#include<stdio.h>
#include<string.h>
int main()
{
    char a[1000],b[1000];
    int num[1100];
    int n,i,j,a1[1100],b1[1100],t,k;
    scanf("%d",&n);     //输入测试样例数
      for(k=1;k<=n;k++)
       { scanf("%s",a);   //读入测试两个大数
         scanf("%s",b);
               int a1=strlen(a);
                  for(j=0,i=a1-1;i>=0;i--)
                   a1[j++]=a[i]-'0';
               int b1=strlen(b);
                  for(j=0,i=b1-1;i>=0;i--)
                   b1[j++]=b[i]-'0';
             for(i=0;i<1000;i++)   
                  {
                        a1[i]+=b1[i];   //a1[i]计算得出大数和
                         if(a1[i]>=10)
                           a1[i]-=10;
                           a1[i+1]++;  //位数大于10则进位
                  }
               printf("Case 1:\n");      //输出 Case 1:   a + b =
               printf("%s + %s = ",a,b);
            
               for(i=1100;(i>=0)&&(a1[i]==0);i--);// 从高位到低位跳过高位的0
                    if(i>=0)
                     for(;i>=0;i--)
                      printf("%d",a1[i]);
                     else printf("0");
                        printf("\n\n");
      
 
       }
         return 0;
 
}
 
提示

--------------------Configuration: 1 - Win32 Debug--------------------
Compiling...
1.c
K:\计算机\VC6.0\Microsoft Visual Studio\MyProjects\数组实验\1.c(12) : error C2143: syntax error : missing ';' before 'type'
K:\计算机\VC6.0\Microsoft Visual Studio\MyProjects\数组实验\1.c(13) : warning C4047: '=' : 'int ' differs in levels of indirection from 'int *'
K:\计算机\VC6.0\Microsoft Visual Studio\MyProjects\数组实验\1.c(15) : error C2143: syntax error : missing ';' before 'type'
K:\计算机\VC6.0\Microsoft Visual Studio\MyProjects\数组实验\1.c(16) : warning C4047: '=' : 'int ' differs in levels of indirection from 'int *'
执行 cl.exe 时出错.

1.obj - 1 error(s), 0 warning(s)

表示不会修改了

依然没有通过编译
--------------------Configuration: 大数加法 - Win32 Debug--------------------
Compiling...
大数加法.c
D:\C++\大数加法.c(18) : error C2143: syntax error : missing ';' before 'type'
D:\C++\大数加法.c(19) : error C2065: 'al' : undeclared identifier
D:\C++\大数加法.c(21) : error C2143: syntax error : missing ';' before 'type'
D:\C++\大数加法.c(22) : error C2065: 'bl' : undeclared identifier
执行 cl.exe 时出错.

大数加法.obj - 1 error(s), 0 warning(s)
________________________________________________________________
这是最简化的
#include <stdio.h>
#include <string.h>
#define MAX_LEN 200
int an1[MAX_LEN+10];
int an2[MAX_LEN+10];
char szLine1[MAX_LEN+10];
char szLine2[MAX_LEN+10];
int main(void)
{
    scanf("%s", szLine1);
    scanf("%s", szLine2);
    int i, j;
    int nLen1 = strlen( szLine1);
    for( j = 0, i = nLen1 - 1;i >= 0 ; i --)
        an1[j++] = szLine1[i] - '0';
    int nLen2 = strlen(szLine2);
    for( j = 0, i = nLen2 - 1;i >= 0 ; i --)
        an2[j++] = szLine2[i] - '0';
    for( i = 0;i < MAX_LEN ; i ++ )
    {  an1[i] += an2[i]; //逐位相加
        if( an1[i] >= 10 )
        { //看是否要进位
            an1[i] -= 10;
            an1[i+1] ++; //进位
        }
    }
    for( i = MAX_LEN; (i >= 0) && (an1[i] == 0); i -- ) ;
    if(i>=0)
        for( ; i >= 0; i--)
            printf("%d", an1[i]);
    else      printf("0");
    return 0;
}

老实说BUG在这里

for( i = MAX_LEN; (i >= 0) && (an1[i] == 0); i -- ) ;
    if(i>=0)
        for( ; i >= 0; i--)
            printf("%d", an1[i]);
    else      printf("0");

我也是用VC++6.0
1L是做北大ACM1002题编写的1000位以内大数加法
为了达到
Case 1:
x + y = z

Case 2:
x + y = Z
这种效果
我编写成

#include<stdio.h>
#include<string.h>
int main()
{
    char a[1100],b[1100];
    int num[1100];
    int n,i,j,a1[1100],b1[1100],t,k,a2,b2;                           //为区分源代码a1和a1[]把a1,b1用a2,b2代替
    scanf("%d",&n);     //输入测试样例数
      for(k=1;k<=n;k++)
       { scanf("%s",a);   //读入测试两个大数
         scanf("%s",b);
                    a2=strlen(a);                                    //为区分源代码a1和a1[]把a1,b1用a2,b2代替
                  for(j=0,i=a2-1;i>=0;i--)
                   a1[j++]=a[i]-'0';
                    b2=strlen(b);                                   //为区分源代码a1和a1[]把a1,b1用a2,b2代替
                  for(j=0,i=b2-1;i>=0;i--)
                   b1[j++]=b[i]-'0';
             for(i=0;i<1100;i++)                                     //避免a1[]溢出
                  {
                        a1[i]+=b1[i];   //a1[i]计算得出大数和
                         if(a1[i]>=10)
                           a1[i]-=10;
                           a1[i+1]++;  //位数大于10则进位
                  }
               printf("Case 1:\n");      //输出 Case 1:   a + b =
               printf("%s + %s = ",a,b);
            
               for(i=1100;(i>=0)&&(a1[i]==0);i--);// 从高位到低位跳过高位的0
                    if(i>=0)
                     for(;i>=0;i--)
                      printf("%d",a1[i]);
                     else printf("0");
                        printf("\n\n");
      

       }
         return 0;

}
编译和组建通过了 但是输入1回车1 2回车。运行后会出现1 + 2=2-1717986919-1717986919-1717986919-1717986919循环。。这怎么回事呢。
下面回的精简版是老师PPT上布置的作业让我找BUG

哎呀 懂了 太谢谢你了
这么多次帮我指出错误改正