HDU OJ 1002 A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 372473    Accepted Submission(s): 72570

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

21 2112233445566778899 998877665544332211

 

Sample Output

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

 

Author

Ignatius.L

 

Recommend

We have carefully selected several similar problems for you:  1008 1005 1093 1092 1009 

代码实现：

import java.math.BigInteger;import java.util.Scanner;public class Main {    //  注意提交的时候要 改为Main    public static void main(String args[]){        Scanner s=new Scanner(System.in);        int n=s.nextInt();        for(int i=1;i<=n;i++){            BigInteger a=s.nextBigInteger();            BigInteger b=s.nextBigInteger();            BigInteger c=a.add(b);             if(i>1) System.out.println();            System.out.println("Case "+(i)+":");            System.out.println(a.toString()+" + "+b.toString()+" = "+c.toString());        }    }}// 建议 用java做 因为 BigInteger 理论上可以表示无限大的数只要内存够 // 如果用C的话 就要考虑 用字符串的相加来做。

提交代码