需要帮助制作我的第一个Java代码。 （硬币变更）

我的作业要求我读取用户输入的数字并输出数字所产生的硬币。 例如，如果用户输入“37”，则程序应响应（1 1/4，1和1便士）。 我最有可能的代码根本没有任何意义，我不知道我需要做些什么来解决它。

import java.util.Scanner; public class Change { public static void main(String[] args) { Scanner sc = new Scanner (System.in); Int n = sc.nextInt("Enter a positive integer" ); int number1, number2; // Division operands int quotient; // Result of division if (QtrCnt > 0) if (QtrCnt > 1) System.out.println(QtrCnt + " quarters"); else System.out.println(QtrCnt + " quarter"); } if (DimeCnt > 0) { if (DimeCnt > 1) System.out.println(DimeCnt + " dimes"); else System.out.println(DimeCnt + " dime"); } if (NicklCnt > 0) { if (NicklCnt > 1) System.out.println(NicklCnt + " nickles"); else System.out.println(NicklCnt + " nickle"); } if (PennyCnt > 0); { if (PennyCnt > 1); System.out.println(PennyCnt + " pennies"); System.out.println(PennyCnt + " penny"); } int q = 25; int d = 10; int n = 5; int p = 1; if (a =25); { int numQuarters = money/ 25; } money -= numQuarters * 25; QtrCnt = (num1 - num1 % 25) / 25; num1 = num1 - QtrCnt * 25; String(money >=10); { int numDimes = money/ 10; } money -= numDimes * 10; DimeCnt = (num1 - num1 % 10) / 10; num1 = num1 - DimeCnt * 10; String (money >=5); { int numNickles = money/ 5; } money -= numNickles * 5; NicklCnt = (num1 - num1 % 5) / 5; num1 = num1 - NicklCnt * 5; String (money >=1); { int numPennies = money/ 1; } money -= numPennies * 1; PennyCnt = (num1 - num1 % 1) / 1; num1 = num1 - PennyCnt * 1; } } } 

Answer 1

划分和下限（你应该重新开始）

 public static int getQuarters(int cents) { return Math.floor(cents / 25.0); } 

这是你如何做到的：

 public static void main(String[] args) { int cents = 46; // 46 just for an example int left = cents; // a variable that represents how many cents are left int quarters = getQuarters(cents); // how many quarters fit into 46 (1) int left -= quarters * 25; // now we have 21 cents left we need to work out int dimes = getDimes(left); // you can implement getDimes yourself (look at getQuarters). This will now return 2, since 2 dimes can go into 21 cents. left -= dimes * 10; // we now have only 1 cent to account for int nickels = getNickels(left) // Returns 0 (no nickels can fit into 1 cent) left -= nickels * 5; // we still have 1 cent left int pennies = left; // how many pennies are left over (always < 5) System.out.println(cents + " cents = " + quarters + " Quarters, " + dimes + " Dimes, " + nickels + " Nickels, and " + pennies + " Pennies."); // print the output } 

请记住在Java类中包含getQuarters方法

例子：

• getQuarters(25) -> 1
• getQuarters(24) -> 0
• getQuarters(49) -> 1
• getQuarters(51) -> 2