年度涨价循环？卡在嵌套循环上

Question

我正在尝试编写一个程序，该程序使用扫描仪键盘输入这些值并使用循环使它们增加（2006-2010 年，价格上涨，90-100 等）

输出应该是这样的：

Enter the current Mars bar price in cents: 90
Enter the expected yearly price increase: 15
Enter the start year: 2006
Enter the end year: 2010
Price in 2006: 90 cents
Price in 2007: 105 cents
Price in 2008: 120 cents
Price in 2009: 135 cents
Price in 2010: 150 cents

这是我目前的代码：

import java.util.Scanner;
public class Problem_2 {

    public static void main(String[] args) {

        // TODO, add your application code
        Scanner keyboard = new Scanner(System.in);
        System.out.print("Enter the current Mars bar price in cents: ");
        int m = keyboard.nextInt();
        System.out.print("Enter the expected yearly price increase: ");
        int p = keyboard.nextInt();
        int a = m+p;

        int count = 1;
        System.out.print("Enter the start year: ");
        int start = keyboard.nextInt();
        System.out.print("Enter the end year: ");
        int end = keyboard.nextInt();

        for (int i = start; i <= end; i++){
            System.out.print("Price in " +i+ ": " +m);start++;
    }
}

它允许我将年份增加到设定的数量（再次像 2006-2010 年一样），但我坚持让价格随着设定的数量增加，我认为这将是一个嵌套循环但是不知道怎么写。

（我假设是一个嵌套循环，因为这是我们现在正在学习的内容。）

如果有人对我可以写什么有任何建议，我将不胜感激！

Answer 1

我认为您只需要在循环的每次迭代中将 expected yearly price increase 与 current price 相加即可，如下所示：

import java.util.Scanner; 
public class Problem_2 {

  public static void main(String[] args) {

    // TODO, add your application code
    Scanner keyboard = new Scanner(System.in);
    System.out.print("Enter the current Mars bar price in cents: ");
    int m = keyboard.nextInt();
    System.out.print("Enter the expected yearly price increase: ");
    int p = keyboard.nextInt();

    int count = 1;
    System.out.print("Enter the start year: ");
    int start = keyboard.nextInt();
    System.out.print("Enter the end year: ");
    int end = keyboard.nextInt();

    for (int i = start; i <= end; i++){
        System.out.print("Price in " +i+ ": " +m);
        m += p; // sum expected price for each iteration

  }
}