感觉应该没问题了，为什么还是WRONG ANSWER？
/*
    Name:
    Copyright:
    Author:
    Date: 05-11-14 22:29
    Description:
    A + B Problem II
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 223072    Accepted Submission(s): 42802


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
 

Author
Ignatius.L

*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>

#define MAX 1001

int Add(char C[], char *A, char *B);//C = A +  B，确保A,B都是正数

int main(void)
{
    char A[MAX], B[MAX], C[MAX];
    int i, n, leftC;
   
    scanf("%d", &n);
    for (i=1; i<=n; i++)
    {
        scanf("%s%s", A, B);
        leftC = Add(&C, &A, &B);
        printf("Case %d:\n", i);
        printf("%s + %s = ", A, B);
        while (leftC < MAX)
            printf("%d", C[leftC++]);
        printf("\n");
        if (i < n)
            printf("\n");
    }

    return 0;
}

int Add(char C[], char *A, char *B)//C = A +  B，确保A,B都是正数
{
    int lenA = strlen(A) -1;
    int lenB = strlen(B) -1;
    int i, leftC = MAX;
   
    for (i=0; i<MAX; i++) //初始化C
        C[i] = 0;
        
    while (lenA >= 0 && lenB >= 0)
    {
        C[--leftC] += A[lenA--] + B[lenB--];
        if (C[leftC] >= 10)
        {
            C[leftC-1] = C[leftC] / 10;
            C[leftC] %= 10;
        }
    }
    while (lenA >= 0)
    {
        C[--leftC] += A[lenA--] - '0';
        if (C[leftC] >= 10)
        {
            C[leftC-1] = C[leftC] / 10;
            C[leftC] %= 10;
        }
    }
    while (lenB >= 0)
    {
        C[--leftC] += B[lenB--] - '0';
        if (C[leftC] >= 10)
        {
            C[leftC-1] = C[leftC] / 10;
            C[leftC] %= 10;
        }
    }
   
    return leftC;
}

已经解决了！
#include<stdio.h>
#include<stdlib.h>
#include<string.h>

#define MAX 1001

int Add(char C[], char * A, char * B);//C = A +  B，确保A,B都是正数

int main(void)
{
    char A[MAX], B[MAX], C[MAX];
    int i, n, leftC;
   
    scanf("%d", &n);
    for (i=1; i<=n; i++)
    {
        scanf("%s%s", A, B);
        printf("Case %d:\n", i);
        printf("%s + %s = ", A, B);
        leftC = Add(C, A, B);
        while (leftC < MAX)
            printf("%d", C[leftC++]);
        printf("\n");
        if (i < n)
            printf("\n");
    }

    return 0;
}

int Add(char C[], char * A, char * B)//C = A +  B，确保A,B都是正数
{
    int lenA = strlen(A) - 1;
    int lenB = strlen(B) - 1;
    int i, leftC = MAX;
   
    for (i=0; i<MAX; i++) //初始化C
        C[i] = 0;

    while (lenA >= 0 && lenB >= 0)
    {
        C[--leftC] += A[lenA--] + B[lenB--] - '0' - '0';
        if (C[leftC] >= 10)
        {
            C[leftC-1] = 1;
            C[leftC] -= 10;
        }
    }
    while (lenA >= 0)
    {
        C[--leftC] += A[lenA--] - '0';
        if (C[leftC] >= 10)
        {
            C[leftC-1] = 1;
            C[leftC] -= 10;
        }
    }
    while (lenB >= 0)
    {
        C[--leftC] += B[lenB--] - '0';
        if (C[leftC] >= 10)
        {
            C[leftC-1] = 1;
            C[leftC] -= 10;
        }
    }

    if (C[leftC-1] > 0) //有进位
        leftC--;
   
    return leftC;
}

超长数相加，学习了。

试验下，如可行，将深入学习