1C - A + B Problem II

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

// too young and didn't notice that 
// "you should not process them by using 32-bit integer"
// and "Output a blank line between two test cases".

 1 #include<stdio.h>
 2 int main()
 3 {
 4     int a, b, t, i;
 5     scanf("%d", &t);
 6     for(i=1;i<=t;i++)
 7     {
 8         scanf("%d %d", &a, &b);
 9         printf("Case %d:\n", i);
10         printf("%d + %d = %d\n", a, b, a+b);
11         printf("\n");
12     }
13     return 0;
14 }

Wrong Answer

// long int is also 32-bit integer.
代码省略
// 不用判断哪个数较长，只要记录答案的长度. 三个数组都要初始化为0！！！

 1 #include<stdio.h>
 2 #include<string.h>
 3 
 4 int reverse_add(int *a, int *b, int *c, int al, int bl)
 5 {
 6     int i, sum=0, k, max;
 7     if(al<bl) max=bl;
 8     else      max=al;
 9     k=0;
10     for(i=0; i<max; i++)
11     {
12         *(c+i)=(*(a+i)+*(b+i)+k)%10;
13         k=(*(a+i)+*(b+i)+k)/10;
14     }
15     if(k!=0) 
16     {
17         *(c+i)=1;
18         return max+1;
19     }
20     else return max;
21 }
22 
23 int main()
24 {
25     char s1[1001], s2[1001];
26     int t, k, i, length_a, length_b, rmax;
27     scanf("%d", &t);
28     for(k=1;k<=t;k++)
29     {
30         int a[1001]={0}, b[1001]={0}, c[1001]={0}; /* 三个数组都要初始化 */
31         scanf("%s %s", s1, s2);
32         length_a=strlen(s1); length_b=strlen(s2);
33         for(i=0; i<length_a; i++) a[i]=s1[length_a-1-i]-'0';
34         for(i=0; i<length_b; i++) b[i]=s2[length_b-1-i]-'0';
35         rmax=reverse_add(a, b, c, length_a, length_b);
36         printf("Case %d:\n", k);
37         printf("%s + %s = ", s1, s2);
38         for(i=0; i<rmax; i++) printf("%d", c[rmax-1-i]);
39         printf("\n");
40         if(t>1&&k<t) printf("\n");
41     }
42     return 0;
43 }

AC