杭电1002题
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

这个是我的程序代码
#include <stdio.h>
#include <string.h>
int main ()
{
    char a[21][1001],b[21][1001];
    int c[21][2001]={{0}};
    int k[21]={0};
    int n,i,p,q,t,m;
    scanf ("%d",&n);
    for (i=1;i<=n;i++)
    {
        scanf ("%s %s",a[i],b[i]);
        p=strlen(a[i])-1;
        q=strlen(b[i])-1;
        while ((p>=0)&&(q>=0))
        {
            if(((a[i][p]-'0')+(b[i][q]-'0')+c[i][k[i]])>=10)
            {
                c[i][k[i]]=(a[i][p]-'0')+(b[i][q]-'0')+c[i][k[i]]-10;
                c[i][k[i]+1]++;
            }
            else
            c[i][k[i]]=a[i][p]-'0'+b[i][q]-'0'+c[i][k[i]];
            k[i]++;
            p--;
            q--;
        }

        if (p>=0)
        {
            for (t=p;t>=0;t--)
            {
                c[i][k[i]]=(a[i][t]-'0')+c[i][k[i]];
                k[i]++;
            }
        }
        else if (q>=0)
        {
            for (t=q;t>=0;t--)
            {
                c[i][k[i]]=(b[i][t]-'0')+c[i][k[i]];
                k[i]++;
            }
        }
        if (c[i][k[i]]!=0)
        k[i]++;
  }
        for (i=1;i<=n;i++)
        {
          printf ("Case %d:\n",i);
          printf ("%s + %s = ",a[i],b[i]);
          for (m=k[i]-1;m>=0;m--)
          printf("%d",c[i][m]);
          if(i<n)
          printf ("\n\n");
        }
    return 0;
}

字符串传递的就是首地址值吧，应该不需要取地址符也可以。

应该怎么改格式啊？我运行了觉得跟给出的格式一样呀😳

c数组是我后来添加的，开始写时候i从1计数的，后来也没改，这个不重要

在杭电上提交了一下，显示Output Limit Exceeded错误。
(┬＿┬)

我试了一下a[i]输出的字符串就是我输入的字符串。

我把我原来那个程序改了一下，现在可以9999999999 + 99得到正确的结果了。
但是提交以后还是说是格式错误。

我用cb检测出来是对的呀。而且如果我对a b中所有元素先赋初值为0，然后从1开始赋给其他值，为了方便从一开始，这样怎么不行。我现在这个程序中没有给a b赋初值而已，最后a0中可能是垃圾数字，为什么从1之后得到的会不是本来输入的数字。