（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 code:

 1  #include  < iostream >
 2  using   namespace  std;
 3  void  add( char  a[], char  b[])
 4  {
 5     char  sum[ 1010 ] = { '   ' };
 6     int  flg = 0 ;
 7     int  temp  = 0 ;
 8     int  len_a  = strlen(a);
 9     int  len_b  = strlen(b);
10     int  i = len_a;
11     int  j = len_b;
12     for  (;i > 0 ;i -- )
13    {
14       if  (j > 0 )
15      {
16        temp  = a[i - 1 ] + b[j - 1 ] + flg - 96 ;
17        j -- ;
18      }
19       else  temp  =  a[i - 1 ] + flg - 48 ;
20       if  (temp >= 10 )
21      {
22       flg = 1 ;
23      }
24       else  flg  = 0 ;
25      temp  = temp % 10 ;
26      sum[i] = temp + 48 ;
27    }
28    if  (flg == 1 )sum[ 0 ] = 49 ;
29   i = 0 ;
30    while  (i <= len_a)
31   {
32      if  (sum[i] != '   ' )cout << sum[i];
33     i ++ ;
34   }
35     cout << endl;
36   }
37  void  main()
38  {
39     int  N;
40     while  (cin  >> N )
41    {
42       for  ( int  i = 1 ;i <= N;i ++ )
43      {
44         char  a[ 1000 ];
45         char  b[ 1000 ];
46        cin  >> a;
47        cin  >> b;
48         int  len_a  = strlen(a);
49         int  len_b  = strlen(b);
50        cout  << " Case  " << i << " :\n " << a << "  +  " << b << "  =  " ;
51         if  (len_a >= len_b)
52        {
53          add(a,b);
54        }
55         else  add(b,a);
56         if  (i != N)cout << endl;
57       }
58    }

59 }