A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 230395 Accepted Submission(s):
44208
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
链接： http://acm.hdu.edu.cn/showproblem.php?pid=1002 题意：就是简单的A+B，就只要注意最后一组数据后没多余的空行就ok了，用c/c++比较麻烦，用java轻而易举就搞定了，so~，c的代码就不贴了。
import java.util.*;
2 import java.math.*;
3 public class Main {
4 public static void main(String[] args) {
5 int T;
6 BigDecimal a,b,c; //定义两个大数型变量a，b
7 Scanner cin = new Scanner(System.in);
8 T = cin.nextInt();
9 for (int i=1; i<=T; i++){
10 a = cin.nextBigDecimal();
11 b = cin.nextBigDecimal();
12 //c = a.add(b); //java的加法和c/c++的不同，四则运算都要写成这种形式
13 System.out.println("Case "+i+":");
14 System.out.println(a+" + "+b+" = "+(a.add(b)));
15 if (i != T) //最后一组数据后没有多余的空行。
16 System.out.println();
17 }
18
19 }
20 }