当我插入350 The时，输出应该是102.80。

Question

请参阅这张图以了解清楚的电块使用情况。

当我插入350 too的数字时，输出应该是102.80，但是我的输出出现RM 180.600006，小数位太长，我已经被放入%2f，但仍然显示为长小数点。在这里输出。用电量将分成几个区块。

#include <stdio.h>
#include <stdlib.h>
/* run this program using the console pauser or add your own getch, system("pause") or input loop */

#include <stdio.h>

//For the first 200 kWh (1-200 kWh) per month = RM21.80
//For the next 100 kWh (201-300 kWh) per month = RM33.40
//For the next 300 kWh (301-600 kWh) per month = RM51.60
//For the next 300 kWh (601-900 kWh) per month = RM54.60
//For the next kWh (901 kWh onwards) per month = RM57.10

int main()
{
   float Electric_Consumption, Bill;

   printf("Please insert the Current Electric Consumption: ");

   scanf("%f", &Electric_Consumption);

   if (Electric_Consumption >=1 && Electric_Consumption <= 200)
   {

    Bill = (Electric_Consumption*(21.80/100));
    printf("Your bill need to pay is RM %f", Bill);


   }

   else if(Electric_Consumption >=201 && Electric_Consumption <= 300)
   {

    Bill = (Electric_Consumption*(33.40/100));
    printf("Your bill need to pay is RM %f", Bill);

   }

   else if(Electric_Consumption >=301 && Electric_Consumption <= 600)
   {

   Bill = (Electric_Consumption*(51.60/100));
    printf("Your bill need to pay is RM %f", Bill);

   }

   else if(Electric_Consumption >=601 && Electric_Consumption <= 900)
   {

    Bill = (Electric_Consumption*(54.60/100));
    printf("Your bill need to pay is RM %f", Bill);

   }

   else if(Electric_Consumption >=900)
   {

    Bill = (Electric_Consumption*(57.10/100));
    printf("Your bill need to pay is RM %f", Bill);

   }

   else
   { printf("You inputed wrong electric consumption!");
   }


    return 0;
    }

Answer 1

您的答案是错误的，因为每个累积kW的账单价格都会发生变化。换句话说，(Electric_Consumption*(33.40/100))计算的是每千瓦的33.40/100，而不是301到600千瓦之间的千瓦。您需要更改公式以添加上一个千瓦成本，然后从Electric_Consumption中减去这些千瓦。

小数太多的原因是您需要对格式字符串使用"%.2f"。

Answer 2

这

//For the first 200 kWh (1-200 kWh) per month = RM21.80
//For the next 100 kWh (201-300 kWh) per month = RM33.40
//For the next 300 kWh (301-600 kWh) per month = RM51.60
//For the next 300 kWh (601-900 kWh) per month = RM54.60
//For the next kWh (901 kWh onwards) per month = RM57.10

意思是，如果有人使用945 kWh，那么他们需要支付

• 21.80前200 kWh为21.8 * 200
• 接下来的100 kWh是33.40 * 100
• 51.60下一个300 kWh，即51.60 * 300
• 54.60下一个300 kWh，即54.60 * 300
• 其余的是57.1，也就是57.10 * 45

您的代码对所有kWh都使用了固定的价格，并且只使用金额来确定此固定价格。这是错的。

这可以通过递归解决：

 double calculate_price(double kWh) {
     if (kWh > 900) return (kWh-900)*57.10 + calculate_price(kWh-900);
     else if (kWh > 600) return (kWh-600)*54.60 + calculate_price(kWh-600);
     else if .... same for the rest....
     else // kWh <= 200 
        return 21.80*kWh;
}

Answer 3

一种通过预先计算完整切片价格的解决方案。

    #include <stdio.h>

//For the first 200 kWh (1-200 kWh) per month = RM 21.80
//For the next 100 kWh (201-300 kWh) per month = RM 33.40
//For the next 300 kWh (301-600 kWh) per month = RM 51.60
//For the next 300 kWh (601-900 kWh) per month = RM 54.60
//For the next kWh (901 kWh onwards) per month = RM 57.10

int main(void)
{
    float Electric_Consumption;
    printf("Please insert the Current Electric Consumption: ");
    scanf("%f", &Electric_Consumption);

    float price_full_slice[] = {21.8*2,
                                21.8*2 + 33.4,
                                21.8*2 + 33.4 + 51.6*3,
                                21.8*2 + 33.4 + 51.6*3, 54.6*3};

    float price = 0;
    if(Electric_Consumption>900)
        price = price_full_slice[3] + (Electric_Consumption-900) * 57.1/100;

    if(Electric_Consumption>600 && Electric_Consumption<=900)
        price = price_full_slice[2] + (Electric_Consumption-600) * 54.6/100;

    if(Electric_Consumption>300 && Electric_Consumption<=600)
        price = price_full_slice[1] + (Electric_Consumption-300) * 51.6/100;

    if(Electric_Consumption>200 && Electric_Consumption<=300)
        price = price_full_slice[0] + (Electric_Consumption-200) * 33.4/100;

    if(Electric_Consumption<=200)
        price = Electric_Consumption * 21.8/100;

    printf("%.2f", price);

    return 0;
}