A + B Problem II

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 <iostream>

 2 #include <cstring>

 3 using namespace std;

 4 int main()

 5 {

 6     int len1,len2,p;

 7     int i,j,t;

 8     char a[1000],b[1000],c[1001];

 9     cin>>t;

10     for(j=1;j<=t;j++)

11     {

12         cin>>a>>b;

13         cout<<"Case"<<" "<<j<<":"<<endl;

14         cout<<a<<" "<<"+"<<" "<<b<<" "<<"="<<" ";

15         len1=strlen(a)-1;

16         len2=strlen(b)-1;

17         p=0;

18         for(i=0;len1>=0||len2>=0;i++,len1--,len2--)

19         {

20             if(len1>=0&&len2>=0)

21                c[i]=a[len1]+b[len2]-'0'+p;

22             if(len1>=0&&len2<0)

23                c[i]=a[len1]+p;

24             if(len2>=0&&len1<0)

25                c[i]=b[len2]+p;

26             p=0;

27             if(c[i]>'9')

28             {

29                c[i]=c[i]-10;

30                p=1;

31             }

32         }

33         if(p==1)

34            cout<<"1";

35         while(i--)

36            cout<<c[i];

37         if(j<t)

38            cout<<endl<<endl;

39         else

40             cout<<endl;

41     }

42     return 0;

43 }