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题目：
A + B Problem II |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 2372 Accepted Submission(s): 917 |
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 |
题目分析：
简单题。这种题比赛的时候直接用java写就好，用C/C++来做其实也是那么个意思。java的BigInteger的相关类底层就是那么实现的。所以没必要写那么1、200,行代码。无论是开发也好，还是算法也好，个人觉得都应该选择最合适的方法来解决问题，而不要拘泥于某种技术。当然，C/C++关于大数的思想还是得看一下的，对于初学者来说，理解后自己敲一遍也是有必要的。但是过了初学阶段的话，个人觉得直接用java来做就好了。
代码如下：
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int i;
for(i = 1 ; i <= n ; ++i){
System.out.println("Case "+ i +":");
BigInteger a = scanner.nextBigInteger();
BigInteger b = scanner.nextBigInteger();
System.out.println(a+" + " + b + " = " + a.add(b));
if(i != n){//最后一个不要出现空行,否则会PE
System.out.println();
}
}
}
}
(hdu step 2.3.1)A + B Problem II(大数加法)
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原文地址：http://blog.csdn.net/hjd_love_zzt/article/details/43446355
