~~~ # A + B Problem II Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 273307????Accepted Submission(s): 52784 ~~~ 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 ~~~ ~~~ import java.math.BigDecimal; import java.util.Scanner; // 代碼提交時改為Main且去掉當前文件的包名 public class Main { public static void main(String[] args) { /** * 從題目中可以看出來，數據比較大，只是用普通的整形或者別的恐怕是無能為力，不過BigDecimal對于數據較大的數處理十分的方便 * 而且封裝好了函數只需要調用就可以了 */ Scanner scanner = new Scanner(System.in); int num = scanner.nextInt(); // 這里的二維數組也需要有些講究，一般一行中查找不同的列的速度要比一列中查找不同的行的數據要快 BigDecimal[][] bigNum = new BigDecimal[num][2]; for(int i = 0; i < num; i++) { bigNum[i][0] = scanner.nextBigDecimal(); bigNum[i][1] = scanner.nextBigDecimal(); } // 輸出的時候注意空格，以及的空行 for(int j = 0; j < num; j++) { if(j != 0) System.out.println(); System.out.println("Case "+(j+1)+":"); System.out.println(bigNum[j][0] +" + "+ bigNum[j][1]+" = "+ bigNum[j][0].add(bigNum[j][1])); } } } ~~~