基本上我正在尝试一个关于如何计算费用的问题。但是，我的代码非常长。我最终使用了大量的'if'和'else'代码。我想知道是否有更好的方法来处理这个问题。
我最初尝试使用循环来保持费用，但是当我的时间跨越关键时刻时，我遇到了问题。我加上TimeIn的时间以及我的费用，但是当它超过7am标记时，我需要将其重置为早上7点，以便我的费用能够正常收费。那时候，当我的代码变得很长时，我放弃了。
EG。循环增加从0630到0730，费用适当增加，0700的前30分钟将被跳过，费用被错误收费。
我试过的问题：
先生。吴每天乘出租车去上班多年。然而，近几年出租车费用一直在快速增长。因此，他正在考虑开车上班。 开车费用之一是停车费。吴先生工作场所停车场的停车费率如下表所示。
Weekday Saturday Sunday
4am ~ 7am $2.00 per hour $2.50 per hour $5
7am ~ 6pm $1.20 per 30 minute $1.50 per 30 minutes per
6pm ~ midnight $5.00 per entry $7.00 per entry entry
Special note:
1. The car park opens at 4am and closes at midnight. All vehicles must leave
by midnight.
2. There is a grace period of 10 minutes on any day (i.e., it is completely
free to park for 10 minutes or less regardless of day and time.)
3. There is a 10% surcharge for parking more than 10 hours on a weekday and
20% for Saturday. There is no surcharge for Sunday.
4. There is an additional $3.00 fee for exiting after 10pm on any day.
(Surcharge is not applicable on this fee.)
您的程序应该读取一个整数，这是1到7之间的整数 表示星期几（1表示星期一，7表示星期日）。它也应该 读取两个数字，表示24小时格式的进入和超时。它 然后应计算并显示停车费（小数点后两位）。 您可以假设输入有效（即，日期在指定范围内， 进入和超时都是在凌晨4点到午夜之间的24小时格式和超时 不早于时间。）
我的代码非常长：
#include <stdio.h>
#include <math.h>
double computeFee(int, int, int);
int main(void){
int day, timeIn, timeOut;
scanf("%d %d %d", &day, &timeIn, &timeOut);
printf("Enter day: %d\n", day);
printf("Enter time-in: %d\n", timeIn);
printf("Enter time-out: %d\n", timeOut);
printf("Parking fee is $%.2lf\n", computeFee(day, timeIn, timeOut));
return 0;
}
double computeFee(int day, int timeIn, int timeOut){
double fee = 0;
double TimeIn, TimeOut;
TimeIn = 60*(floor(timeIn/100)) + (timeIn%100);
TimeOut = (60*(floor(timeOut/100)) + (timeOut%100));
if((day>=1)&&(day<=5)){
if(TimeIn<420){
if(TimeOut<420){
fee += 2*ceil((TimeOut - TimeIn)/60);
}
else{
fee += 6;
if(TimeOut>=1080){
fee += 31.4;
}
else{
fee += 1.2*ceil((TimeOut - 420)/30);
}
}
}
if(TimeIn>=420){
if(TimeIn>=1080){
fee = 5;
}
else{
if(TimeOut>=1080){
fee += (1.2*ceil((1080 - TimeIn)/30) + 5);
}
else{
fee += (1.2*ceil((TimeOut - TimeIn)/30));
}
}
}
if((TimeOut-TimeIn)>=600){
fee *= 1.1;
}
}
if(day == 6){
if(TimeIn<420){
if(TimeOut<420){
fee += (2.5*ceil((TimeOut - TimeIn)/60));
}
else{
fee += 7.5;
if(TimeOut>=1080){
fee += 40;
}
else{
fee += (1.5*ceil((TimeOut - 420)/30));
}
}
}
if(TimeIn>=420){
if(TimeIn>=1080){
fee = 7;
}
else{
if(TimeOut>=1080){
fee += (1.5*ceil((1080 - TimeIn)/30) + 7);
}
else{
fee += (1.5*ceil((TimeOut - TimeIn)/30));
}
}
}
if((TimeOut-TimeIn)>=600){
fee *= 1.2;
}
}
if(day == 7){
fee = 5;
}
if((timeOut/100)>=22){
fee += 3;
}
if((timeOut - timeIn)<=10){
fee = 0;
}
return fee;
}
如何计算费用的例子：
Example 1: Tuesday, 4:29am to 7:50am.
• 4:29am to 7am is charged as 3 1-hour slots: $2.00 * 3 = $6.00
• 7am to 7:50am is charged as 2 30-minute slots: $1.20 * 2 = $2.40
• Total fee = $6.00 + $2.40 = $8.40
Example 2: Saturday, 7:01am to 7:49pm.
• 7:01am to 6pm is charged as 22 30-minute slots: $1.50 * 22 = $33.00
• 6pm to 7:49pm is charged as one entry: $7.00
• 20% Surcharge for parking more than 10 hours: ($33.00 + $7.00) * 20% =
$8.00
• Total fee = $33.00 + $7.00 + $8.00 = $48.00
Example 3: Sunday, 3pm to 10:01pm.
• 3pm to 10:01pm is charged as one entry: $5.00
• Additional fee for exiting after 10pm: $3.00
• Total fee = $5.00 + $3.00 = $8.00
Example 4: Thursday, 11:49pm to 11:59pm.
• Grace period
• Total fee = $0.00
Example 5: Monday, 12pm to 10:01pm.
• 12pm to 6pm is charged as 12 30-minute slots: $1.20 * 12 = $14.40
• 6pm to 10:01pm is charged as one entry: $5.00
• 10% Surcharge for parking more than 10 hours: ($14.40 + $5.00) * 10% =
$1.94
• Additional fee for exiting after 10pm: $3.00
• Total fee = $14.40 + $5.00 + $1.94 + $3.00 = $24.34
感谢您阅读我的长篇问题。并提前感谢您的帮助。
注意：我还没有学过数组和其他任何东西。到目前为止只学习循环和选择语句（阅读K＆amp; R编程教程，直到第17章，使用Online GeekforGeek作为编译器）。但是，我仍然会非常欣赏使用其他方法的解决方案。