HDOJ 1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 124589 Accepted Submission(s): 23993

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

 1 #include <iostream>
 2 using namespace std;
 3 
 4 void sum(char (&a)[1001],char (&b)[1001]);
 5 int main()
 6 {
 7     int T, count_T;
 8     char a[1001], b[1001];
 9     cin >> T;
10     for(count_T=1; count_T <= T; count_T++) 
11     {
12         cin >> a >> b;
13 
14         if(count_T==1) 
15         {
16             cout << "Case " << count_T << ":" << endl;
17         }
18         else 
19         {
20             cout << "\nCase " << count_T << ":" << endl;
21         }
22 
23         cout << a << " + " << b << " = ";
24 
25         sum(a,b);
26     }
27     return(0);
28 }
29 void sum(char (&a)[1001],char (&b)[1001])
30 {
31     char c[1002];    
32     int la, lb;
33     la=strlen(a)-1;
34     lb=strlen(b)-1;
35     int i;
36     int flag=0;
37     for(i=0;la>=0 && lb>=0;la--,lb--,i++)
38     {
39         flag+=(a[la]-'0')+(b[lb]-'0');
40         c[i]=flag%10;
41         flag/=10;
42     }
43     for(la=la;la>=0;la--,i++)
44     {
45         flag+=(a[la]-'0');
46         c[i]=flag%10;
47         flag/=10;
48     }
49     for(lb=lb;lb>=0;lb--,i++)
50     {
51         flag+=(b[lb]-'0');
52         c[i]=flag%10;
53         flag/=10;
54     }
55     if(flag)
56     {
57         c[i]=flag;
58         i++;
59     }
60     for(i--;i>=0;i--)
61     {
62         cout<<(char)(c[i]+'0');
63     }
64     cout<<endl;
65 }