（HDOJ 1002）A + B Problem II

                                                         A + B Problem II

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

 

Author

Ignatius.L

 

AC程序代码：

#include <iostream>
using namespace std;
void add(char a[],char b[])
{
  char sum[1010]={' '};
  int flg=0;
  int temp =0;
  int len_a =strlen(a);
  int len_b =strlen(b);
  int i=len_a;
  int j=len_b;
  for (;i>0;i--)
  {
    if (j>0)
    {
      temp =a[i-1]+b[j-1]+flg-96;
      j--;
    }
    else temp = a[i-1]+flg-48;
    if (temp>=10)
    {
     flg=1;
    }
    else flg =0;
    temp =temp%10;
    sum[i]=temp+48;
  }
 if (flg==1)sum[0]=49;
 i=0;
 while (i<=len_a)
 {
   if (sum[i]!=' ')cout<<sum[i];
   i++;
 }
   cout<<endl;
 }
void main()
{
  int N;
  while (cin >>N )
  {
    for (int i=1;i<=N;i++)
    {
      char a[1000];
      char b[1000];
      cin >>a;
      cin >>b;
      int len_a =strlen(a);
      int len_b =strlen(b);
      cout <<"Case "<<i<<":\n"<<a<<" + "<<b<<" = ";
      if (len_a>=len_b)
      {
        add(a,b);
      }
      else add(b,a);
      if (i!=N)cout<<endl;
     }
   }
}