将两个字符串相加，并考虑其的正负和进位情况。
如 输入 1 9999
   输出 10000
 若输入 12345  3456678888
   输出 3456691233
 若输入 -123 23
   输出 -100
 若输入 -123 -77
   输出  -200
求大神回帖

字符串长度没有限制。

 这个还是有些问题！！！
  求大神在看一下，能否在完全一些

自己和同学研究的一个题目，问过老师了，老师给的答案只是无符号加减！！！
  所以在论坛找大神解决了

[此贴子已经被作者于2017-1-10 15:05编辑过]

#include "stdio.h"
#include "string.h"
void fun1(char *s1,char *s2,char *sum)
 { char ch;
     int i,j,t;
     for(i=0;i<100;i++)
       sum[i]='0';
     i=strlen(s1)-1;
     j=strlen(s2)-1;
     if(i>j) t=i;
     else t=j;
     sum[t+2]='\0';
     for(t++;i>=0&&j>=0;i--,j--)
      { ch=s1[i]+s2[j]-'0'-'0';
          if(ch+sum[t]<=57)
            sum[t--]=sum[t]+ch;
          else
          { sum[t-1]='1';
                 sum[t]=sum[t]+ch-10;
                 t-=1;
          }
      }
      if(i>0)
       for(;i>=0;i--)
         { if(sum[t]+s1[i]-'0'<='9')
              sum[t--]=sum[t]+s1[i]-'0';
           else
             { sum[t-1]='1';
                 sum[t--]=sum[t]+s1[i]-'0'-10;
             }
         }
       else
        for(;j>=0;j--)
         { if(sum[t]+s2[j]-'0'<='9')
              sum[t--]=sum[t]+s2[j]-'0';
           else
             { sum[t-1]='1';
                 sum[t--]=sum[t]+s2[j]-'0'-10;
             }
         }      
 }

void main()
{  
   char s1[100],s2[100],sum[120]="0";
   int i,j;
   printf(" input s1:");
   gets(s1);
   printf(" input s2:");
   gets(s2);
   if(s1[0]!='-'&&s2[0]!='-')
    {  fun1(s1,s2,sum);
       printf(" output sum:");
       if(sum[0]!='0')
       puts(sum);
       else puts(sum+1);
    }
   else if(s1[0]=='-'&&s2[0]=='-')
       {  fun1(s1+1,s2+1,sum+1);
             sum[0]='-';
             printf(" output sum:");
             if(sum[1]!='0')
               puts(sum);
             else
              { sum[1]='-';
                   puts(sum+1);
              }
       }
}

这个程序只能完成同号相加

#include "stdio.h"
 #include "string.h"
 void fun1(char *s1,char *s2,char *sum)
  { char ch;
      int i,j,t;
      for(i=0;i<100;i++)
        sum[i]='0';
      i=strlen(s1)-1;
      j=strlen(s2)-1;
      if(i>j) t=i;
      else t=j;
      sum[t+2]='\0';
      for(t++;i>=0&&j>=0;i--,j--)
       { ch=s1[i]+s2[j]-'0'-'0';
           if(ch+sum[t]<=57)
             sum[t--]=sum[t]+ch;
           else
          { sum[t-1]='1';
                  sum[t]=sum[t]+ch-10;
                  t-=1;
           }
       }
       if(i>=0)     // 改为 if(i>=0)  原来的 if(i>0) 没有考虑到高一位的问题
       for(;i>=0;i--)
          { if(sum[t]+s1[i]-'0'<='9')
               sum[t--]=sum[t]+s1[i]-'0';
            else
             { sum[t-1]='1';
                  sum[t--]=sum[t]+s1[i]-'0'-10;
              }
          }
        else
        for(;j>=0;j--)
          { if(sum[t]+s2[j]-'0'<='9')
               sum[t--]=sum[t]+s2[j]-'0';
            else
             { sum[t-1]='1';
                  sum[t--]=sum[t]+s2[j]-'0'-10;
              }
          }      
 }
 
void main()
 {  
    char s1[100],s2[100],sum[120]="0";
    int i,j;
    printf(" input s1:");
    gets(s1);
   printf(" input s2:");
    gets(s2);
    if(s1[0]!='-'&&s2[0]!='-')
     {  fun1(s1,s2,sum);
        printf(" output sum:");
        if(sum[0]!='0')
        puts(sum);
        else puts(sum+1);
     }
    else if(s1[0]=='-'&&s2[0]=='-')
        {  fun1(s1+1,s2+1,sum+1);
              sum[0]='-';
              printf(" output sum:");
              if(sum[1]!='0')
                puts(sum);
              else
              { sum[1]='-';
                    puts(sum+1);
               }
        }
 }