A+B Problem II （java+高精度问题）

看了一天的java，终于可以用它来写一个带有输入的程序了，

因为刚开始学的java都是很简单的输出，并没有涉及到输入，一直都很疑惑java该怎么输入

我用的编译环境是NetBeans，今天才知道原来直接在运行框内输入啊，好神奇

（一直用的是C/C++的编译器，习惯了弹出运行框。。。）

……

不说废话了。。下面是部分读入的代码

//用Scanner类定义对象进行控制台读入,Scanner类在java.util.*包中 Scanner cin=new Scanner(System.in);// 读入 while(cin.hasNext()) //等同于!=EOF { int n; BigInteger m; n=cin.nextInt(); //读入一个int; m=cin.nextBigInteger();//读入一个BigInteger; System.out.print(m.toString()); }

至于这道对于java大数类来说的简单题，代码如下：

import java.util.Scanner; import java.math.*; public class Main { public static void main(String[] args) { Scanner cin = new Scanner(System.in); int k=1,n; n=cin.nextInt(); while(n!=0) { BigInteger num1,num2; num1=cin.nextBigInteger(); num2=cin.nextBigInteger(); System.out.println("Case "+ k++ +":"); System.out.println(num1 + " + " + num2 +" = " + num1.add(num2)); n--; } } }

1873: A+B Problem II

时间限制: 3 Sec   内存限制: 64 MB 提交: 15   解决: 4 您该题的状态：已完成[提交][状态][讨论版]

题目描述

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

输入

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.

输出

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.

样例输入

2 1 2 112233445566778899 998877665544332211

样例输出

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