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

 

1 #include
           
             2 #include
            
              3 using namespace std; 4  5 #define P(x) \ 6 cout << #x " " << x << ":" << endl; 7  8 #define PE(a, b, c, sum) \ 9 cout << a << " + " <<  b << " = " << (c ? "1" : "") << sum << endl;10 int main()11 {12     int T;13     cin >> T;14     int Case = 1;15     while(T--)16     {17         P(Case);18         ++Case;19         string a, b, sum;20         cin >> a >> b;21         //cout << a << " + " <<  b << " = " ;22         int l_a = a.size()-1, l_b = b.size()-1, c = 0;23         sum = (l_a > l_b ? a : b);24         int i, j;25         for(i = l_a, j = l_b; i >= 0 && j >= 0; --i, --j)26         {27             int s  = a[i] - '0' + b[j] - '0' + c;28             sum[i>j?i:j] = s % 10 + 48;29             c = s / 10;30         }31         while(i >= 0)32         {
         //处理进位33             int s = sum[i] - '0' + c;34             sum[i] = s % 10 + 48;35             c = s / 10;36             i--;37         }38         while(j >= 0)39         {
         //处理 进位40             int s = sum[j] - '0' + c;41             sum[j] = s % 10 + 48;42             c = s / 10;43             j--;44         }45         // cout << (c ? "1" : "") << sum << endl;46         PE(a, b, c, sum);47         if(T)48             cout << endl;49     }50     return 0;51 }
            
           

 

#include
           
            #include
            
             using namespace std;const int maxn = 500;#define P(x) \cout << #x ": " << x << endl;int main(){    string a;    cin >> a;    //string str;    //string str("shi");    string str(maxn, 'a');    //str = ""; //长度也变为了0    for(int i = 0; i != a.size(); ++i)  //the same above    //for(int i = a.size()-1; i >= 0; --i)        str[i] = a[i];    P(a);    P(str);    return 0;}/*str: ""output: ""str:"shi"output: san由此可见，string 对象是有默认长度的(根据其赋值对象的长度决定)与字符数组类似的，只不过不需明确指明，可根据赋值对象动态的适应并改变它。字符赋值，其实质是在原有的基础上修改，是有长度限制的。但是后面以后str = "";以后长度也都变为了0.由此可见，使用字符赋值的关键是如何确定他的可能长度。*/