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
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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
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