C语言版大数运算加法

#include                                                                       /*用于大数加法运算*/
#include
int main(void)
{
 char a[10000],b[10000],*pa,*pb;                                                     /*数组a,数组b用来输入两个字符型的大数*/
 int c[10000]={0},d[10000]={0},e[10001]={0};                                         /*数组c,数组d用来存储字符型转化成整型的两个数，数组e用来存储两个数相加的结果*/
 int i,j,k,l,temp,a1,b1;
 int point_a=-1,point_b=-1,point_max,len_max,len_a,len_b,flag_a=1,flag_b=1;          /*point_a、point_b分别记录两个数的小数点位置，point_max记录两个数中小数点位置最远的那个*/
                                                                                     /*len_a,len_b分别记录两个数的长度，len_max用于记录两数对齐后最远的那个*/
 printf("请输入第一个数A:\n");                                                        /*flag_a,flag_b分别用来标记两个数是否有小数点*/                          
 gets(a);
 printf("请输入第二个数B:\n");                                                                          
 gets(b);

 len_a=strlen(a);
 len_b=strlen(b);
 pa=a;
 pb=b;
 
 for(i=0;*(pa+i)!='\0';i++)                                                        /*判断是否带小数点,并标记小数点的位置*/
 {
  if(*(pa+i)=='.')
  {
   point_a=i;
   break;
  }
 }
 if(point_a>=0)
     flag_a=0;
    
 for(j=0;*(pb+j)!='\0';j++)
 {
  if(*(pb+j)=='.')
  {
   point_b=j;
   break;
  }
 }
 if(point_b>=0)                                                                  
     flag_b=0;                                                               
                                                                     /*注意两个数对齐的位置*/
 if(flag_a==1 && flag_b==1)                                                       /*两个数都不带小数点，并将它们转化成整型存储在数组中*/
 {
  if(len_a==len_b)
  {
   for(i=0;i       c[i]=*(pa+i)-'0';
   for(j=0;j       d[j]=*(pb+j)-'0';
  }
  else if(len_a>len_b)
  {
   temp=len_a-len_b;
   for(i=0;i       c[i]=*(pa+i)-'0';
   for(j=temp;j       d[j]=*(pb+j-temp)-'0';
  }
  else
  {
   temp=len_b-len_a;
   for(i=temp;i       c[i]=*(pa+i-temp)-'0';
   for(j=0;j       d[j]=*(pb+j)-'0';
  }
  len_max=len_a>len_b?len_a:len_b;
 }
 else if(flag_a==1 && flag_b==0)                                                /*第一个数带小数点，并将两个数转化成整型存储在数组中*/
 {
  if(len_a>=point_b)
  {
   for(i=0;i       c[i]=*(pa+i)-'0';
   temp=len_a-point_b;
   b1=temp;
      for(j=temp;j      {
       if(*(pb+j-temp)=='.')
          continue;
       d[b1]=*(pb+j-temp)-'0';
       b1++;
   }
   len_max=len_b+temp-1;
   point_max=point_b+temp;
  }
  else
  {
   b1=0;
   temp=point_b-len_a;
   for(i=temp;i       c[i]=*(pa+i-temp)-'0';
   for(j=0;j   {
    if(*(pb+j)=='.')
        continue;
    d[b1]=*(pb+j)-'0';
    b1++;
   }
   len_max=len_b-1;
   point_max=point_b;
  }
 }
 else if(flag_a==0 && flag_b==1)                                             /*第二个数带小数点，并将它们转化成整型存储在数组中*/
 {
  if(len_b>=point_a)
  {
   temp=len_b-point_a;
   a1=temp;
   for(i=temp;i   {
    if(*(pa+i-temp)=='.')
       continue;
    c[a1]=*(pa+i-temp)-'0';
    a1++;
   }
   for(j=0;j       d[j]=*(pb+j)-'0';
   len_max=len_a+temp-1;
   point_max=point_a+temp;
  }
  else
  {
   temp=point_a-len_b;
   a1=0;
   for(i=0;i   {
    if(*(pa+i)=='.')
        continue;
    c[a1]=*(pa+i)-'0';
    a1++;
   }
   for(j=temp;j       d[j]=*(pb+j-temp)-'0';
   len_max=len_a-1;
   point_max=point_a;
  }
 }
 else                                                                        /*两个数都带小数点，将它们转化成整型存储在数组中*/
 {
  if(point_a>=point_b)
  {
   temp=point_a-point_b;
   a1=0;
   b1=temp;
   for(i=0;i   {
    if(*(pa+i)=='.')
        continue;
    c[a1]=*(pa+i)-'0';
    a1++;
   }
   for(j=temp;j   {
    if(*(pb+j-temp)=='.')
        continue;
    d[b1]=*(pb+j-temp)-'0';
    b1++;
   }
   len_max=(len_a-1>len_b+temp-1)?len_a-1:len_b+temp-1;
   point_max=point_a;
  }
  else
  {
   temp=point_b-point_a;
   a1=temp;
   b1=0;
   for(i=temp;i   {
    if(*(pa+i-temp)=='.')
        continue;
    c[a1]=*(pa+i-temp)-'0';
    a1++;
   }
   for(j=0;j   {
    if(*(pb+j)=='.')
        continue;
    d[b1]=*(pb+j)-'0';
    b1++;
   }
   len_max=(len_a+temp-1>len_b-1)?len_a+temp-1:len_b-1;
   point_max=point_b;
  }
 }
  
 l=0;                                                                 /*将两个数从右至左相加，并把结果存在数组e中*/
 for(k=len_max-1;k>=0;k--)
 {
  e[k]=(c[k]+d[k]+l)%10;
  l=(c[k]+d[k]+l)/10;
 }
 
 printf("A+B=");                                                      /*打印结果*/
 if(l!=0)                                                           
  printf("1");
 for(k=0;k {
  if((flag_a==0 || flag_b==0) && k==point_max)
   printf(".");
  printf("%d",e[k]);
 }
 
 return 0;
}