Java練習題 - 求s = a+aa+aaa+aaaa+aa...a值

題目：基本 求s = a+aa+aaa+aaaa+aa...a值
EX:2+22+222+2222+22222(使用迴圈 總共五個數相加)

import java.util.*;

public class Java_Num
{

 public static void main(String[] args)
 {
  Scanner scan = new Scanner(System.in);
  System.out.print("請輸入一個數：");
  int a = scan.nextInt();
  int s = 0;
  for (int i = 1; i <= 5; i++)
  {
   int temp = 0;
   for (int j = 0; j < i; j++)
    temp = temp * 10 + a;
   s += temp;
  }
  System.out.println("s：" + s);
 }

}

    進階 另外加一個變數 可控制相加數目 EX: a=2 , b=6
                        s=2+22+222+2222+22222+222222

import java.util.*;

public class Java_Num
{

 public static void main(String[] args)
 {
  Scanner scan = new Scanner(System.in);
  System.out.print("請輸入一個數：");
  int a = scan.nextInt();
  System.out.print("請輸入相加數目：");
  int b = scan.nextInt();
  int s = 0;
  for (int i = 1; i <= b; i++)
  {
   int temp = 0;
   for (int j = 0; j < i; j++)
    temp = temp * 10 + a;
   s += temp;
  }
  System.out.println("s：" + s);
 }

}