我在计算手机套餐的节省金额时遇到了书籍问题。我想指出,我已经完成了第一部分的代码。这是我遇到麻烦的第二部分。好的,这里是书中的问题(用于上下文):
Part 1
A mobile phone service provider has three different subscription packages for its customers:
Package A: For $39.99 per month 450 minutes are provided. Additional minutes are $0.45 per minute.
Package B: For $59.99 per month 900 minutes are provided. Additional minutes are $0.40 per minute.
Package C: For 69.99 per month unlimited minutes provided.
Write a program that calculates a customer’s monthly bill. It should ask which package the customers has purchased ans how many minutes were used. It should then display the total amount due.
Part 2
Modify the program in part 1 so that it also displays how much money Package A customers would save if they purchased packages B or C, and how much money Package B customers would save if they purchased Package C. If there would be no saving, no message should be printed.
第二部分,我如何计算计划的节省?我似乎无法正确实现它们。我最初的想法是从每月总账单中减去总额外分钟数的超额成本,但我无法像我想要的那样工作。这是我的代码:
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
double minUsed, minLeft, extraMinCost, monthTotal, planSaveB, planSaveC;
char choice;
const double planCostA = 39.99, planCostB = 59.99, planCostC = 69.99, monthlyMinA = 450, monthlyMinB = 900;;
cout<<"Enter your monthly package plan: Ex. A, B or C"<<endl<<endl;
cin>>choice;
cout<<endl;
cout<<"Enter the amount of minutes you used: "<<endl;
cin>>minUsed;
cout<<endl;
if (choice == 'a' || choice == 'A')
{
minLeft=monthlyMinA-minUsed;
if (minLeft < 0)
{
extraMinCost=minLeft*(-0.45);
monthTotal=planCostA+extraMinCost;
planSaveB = (planCostB-extraMinCost)-(-planCostB);
planSaveC = planCostC - (planCostC-extraMinCost);
cout << "Your total bill amount is: " << setprecision(2) << fixed << "$" << monthTotal << endl<<endl;
cout << "You could save " << setprecision(2) << fixed << "$" << planSaveB << " if you switch to plan B or ";
cout << "save " << setprecision(2) << fixed << "$" << planSaveC << " If you switch to plan C." << endl << endl;;
}
else if (minLeft >= 0)
{
monthTotal = planCostA;
cout << "Your total bill amount is: " << setprecision(2) << fixed << "$" << monthTotal << endl;
}
}
/*else if (choice == 'b'|| choice == 'B')
{
minLeft=monthlyMinB-minUsed;
if (minLeft < 0)
{
extraMinCost=minLeft*(-0.40);
monthTotal=planCostB+extraMinCost;
}
else
monthTotal=planCostB;
cout<<"Your total bill amount is: "<<setprecision(2)<<fixed<<"$"<<monthTotal<<endl;
}
else if (choice == 'c' || choice == 'C')
{
monthTotal=planCostC;
cout<<"Your total bill amount is: "<<setprecision(2)<<fixed<<"$"<<monthTotal<<endl;
cout<<"Current plan has unlimited minutes!"<<endl;
}*/
cout<<choice;
return 0;
}
相关部分是我代码的选择部分。当我能把那个包的一部分弄好时,我会修改其余部分。
最佳答案
我想你可能想写一个像
这样的函数float getPlanACost( int minutes ) { ... }
返回如果客户有计划 A 并使用“minutes”分钟数,他们将被收取的金额。然后为 B 计划和 C 计划编写类似的函数。使用这些函数打印您的每月账单,并根据要求计算可用的储蓄。
软件开发的很大一部分是将一个大问题分解成多个小问题。
关于c++ - 我应该如何处理编写一段确定订阅包节省的代码的逻辑,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/28685501/