每日一练ACM 2019.04.13

2019.04.13

第1002题：A+B Proble Ⅱ

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

题目解析：
输入的第一行：T为实例的数量，再输入T行的例子，每行有2个正整数（他们很大，所以不能用32位的整数来表示），要求计算两个数的和并输出，保证输入的每个整数的长度不超过1000。

解:

package Acm;

import java.math.BigInteger;
import java.util.Scanner;

public class test3 {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner input=new Scanner(System.in);
        int n=input.nextInt();
        for(int i=0;i<n;i++)
        {
            BigInteger a=input.nextBigInteger();
            BigInteger b=input.nextBigInteger();
            if(i==n-i)
            {    
                System.out.print("case"+(i+1)+"
"+a+"+"+b+"="+a.add(b));
                
            }
            else
            {
                System.out.println("case"+(i+1)+"
"+a+"+"+b+"="+a.add(b));
            }
        }
    }

}