3.7（金融應(yīng)用：整錢兌零）修改程序清單2-10，使之只顯示非零的幣值單位，用單詞的單數(shù)形式顯示一個單位，復(fù)數(shù)形式顯示多于一個的單位的值

• 題目
• • 題目概述
  • 程序清單2-10（非本題代碼）
  • 破題/思路：這道題我思路可能理解比較難，可以自行搜索看其他博客如何處理
• 代碼

題目

題目概述

3.7（金融應(yīng)用：整錢兌零）修改程序清單2-10，使之只顯示非零的幣值單位，用單詞的單數(shù)形式顯示一個單位，復(fù)數(shù)形式顯示多于一個的單位的值
1美元和1美分：1 dollar and 1 penny
2美元和3美分：2 dollars and 3 pennies

程序清單2-10（非本題代碼）

import java.util.Scanner;public class QingDan {public static void main(String[] args) {// Create a ScannerScanner input = new Scanner(System.in);// Receive the amountSystem.out.println("Enter an amount in int, for example 11.56");double amount = input.nextDouble();int remainingAmount = (int)(amount * 100);// Find the number of one dollarsint numberOfOneDollars = remainingAmount / 100;remainingAmount = remainingAmount % 100;// Find the number of quarters in the remaining amountint numberOfQuarters = remainingAmount / 25;remainingAmount = remainingAmount % 25;// Find the number of dimes in the remaining amountint numberOfDimes = remainingAmount / 10;remainingAmount = remainingAmount % 10;// Find the number of nickels in the remaining amountint numberOfNickels = remainingAmount / 5;remainingAmount = remainingAmount % 5;// Find the number of pennies in the remaining amountint numberOfPennies = remainingAmount;// Display resultsSystem.out.println("Your amount" + remainingAmount + " consists of");System.out.println(" " + numberOfOneDollars + " dollars");System.out.println(" " + numberOfQuarters + " quarters");System.out.println(" " + numberOfDimes+ " dimes");System.out.println(" " + numberOfNickels + " nickels");System.out.println(" " + numberOfPennies+ " pennies");} }

破題/思路：這道題我思路可能理解比較難，可以自行搜索看其他博客如何處理

只需要理解它（清單2-10）的大體框架即可，這種改寫題因?yàn)榻o出了很多代碼相對也比較簡單
理解難點(diǎn)在中間幣值的轉(zhuǎn)換，自己帶幾個數(shù)進(jìn)去就能逐漸掌握每行代碼想表達(dá)的意思了

對于3.7，我的思路是先獲取到數(shù)據(jù)并進(jìn)行處理轉(zhuǎn)換后，在輸出部分每種幣值分三種情況（為0、為1、大于1）使用print()函數(shù)輸出內(nèi)容即可

如何在輸出的兩個幣值之間加and、留空格？
因?yàn)槊總€單位輸出都存在不確定性，不能直接print()函數(shù)中簡單地添加and并留出空格
筆者考慮到可以用boolean打標(biāo)進(jìn)行判斷
假設(shè)dollar前先標(biāo)記為true，那么后續(xù)所有輸出的地方都將這個標(biāo)記賦值為false
那么一旦讀到這個標(biāo)記是false，說明前面已經(jīng)有輸出，即前面留出空格并加上and

代碼

import java.util.Scanner;public class Test3_7 {public static void main(String[] args) {// 接收數(shù)據(jù)Scanner input = new Scanner(System.in);System.out.println("Enter an amount in int, for example 11.56");double amount = input.nextDouble();// 轉(zhuǎn)換單位+打標(biāo)int remainingAmount = (int)(amount * 100);boolean bool = true;// Find the number of one dollarsint numberOfOneDollars = remainingAmount / 100;remainingAmount = remainingAmount % 100;if(numberOfOneDollars > 1) {System.out.print(numberOfOneDollars + " dollars");bool = false;}else if(numberOfOneDollars == 1) {System.out.print(numberOfOneDollars + " dollar");bool = false;}// Find the number of quarters in the remaining amountint numberOfQuarters = remainingAmount / 25;remainingAmount = remainingAmount % 25;if(numberOfQuarters > 1 && bool == true) {System.out.print(numberOfQuarters + " quarters");bool = false;}else if(numberOfOneDollars == 1 && bool == true) {System.out.print(numberOfQuarters + " quarter");bool = false;}else if(numberOfQuarters > 1 && bool == false)System.out.println(" and " + numberOfQuarters + " quarters");else if(numberOfQuarters == 1 && bool == false)System.out.println(" and " + numberOfQuarters + " quarter");// Find the number of dimes in the remaining amountint numberOfDimes = remainingAmount / 10;remainingAmount = remainingAmount % 10;if(numberOfDimes > 1 && bool == true) {System.out.print(numberOfDimes + " dimes");bool = false;}else if(numberOfDimes == 1 && bool == true) {System.out.print(numberOfDimes + " dime");bool = false;}else if(numberOfDimes > 1 && bool == false)System.out.println(" and " + numberOfDimes + " dimes");else if(numberOfDimes == 1 && bool == false)System.out.println(" and " + numberOfDimes + " dimes");// Find the number of nickels in the remaining amountint numberOfNickels = remainingAmount / 5;remainingAmount = remainingAmount % 5;if(numberOfNickels > 1 && bool == true) {System.out.print(numberOfNickels + " nickels");bool = false;}else if(numberOfNickels == 1 && bool == true) {System.out.print(numberOfNickels + " nickel");bool = false;}else if(numberOfNickels > 1 && bool == false)System.out.println(" and " + numberOfNickels + " nickels");else if(numberOfNickels == 1 && bool == false)System.out.println(" and " + numberOfNickels + " nickel");// Find the number of pennies in the remaining amountint numberOfPennies = remainingAmount;if(numberOfPennies > 1 && bool == true)System.out.print(numberOfPennies + " pennies");else if(numberOfPennies == 1 && bool == true)System.out.print(numberOfPennies + " penny");else if(numberOfPennies > 1 && bool == false)System.out.println(" and " + numberOfPennies + " pennies");else if(numberOfPennies == 1 && bool == false)System.out.println(" and " + numberOfPennies + " penny");} }

總結(jié)

以上是生活随笔為你收集整理的Java黑皮书课后题第3章：3.7（金融应用：整钱兑零）修改程序清单2-10，使之只显示非零的币值单位，用单词的单数形式显示一个单位，复数形式显示多于一个的单位的值的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

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