菜鸟上路，杭电OJ1002之大数相加

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.
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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.
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Sample Input

   
   
    
    2
1 2
112233445566778899 998877665544332211
   
   

   
   
    
    Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 2222222222222221110
   
   

#include <iostream>
#include <stack>
#include <string>
using namespace std;

int main()
{
    string s1,s2;
    int n;
    int i=0;
	int j=1;
    cin>>n;
    stack<char> A;
    stack<char> B;
    stack<char> C;
    int c=0;
    int num=0;
    while(n--)
    {
		c=0;
		i=0;
        cin>>s1;
        cin>>s2;
		while(!A.empty())
			A.pop();
		while(!B.empty())
			B.pop();
		if(j!=1)
			cout<<endl;
        while(s1[i]!='')
            A.push(s1[i++]);

        i=0;
        while(s2[i]!='')
            B.push(s2[i++]);
        while(!A.empty()||!B.empty())
        {
            if(!A.empty()&&B.empty())
            {
               num=A.top()-'0'+c;
               A.pop();
            }
            else if(A.empty()&&!B.empty())
            {
                num=B.top()-'0'+c;
                B.pop();
            }
            else if(!A.empty()&&!B.empty())
            {
                num=A.top()-'0'+B.top()-'0'+c;
                A.pop();
                B.pop();
            }
            if(num >=10)
            {
                c=1;
                C.push(num%10+'0');
            }
            else
            {
                c=0;
                C.push(num+'0');
            }
           
        }
		/*if(C.top()=='0')
			C.push('1');*/
		cout<<"Case "<<j++<<":"<<endl;
		cout<<s1<<" + "<<s2<<" = ";
		 while(!C.empty())
         {
             cout<<C.top();
             C.pop();
          }
		 cout<<endl;
    }

}

if(j!=1)
			cout<<endl;

（编辑：源码网）

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