杭电 POJ 1002

最新推荐文章于 2021-02-17 13:26:15 发布

原创最新推荐文章于 2021-02-17 13:26:15 发布 · 263 阅读

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这道题，看上去很简单，但是有陷阱啊，我A了两次才通过，第一次是用long型的，发现错误，不接受，改用BigInteger，然后第二次是最后的输出结果格式错误，题目里说了，两次输出结果之间空一行，但是最后一行是不用空的，所以得小心。

题目如下：

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

AC代码如下：

import java.math.BigInteger;
import java.util.Scanner;
 
public class Main {  
    public static void main(String[] args){  
    	Scanner cin = new Scanner(System.in);

    	int T = cin.nextInt();
    	int count = 1;
    	while (T-- != 0){
    		BigInteger a = cin.nextBigInteger();
    		BigInteger b = cin.nextBigInteger();
    		BigInteger c = a.add(b);
    		
    		System.out.println("Case " + count++ + ": \n" + a + " + " + b + " = " + c);
    		if (T != 0) {
    			System.out.println();    			
    		}
    	}
    }
}

总结今天做的一些A+B的题目，本质上都需要把输入进行分割来处理。java的一些包装类正好提供这些功能，无形中降低了难度。java还得抓紧熟悉。