hdu 1002 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): 147889 Accepted Submission(s): 27893

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

1 2

112233445566778899 998877665544332211

Sample Output

Case 1:

1 + 2 = 3

Case 2:

112233445566778899 + 998877665544332211 = 1111111111111111110

由于处理的数比较大超过了long long 型，所以要变为数组处理比较好！

解题代码：

View Code

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 using namespace std;
 5 char a[10003], b[10003], c[10003];
 6 
 7 void deal(int lena, int lenb)
 8 {
 9     if (lena > lenb)
10     {
11           lenb --;
12         for (int i = lena - 1; i >= 0; i--)
13         {
14             //cout << b[lenb]<<endl;
15             if(lenb >= 0)
16             {
17                 b[i] = b[lenb];
18                 b[lenb] = '0';
19                 lenb --;
20             }
21             else b[i] = '0';
22         }
23         return;
24     }
25     if (lena < lenb)
26     {
27         lena--;
28         for (int i = lenb - 1; i >= 0; i --)
29         {
30             if (lena >= 0)
31             {
32                 a[i] = a[lena];
33                 a[lena] = '0';
34                 lena --;
35             }
36             else a[i] = '0';
37         }
38         return;
39     }
40     return ;
41 }
42 int main()
43 {
44     int n, num = 1;
45     int lena, lenb;
46     cin >> n;
47     while (n--)
48     {
49         cin >> a >> b;
50         lena = strlen (a);
51         lenb = strlen (b);
52         printf ("Case %d:\n", num++);
53         cout << a <<" + " << b<<" = ";        
54         deal(lena, lenb);
55         int d = 0;
56            int j = 0;
57           for (int i = (lena > lenb ? lena : lenb) - 1; i >= 0; i--)
58           {
59                 int temp = a[i] - 48 + b[i] - 48 + d;
60                 d = temp/10;
61                 c[j] = temp%10 + 48;
62                 j++;
63            }
64            c[j] = d;
65            if (d)
66            {
67                 c[j] = d + 48;
68                c[j+1] = 0; 
69            }
70         for (int i =j-1; i >= 0; i --)
71                cout << c[i];
72           cout << endl;
73         if (n)
74             cout <<endl;
75     }
76     return 0;
77 }