ACM 1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 98171    Accepted Submission(s): 18601

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

 

 

  转的答案 ac了

 

 import java.math.BigDecimal;

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		int t, i;
		String num1, num2;
		BigDecimal big1, big2;
		Scanner in = new Scanner(System.in);
		t = in.nextInt();
		for (i = 0; i < t; i++) {
			num1 = in.next();
			num2 = in.next();
			big1 = new BigDecimal(num1);
			big2 = new BigDecimal(num2);
			System.out.println("Case " + (i + 1) + ":");
			System.out.println(big1 + " + " + big2 + " = " + big1.add(big2));
			if (i != t - 1)
				System.out.println();
		}
	}
}

 

 

我的答案 wrong 不知道是为什么

 

package endual;

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.Stack;

public class Main {

	public static void main(String[] args) {

		Scanner cin = new Scanner(System.in);
		int n = 0;
		n = cin.nextInt();
		List<String> list = new ArrayList<String>();
		String a, b;

		for (int i = 0; i < n; i++) {

			a = cin.next();
			b = cin.next();
			
			String tepa = a; 
			String tepb = b;
			
			int aLen = a.length();
			int bLen = b.length();
			if (aLen < bLen) { // 保证a的长度要大于b的长度
				String tempBtoA = a;
				a = b;
				b = tempBtoA;
			} // 保证a的长度要大于等于b的长度
			String he = jia(a, b);
			list.add(tepa + " + " + tepb + " = " + he);
		} // end for
		System.out.println();
		for (int j = 0; j < list.size(); j++) {
			System.out.println("Case " + (j + 1) + ":");
			System.out.println(list.get(j));
			System.out.println();

		}

	}

	private static String jia(String a, String b) {

		Stack stacka = new Stack();
		Stack stackb = new Stack();
		String ab = a;
		while (ab.length() != 0) {

			char c = ab.charAt(0);
			String subab = ab.substring(1);
			stacka.push(c);
			ab = subab;
		}

		String abc = b;
		while (abc.length() != 0) {

			char c = abc.charAt(0);
			String subabc = abc.substring(1);
			stackb.push(c);
			abc = subabc;
		}

		Stack sum = new Stack();
		int temp = 0;
		int t = 0;
		while (!stackb.isEmpty()) { // 如果两个里面其中有一个是空了，那么停止

			int aInt = Integer.parseInt(stacka.pop().toString());
			int bInt = Integer.parseInt(stackb.pop().toString());
			temp = aInt + bInt + t;
			if (temp < 10) {
				t = 0;
			} else {
				temp = temp - 10;
				t = 1; // 将t设置成为1，此时将sumAB的1取得
			}
			sum.push(temp); // 添加到栈中去
		}// end while ;

		while (!stacka.isEmpty()) {

			int aInt = Integer.parseInt(stacka.pop().toString());
			temp = aInt + t;
			if (temp < 10) {
				sum.push(temp);
				t = 0;
			} else {

				temp = temp - 10;
				t = 1; // 将t设置成为1，此时将sumAB的1取得
			}
			sum.push(temp); // 添加到栈中去

		}
		if (t == 1) {

			sum.push(t);
		}
		String s = "";
		while (!sum.isEmpty()) { // 如果两个里面其中有一个是空了，那么停止

			s = s + sum.pop().toString();

		}
		return s;

	}

}