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)+"\r\n"+a+"+"+b+"="+a.add(b));
- }
- else
- {
- System.out.println("case"+(i+1)+"\r\n"+a+"+"+b+"="+a.add(b));
- }
- }
- }
- }
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