cee ane 代碼 () each space ces put follow

Problem Description I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B. Input The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000. Output For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases. Sample Input 2 1 2 112233445566778899 998877665544332211 Sample Output Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110 AC代碼：

 1
 
 import java.math.BigDecimal;
 2 import java.util.Scanner;
 3 
 4 public class Main {
 5 
 6     public static void main(String[] args) {
 7         Scanner reader = new Scanner(System.in);
 8         int nTotal = reader.nextInt();   //nTotal為哪種情況
 9         for(int nCount = 0; nCount < nTotal; nCount++){
 
10             BigDecimal num1 = reader.nextBigDecimal();
11             BigDecimal num2 = reader.nextBigDecimal();
12             StringBuffer sbTemp = new StringBuffer("Case ");
13             sbTemp.append(nCount+1);
14             sbTemp.append(":\r\n");  //JAVA中的\r\n才是對應的回車換行
15             sbTemp.append(num1.toString());
 
16             sbTemp.append(" + ");
17             sbTemp.append(num2.toString());
18             sbTemp.append(" = ");
19             sbTemp.append(num1.add(num2).toString());
20             System.out.println(sbTemp.toString());
21             if(nCount != nTotal-1)
22                 System.out.println();
23         }
24     }
25 }

A + B Problem II