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
Author
Ignatius.L
这道题考我们的是大数计算问题，也就是说一个超过我们系统中long类型的数，计算加法的时候，需要我们自己处理这里面的逻辑，但是无奈，java为我们想好了这一切，我们java中内置了一个java.math.BigInteger工具类，可以直接计算大数，所有的问题都解决咯。
顺便说一下，本人喜欢java的原因，可能也就是在最开始学编程的时候，没怎么见过像eclipse一样的编译器，后来就喜欢上了它，再后来自己就在学习安卓啦，对java的语法也比较喜欢，就这样咯，换句话说，笔者的c学的不好，而acm也不支持什么别的语言啦，那就java吧，但是有的时候java确实会解题时间长的问题，但是这种情况非常的少，目前已知的很多acm大神也在使用java的~
代码：
import java.math.BigInteger;
import java.util.Scanner;
public class Main1002 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
for (int i = 0; i < n; i++) {
BigInteger a = in.nextBigInteger();
BigInteger b = in.nextBigInteger();
if (i == n - 1)// 输出格式控制
{
System.out.println("Case " + (i + 1) + ":\r\n" + a +
" + " + b + " = " + a.add(b));
} else {
System.out.println("Case " + (i + 1) + ":\r\n" + a +
" + " + b + " = " + a.add(b) + "\r\n");
}
}
}
}
