原题地址：http://acm.hdu.edu.cn/showproblem.php?pid=1002
【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
【AC代码】
1 #include<stdio.h>
2 #include<string.h>
3 #define max(a, b) a > b ? a:b
4
5 char a[1024];
6 char b[1024];
7 char c[1024];
8 int i;
9
10 void reverse(char *a)
11 {
12 int aa = strlen(a);
13 char t;
14 for(i=0; i<aa/2; i++)
15 {
16 t = a[i];
17 a[i] = a[aa-1-i];
18 a[aa-1-i] = t;
19 }
20 }
21 void add(char *a, char *b, char *c)
22 {
23 int cc = 0, aa = strlen(a), bb = strlen(b);
24 int len = max(aa, bb);
25 for(i=0; i<aa; i++)
26 {
27 a[i] = a[i]-‘0‘;
28 }
29 for(i=0; i<bb; i++)
30 {
31 b[i] = b[i]-‘0‘;
32 }
33 for(i=0; i<len; i++)
34 {
35 c[i] = (a[i]+b[i]+cc) % 10 + ‘0‘;
36 cc = (a[i]+b[i]+cc) / 10;
37 }
38 if(cc) c[i++] = cc + ‘0‘;
39 c[i] = ‘\0‘;
40 }
41 void print(char *c)
42 {
43 for(i = strlen(c)-1; i>=0; i--)
44 printf("%c", c[i]);
45 printf("\n");
46 }
47 main()
48 {
49 int n, j;
50 scanf("%d", &n);
51 for(j=1; j<=n; j++)
52 {
53 memset(a, 0, sizeof(a));
54 memset(b, 0, sizeof(b));
55 memset(c, 0, sizeof(c));
56 scanf("%s %s", a, b);
57 printf("Case %d:\n", j);
58 printf("%s + %s = ", a, b);
59 reverse(a), reverse(b);
60 add(a, b, c);
61 print(c);
62 if(j != n) printf("\n");
63 }
64
65 }