HDU 1002 大數運算 java 的強大功能

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 123322    Accepted Submission(s): 23706


Problem DescriptionI have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B. 

InputThe 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. 

OutputFor 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

21 2112233445566778899 998877665544332211

 

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

java的強大運算，還記得當年學c++大數加法的時候整整看了一天這道題的代碼，要轉化成數組進行處理，一年後的今天，學會了大數加法處理這道題。

post code

import java.util.*;import java.math.*;import java.io.*;public class  Main    //注意類名一定要用首字母大寫的Main{    public static void main( String[] args  )    {       Scanner cin=new Scanner(new BufferedInputStream (System.in)  );       BigInteger a,b,c;                    //定義了三個大整數       int ji,num=0;       ji=cin.nextInt();                    //讀入判斷條件       while( ji>=1 )       {           ji--;       num++;        a=cin.nextBigInteger();             //讀入大整數       b=cin.nextBigInteger();       c=a.add(b);       System.out.println("Case "+num+":");          //輸出結果       System.out.println(a + " + " + b + " = "+c);           if(ji!=0)System.out.println("");       }           }}