hdoj1002.txt

北京大学 在线acm在线判题系统 一些题目解答

A + B Problem II
Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) 
Total Submission(s) : 4941   Accepted Submission(s) : 855



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

/*
   农夫三拳@seu
   drizzlecrj.gmail.com
   2007-03-02
*/
#include <iostream>
#include <string>

using namespace std;

void Add(const string& n1, const string& n2)
{
	int len1 = n1.length(), len2 = n2.length();
	int i = len1 - 1, j = len2 - 1;
	int result[1001] = {0};
	int carry = 0;
	int index = 0;
	while(i >= 0 || j >= 0 || carry > 0)
	{
		if(i >= 0)
			result[index] += n1[i] - '0';
		if(j >= 0)
			result[index] += n2[j] - '0';
		result[index] += carry;
		carry = result[index] / 10;
		result[index] -= carry * 10;
		++index, --i, --j;
	}
	for(i = index - 1; i >= 0; i--)
		cout << result[i];
	cout << endl;
}

int main()
{
	int t;
	string n1, n2;
	cin >> t;
	int c = 1;
	while(t--)
	{
		cin >> n1 >> n2;
		if(c != 1)
			cout << endl;
		cout << "Case " << c++ << ":" << endl;
		cout << n1 << " + " << n2 << " = ";

		Add(n1, n2);
	}
	
}