A + B Problem II

         A + B Problem II

                      Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K(Java/Others)
                                             Total Submission(s): 100564 Accepted Submission(s):19068

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 aninteger T(1<=T<=20) which means thenumber of test cases. Then T lines follow, each line consists oftwo positive integers, A and B. Notice that the integers are verylarge, that means you should not process them by using 32-bitinteger. You may assume the length of each integer will not exceed1000.

 

Output

For each test case, you should outputtwo lines. The first line is "Case #:", # means the number of thetest case. The second line is the an equation "A + B = Sum", Summeans the result of A + B. Note there are some spaces int theequation. Output a blank line between two test cases.

 

Sample Input

2 1 2112233445566778899 998877665544332211

 

Sample Output

Case 1: 1 +2 = 3 Case 2: 112233445566778899 + 998877665544332211 =1111111111111111110

 

Author

Ignatius.L

#include<iostream>
#include<cstring>
using namespace std;
#define max 1001
inta[max+10];
int b[max+10];
char x[max+10];
char y[max+10];
int main()
{
   int i,j,k,t,m=1;
   cin>>t;
while(t--)
{
    cin>>x>>y;
  memset(a,0,sizeof(a));
  memset(b,0,sizeof(b));
  int xlen=strlen(x),ylen=strlen(y);
  j=0;
  for(i=xlen-1;i>=0;i--)
   a[j++]=x[i]-'0';
  k=0;
  for(i=ylen-1;i>=0;i--)
   b[k++]=y[i]-'0';
    for(i=0;i<max;i++)
  {
   a[i]+=b[i];
   if(a[i]>9)
   {
   a[i]-=10;
   a[i+1]++;
   }
  }
    cout<<"Case "<<m++<<":"<<endl;
   cout<<x<<"+ "<<y<<"= ";
    bool flag=false;
  for(i=max;i>=0;i--)
  {
   if(flag)
    cout<<a[i];
    elseif(a[i])
   {
   cout<<a[i];
   flag=true;
    }
  }
  if(!flag)
   cout<<"0";
  cout<<endl;
  if(t>=1)
     cout<<endl;
}
}