InputThe 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. 
OutputFor 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
解题思路：
既然是大整数，肯定就不能用实型来储存变量，在这里使用字符串string类型来进行模拟运算。（即：从个位数加起，大于10就进1，以此类推。）
有两个进位的地方需要注意！一个是当短的那个数加完时，另一个是长的那个数加完时（比如：1+9999和543+457特别注意）
下面给出大整数加法函数：

 string add(string a,string b)

 {

     string c="";            //长的那个数进位时直接在结果前面加个‘1’。

     int cur=,d=,e=;

     if (a.size()<b.size())swap(a,b);  //a为长的数，b为短的。swap（a，b）交换a，b。

     int la=a.size();

     int lb=b.size();

     while (lb--)          //把短的那个数计算完。

     {

         la--;

         d=(cur+a[la]-''+b[lb]-'')/;   //判断是否有进位。

         a[la]=(cur+a[la]-''+b[lb]-'')%+'';

         cur=d;

     }

     if (cur==)      //如果短的那个数最后一位数还有进位。

     while (la--)

     {

         e=(a[la]-''+cur)/;

         a[la]=(a[la]-''+cur)%+'';

         cur=e;

     }

     if (cur==)     //如果长的那个数最后一位还有进位。

        a=c+a;

     return a;

 }