero.
  if (interest <= 0) {
    fprintf(stderr, "Interest must be bigger than zero but is %d\n", interest);
    exit(EXIT_FAILURE);
  }

  // some of the casts that have been added for clarity are redundant
  scholarship_money = fund * (1.0 + (float) interest / 100.0);
  printf("scholarship_money:  %.20f\n", scholarship_money);
  ten = (int) floor(scholarship_money / 1000.0);
  // the Fund prefers to award 5 $1000 scholarships...
  if (ten > 5) {
    scholarship_money = scholarship_money - (5000.0);
    ten = 5;
  }
  printf("ten:                %d\n", ten);

  leftovers_ten = scholarship_money - (ten * 1000.0);
  printf("leftovers_ten:      %.20f\n", leftovers_ten);
  five = (int) floor(leftovers_ten / 500.0);
  // ... 10 $500 scholarships ...
  if (five > 10) {
    leftovers_ten = leftovers_ten - (5000.0);
    five = 10;
  }
  printf("five:               %d\n", five);

  leftovers_five = scholarship_money - (float) (ten * 1000 + five * 500);
  printf("leftovers_five:     %.20f\n", leftovers_five);
  // ... and as many $250 as they have money left for
  twofive = (int) floor(leftovers_five / 250.0);
  printf("twofive:            %d\n", twofive);

  printf("%d $1000 scholarships will be awarded.\n", ten);
  printf("%d $500 scholarships will be awarded.\n", five);
  printf("%d $250 scholarships will be awarded.\n", twofive);
  printf("Sum:  %d\n", ten * 1000 + five * 500 + twofive * 250);
  printf("Rest: %.20f\n",
         scholarship_money - (float) (ten * 1000 + five * 500 + twofive * 250));

  exit(EXIT_SUCCESS);
}

您最大的问题之一是，您使用浮点进行货币运算，但我会将其留给您(或您的教授，他稍后可能会或可能不会指出这些问题)。

此代码还有很多其他方式可能会失败，您需要填补所有这些漏洞!