,它提示用戶輸入便士,鎳,硬幣和宿舍的數量,然後顯示其總金額。該應用程序應該包含一個getDollarAmount()方法,該方法有4個int參數,對應於便士,鎳,硬幣和宿舍的數量,並返回一個對應於硬幣美元值的字符串。如何創建AddCoins應用程序?
應用程序的輸出應該類似於:
輸入你個硬幣:
宿舍:3 角錢:2個 尼克爾斯:1個 便士:8
總計:$ 1.08
這是我的嘗試:
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;
}
}
這是我與你的意見幫助第2次嘗試:
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);
}}
我覺得我的努力已經結束:
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);
}}
將您的代碼添加到問題中,而不是將其作爲評論發佈。 – 2014-11-07 01:28:28