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

 

 

 

#include <iostream>
#include <cstring>
using namespace std;
int main()
{
    int len1,len2,p;
    int i,j,t;
    char a[1000],b[1000],c[1001];
    cin>>t;
    for(j=1;j<=t;j++)
    {
        cin>>a>>b;
        cout<<"Case"<<" "<<j<<":"<<endl;
        cout<<a<<" "<<"+"<<" "<<b<<" "<<"="<<" ";
        len1=strlen(a)-1;
        len2=strlen(b)-1;
        p=0;
        for(i=0;len1>=0||len2>=0;i++,len1--,len2--)
        {
            if(len1>=0&&len2>=0)
               c[i]=a[len1]+b[len2]-'0'+p;
            if(len1>=0&&len2<0)
               c[i]=a[len1]+p;
            if(len2>=0&&len1<0)
               c[i]=b[len2]+p;
            p=0;
            if(c[i]>'9')
            {
               c[i]=c[i]-10;
               p=1;
            }
        }
        if(p==1)
           cout<<"1";
        while(i--)
           cout<<c[i];
        if(j<t)
           cout<<endl<<endl;
        else
            cout<<endl;
    }
    return 0;
}