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c++ - 收银机美分计算不起作用？

问题描述

你应该拿回你的美分的部分不想工作，我试图尽可能地优化我的代码以减少代码行。

有什么帮助吗？

#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;
}

标签： c++computer-science

解决方案

这是你的问题：

iMoney_back = (int)iMoney_back % 100;

让我猜猜：您尝试iMoney_back % 100并发现您不能将 mod ( %) 与 double 一起使用。因此，您只需将其转换为 int。

虽然这将编译得很好，但该模运算的结果仍然是一个 int。换句话说，您只是丢失了iMoney_back.

你可能想要fmod()：

iMoney_back = fmod(iMoney_back, 100);

此外，看起来您实际上是.1为 1 美分值而不是.01. 修复这些和美分上的轻微舍入错误将向您展示这是有效的：https ://ideone.com/0cWcuQ

或者，您可以创建iMoney_back_cents一个 int 变量，这样您甚至不需要担心双精度数（这可能会导致一些不精确的问题）：

int iMoney_back_cents = (iMoney_back * 100) + .5;

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