Question

您本应归还的部分并不想工作，我正尽我所能优化代码以减少代码行。

有帮助吗？

#include <iostream>
#include <math.h>
using namespace std;

int main() 
{

   double iAmount_due { 0 };

   double iGiven_money { 0 };

   double iMoney_back; 

   iMoney_back = iGiven_money - iAmount_due;

   int iMoney100 { 0 };
   int iMoney50  { 0 };
   int iMoney20  { 0 };
   int iMoney10  { 0 };
   int iMoney5   { 0 };
   int iMoney2   { 0 };
   int iMoney1   { 0 };
   int iCent25   { 0 }; 
   int iCent10   { 0 };
   int iCent5    { 0 };
   int iCent1    { 0 };

   cout << "Enter the amount due please: " << endl;

   cin >> iAmount_due;

   cout << "Enter the amount given please: " << endl;

   cin >> iGiven_money;


   if (iGiven_money >= iAmount_due) {

      iMoney_back = iGiven_money - iAmount_due;

      cout << "We will give you : " << iMoney_back << " $ back" << endl;

   }

   else {

      cout << "No money back" << endl;

   }


   while (iGiven_money >= iAmount_due) {

       iMoney100 = iMoney_back / 100;
       iMoney_back = (int)iMoney_back % 100;

       cout << "You will get: " << iMoney100 << " X 100 $ " << endl;

       iMoney50 = iMoney_back / 50;
       iMoney_back = (int)iMoney_back % 50;

       cout << "You will get: " << iMoney50 << " X 50 $ " << endl;

       iMoney20 = iMoney_back / 20;
       iMoney_back = (int)iMoney_back % 20;

       cout << "You will get: " << iMoney20 << " X 20 $ " << endl;

       iMoney10 = iMoney_back / 10;
       iMoney_back = (int)iMoney_back % 10;

       cout << "You will get: " << iMoney10 << " X 10 $ " << endl; 

       iMoney5 = iMoney_back / 5;
       iMoney_back = (int)iMoney_back % 5;

       cout << "You will get: " << iMoney5 << " X 5 $ " << endl; 


       iMoney2 = iMoney_back / 2;
       iMoney_back = (int)iMoney_back % 2;

       cout << "You will get: " << iMoney2 << " X 2 $ " << endl; 

       iMoney1 = iMoney_back / 1;
       iMoney_back = (int) iMoney_back % 1;

       cout << "You will get: " << iMoney1 << " X 1 $ " << endl;


       iCent25 = iMoney_back / 0.25;
       iMoney_back = fmod ((int)iMoney_back, 0.25);

       cout << "You will get: " << iCent25 << " X 0.25 $ " << endl;


       iCent10 = iMoney_back / 0.10;
       iMoney_back = fmod ((int)iMoney_back , 0.10);

       cout << "You will get: " << iCent10 << " X 0.10 $ " << endl;


       iCent5 = iMoney_back / 0.5;
       iMoney_back = fmod ((int)iMoney_back , 0.5);

       cout << "You will get: " << iCent5 << " X 0.5 $ " << endl;

       iCent1 = iMoney_back / 0.1;
       iMoney_back = fmod ((int)iMoney_back, 0.1);

       cout << "You will get: " << iCent1 << " X 0.1 $ " << endl;

       break;

   }

   return 0;
}

Answer 1

这是您的问题：

iMoney_back = (int)iMoney_back % 100;

让我猜：您尝试过iMoney_back % 100，但发现不能同时使用mod（%）和double。因此，您只需将其转换为整数即可。

虽然可以很好地进行编译，但是该模块化算术的结果仍然是int。换句话说，你只是失去的小数部分的iMoney_back，所以你的程序，它看起来像没有变化是由于！

您可能想要fmod()：

iMoney_back = fmod(iMoney_back, 100);

此外，看起来您实际上是为{cent {1}}编写的，而不是.1。修正这些错误，并在美分上略有四舍五入的错误，将会向您显示此方法有效：https://ideone.com/0cWcuQ

或者，您可以将一个.01变量设为一个整数，这样您甚至不必担心双精度（这可能会导致不精确的问题）：

iMoney_back_cents

收银机仙计算不起作用？

1 个答案: