Question

import java.io.*;

import java.util。*;

公共课CH04 {

// Constants Declaration Section
//******************************

public static void main(String[] args) throws FileNotFoundException {
    // TODO Auto-generated method stub

    // Variables Declaration Section
    //******************************
    int  quarters;
    int  dimes;
    int  nickles;
    int  numberOfQuarters;
    int  numberOfDimes;
    int remainingAmount; 
    int numberOfNickles;
    int numberOfPennies;
    int maxSize;

    // Variables Initialization Section
    //*********************************
    Scanner data = new Scanner(new File("C:\\testing3.txt"));     //Reads from text file 
    ArrayList<Double> datadata = new ArrayList<Double>();    //Creates Array
    quarters = 25;
    dimes = 10;
    nickles = 5;
    maxSize = 10000;



    // Code Section
    //*************
    while(data.hasNextDouble())    //Retrieves data from text file and places it in array
    {
        datadata.add(data.nextDouble());  //Retrieves data from text file and places it in array
    }

    for (int i = 0; i < datadata.size(); i++)  //Loops through the data
    {
        double value = datadata.get(i);   //returns number found in file as a double 'value' then runs it through the calculator
        System.out.println(value);

        while(value > maxSize) //If value is greater than 10,000, it will be skipped and not read through the calculator
        {

        }

        remainingAmount = (int)Math.ceil((value) * 100);      //multiplies value by 100

        numberOfQuarters = remainingAmount / quarters;   //divides value by 25
        remainingAmount = remainingAmount % quarters;    //returns the remaining value 

        numberOfDimes = remainingAmount / dimes;    //divides the remaining value by 10
        remainingAmount = remainingAmount % dimes;  //returns the remaining value of that

        numberOfNickles = remainingAmount / nickles;  //divides the remaining value from dimes by 5
        remainingAmount = remainingAmount % nickles;  //returns remaining value

        numberOfPennies = remainingAmount;   //divides remaining value by 1


        System.out.print("Your amount of $" + value + " consists of: \n" + numberOfQuarters +            //prints out the amount of quarters, dimes, nickels, pennies
                " quarteres, " + numberOfDimes + " dimes, " + numberOfNickles + " nickles, and " +       //that make up each value found inside the file.
                  numberOfPennies + " pennies");

        System.out.print("\n\n");

    }



    //Output Section
    //**************




    //Resource Cleaning
    //*****************
    data.close();

}

}

**Output:

10001.0 值太大，无法处理10001.0 您的$ 10001.0金额包括： 40004夸脱，0角钱，0镍，0便士

9755.35 您的9755.35美元金额包括： 39021个四分之一，1个硬币，0个镍币和0个便士

875.4 您的$ 875.4金额包括： 3501夸脱，1角钱，1个镍币和0便士

78.99 您的78.99美元包括： 315夸脱，2角钱，0个镍币和4个便士

5.0 您的5.0美元金额包括： 20个四分之一，0角钱，0个镍币和0个便士

0.7 您的0.7美元包括： 2个四分之一，2个角钱，0个镍币和0个便士

使用的值：[10001.0,9755.35,875.4,78.99,5.0,0.7] **

出于某种原因，我输出的斜体部分收到1便士。这让我感到困惑，因为输出的其余部分是正确的。

我的文本值中的值是：

10,001.00 -
9,755.35 -
875.40 -
78.99 -
5.00 -
0.70

Answer 1

我不会写代码本身。我猜这是手工制作的一个例子。调试此类问题的一般经验法则是：查看最后或最开始的'。除法是正确的，因此它必须在一开始就发生，这是你的值被转换为int的地方。如果你检查产品值* 100的结果，那么你会看到7898.999999999999，而不是你想象的7899。您应该使用Math.ceil来获取正确的值，如下所示：

 int result = (int)Math.ceil(value * 100);

Answer 2

您正在处理浮点算术错误。如果你打印出value * 100，你会发现它出现在7898.9999 ...... 浮点运算误差仅在发生时非常小。

您案例的简单解决方案是添加.1然后转换为int。

更改此行：

remainingAmount = (int)(value * 100);

为：

remainingAmount = (int)(value * 100 + .1);

如果测试文件包含的值超过2个小数位，则打破此问题的唯一方法就是这样。但既然你正在处理金钱问题，你就不必担心。

即使数字太大，它仍在处理中（忽略，我修复了它）

2 个答案: