#include <stdio.h>
#include <string.h>
int main()
{
 int i,j,n,k,m;
 char c1[1002]={0},c2[1002]={0},sum[1002]={0},c11[1002]={0},c22[1002]={0};
 scanf("%d",&n);
 m=n;
 while(scanf("%s%s",c1,c2)!=EOF&&n--)
 {

     if(m>=1&&m<=20)
     {
     printf("Case %d:\n",m-n);
 for(i=strlen(c1)-1,j=0;i>=0;i--,j++)
 c11[j]=c1[i]-'0';
 c11[j]=0;
  for(i=strlen(c2)-1,j=0;i>=0;i--,j++)
 c22[j]=c2[i]-'0';
 c22[j]=0;
  for(i=0;i<strlen(c1);i++)
 printf("%c",c1[i]);
 printf(" + ");
for(i=0;i<strlen(c2);i++)
 printf("%c",c2[i]);
printf(" = ");
  for(i=0;i<=strlen(c1)||i<=strlen(c2);i++)
  if((sum[i]+c11[i]+c22[i])>=10)
  {
      sum[i]=sum[i]+c11[i]+c22[i]-10;
      sum[i+1]=sum[i+1]+1;
  }
  else
 sum[i]=sum[i]+c11[i]+c22[i];
  k=i;
 if(sum[k-1])
 for(i=k-1;i>=0;i--)
     printf("%d",sum[i]);
 else
 for(i=k-2;i>=0;i--)
 printf("%d",sum[i]);
printf("\n");
for(i=0;i<strlen(c1);i++)
      c1[i]=0;
for(i=0;i<strlen(c2);i++)
      c2[i]=0;
for(i=0;i<strlen(sum);i++)
    sum[i]=0;
     if(n)
         printf("\n");
 }
 }
 return 0;
}

暂时没看完你的代码，不过想问一下：
for(i=strlen(c1)-1,j=0;i>=0;i--,j++)
c11[j]=c1[i]-'0';
c11[j]=0;
这是不是把c1中的字符串分别减去0的ASCII值再逆序存放到c11中啊？
如果是这样并且后面不存放其他的东西的话，那么最后的c11[j]=0;应该写成c11[j]='\0';
剩下的代码我再看看……呵呵……

[ 本帖最后由 vs_inzaghi 于 2010-7-19 19:27 编辑 ]

恩，是这样的，但我改了还是没有提交成功

好纠结的算法……能给点提示吗？就是说这程序的目的……

题目是这样的
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

貌似你c11[j]=0;中的j值就是你最后存进去的j值，也就是说，你实际上用0这个字符把你最后存进去的一个字符给覆盖掉了……

这个我是为了最前面的数字进位准备的

哪位大侠看看咯，
我把所有积分送出

很水的错误，在while循环结束一次之后你没有初始化，再初始化一次就行了
#include <stdio.h>
#include <string.h>
int main()
{
int i,j,n,k,m;
char c1[1002]={0},c2[1002]={0},sum[1002]={0},c11[1002]={0},c22[1002]={0};


scanf("%d",&n);
m=n;
while(scanf("%s%s",c1,c2)!=EOF&&n--)
{

     if(m>=1&&m<=20)
     {
     printf("Case %d:\n",m-n);
for(i=strlen(c1)-1,j=0;i>=0;i--,j++)
c11[j]=c1[i]-'0';
c11[j]=0;
  for(i=strlen(c2)-1,j=0;i>=0;i--,j++)
c22[j]=c2[i]-'0';
c22[j]=0;
  for(i=0;i<strlen(c1);i++)
printf("%c",c1[i]);
printf(" + ");
for(i=0;i<strlen(c2);i++)
printf("%c",c2[i]);
printf(" = ");
  for(i=0;i<=strlen(c1)||i<=strlen(c2);i++)
  if((sum[i]+c11[i]+c22[i])>=10)
  {
      sum[i]=sum[i]+c11[i]+c22[i]-10;
      sum[i+1]=sum[i+1]+1;
  }
  else
sum[i]=sum[i]+c11[i]+c22[i];
  k=i;
if(sum[k-1])
for(i=k-1;i>=0;i--)
     printf("%d",sum[i]);
else
for(i=k-2;i>=0;i--)
printf("%d",sum[i]);
printf("\n");
for(i=0;i<strlen(c1);i++)
      c1[i]=0;
for(i=0;i<strlen(c2);i++)
      c2[i]=0;
for(i=0;i<strlen(sum);i++)
    sum[i]=0;
     if(n)
         printf("\n");
}
memset(c1,0,sizeof(sum));
memset(c2,0,sizeof(sum));
memset(sum,0,sizeof(sum));
memset(c11,0,sizeof(sum));
memset(c22,0,sizeof(sum));
}
return 0;
}

然后就AC了
╮(╯_╰)╭

太感谢了，我终于提交成功了