,它提示用户输入便士,镍,硬币和宿舍的数量,然后显示其总金额。该应用程序应该包含一个getDollarAmount()方法,该方法有4个int参数,对应于便士,镍,硬币和宿舍的数量,并返回一个对应于硬币美元值的字符串。如何创建AddCoins应用程序?
应用程序的输出应该类似于:
输入你个硬币:
宿舍:3 角钱:2个 尼克尔斯:1个 便士:8
总计:$ 1.08
这是我的尝试:
package ch7e5;
import java.util.Scanner;
public class Ch7E5 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int n1, p, d, n, t;
double Q1, D1, N1, P1;
System.out.println("Enter your total coins:");
System.out.print("Quarters:");
n1 = input.nextInt();
System.out.print("Dimes:");
d = input.nextInt();
System.out.print("Nickles:");
n = input.nextInt();
System.out.print("Pennies:");
p = input.nextInt();
double Q1 = (pennies * 0.01);
private static double calctotal(double Q1, double D1, double P1, double N1) {
double dbltotal;
dbltotal = (Q1 + D1 + P1 + N1);
return dbltotal;
}
}
这是我与你的意见帮助第2次尝试:
package chapter7ex5;
import java.util.Scanner;
public class Chapter7ex5 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("Enter your total coins:");
System.out.print("Quarters:");
int Q1 = input.nextInt();
System.out.print("Dimes:");
int D1 = input.nextInt();
System.out.print("Nickles:");
int N1 = input.nextInt();
System.out.print("Pennies:");
int P1 = input.nextInt();
}
public static double calctotal(int Q1, int D1, int N1, int P1) {
double total;
total=((0.25 * Q1) + (0.1 * D1) + (0.05 * N1) + (0.01 * P1));
return (total);
}}
我觉得我的努力已经结束:
package chapter7ex5;
import java.text.DecimalFormat;
import java.util.Scanner;
public class Chapter7ex5 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("Enter your total coins:");
System.out.print("Quarters:");
int Q1 = input.nextInt();
System.out.print("Dimes:");
int D1 = input.nextInt();
System.out.print("Nickles:");
int N1 = input.nextInt();
System.out.print("Pennies:");
int P1 = input.nextInt();
DecimalFormat fmt = new DecimalFormat("$#,###.##");
System.out.println("Total:"+fmt.format(calctotal(Q1, D1, N1,
P1)));
}
public static double calctotal(int Q1, int D1, int N1, int P1) {
double total;
total=((0.25 * Q1) + (0.1 * D1) + (0.05 * N1) + (0.01 * P1));
return (total);
}}
将您的代码添加到问题中,而不是将其作为评论发布。 – 2014-11-07 01:28:28