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".

#include<stdio.h>
int main()
{
    int a, b, t, i;
    scanf("%d", &t);
    for(i=1;i<=t;i++)
    {
        scanf("%d %d", &a, &b);
        printf("Case %d:\n", i);
        printf("%d + %d = %d\n", a, b, a+b);
        printf("\n");
    }
    return 0;
}

// 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！！！

#include<stdio.h>
#include<string.h>

int reverse_add(int *a, int *b, int *c, int al, int bl)
{
    int i, sum=0, k, max;
    if(al<bl) max=bl;
    else      max=al;
    k=0;
    for(i=0; i<max; i++)
    {
        *(c+i)=(*(a+i)+*(b+i)+k)%10;
        k=(*(a+i)+*(b+i)+k)/10;
    }
    if(k!=0) 
    {
        *(c+i)=1;
        return max+1;
    }
    else return max;
}

int main()
{
    char s1[1001], s2[1001];
    int t, k, i, length_a, length_b, rmax;
    scanf("%d", &t);
    for(k=1;k<=t;k++)
    {
        int a[1001]={0}, b[1001]={0}, c[1001]={0}; /* 三个数组都要初始化 */
        scanf("%s %s", s1, s2);
        length_a=strlen(s1); length_b=strlen(s2);
        for(i=0; i<length_a; i++) a[i]=s1[length_a-1-i]-'0';
        for(i=0; i<length_b; i++) b[i]=s2[length_b-1-i]-'0';
        rmax=reverse_add(a, b, c, length_a, length_b);
        printf("Case %d:\n", k);
        printf("%s + %s = ", s1, s2);
        for(i=0; i<rmax; i++) printf("%d", c[rmax-1-i]);
        printf("\n");
        if(t>1&&k<t) printf("\n");
    }
    return 0;
}