برنامه يک بر روی e تا ۵۰ اقم اعشار

برنامه يک بر روی e تا ۵۰ اقم اعشار :

#include<stdio.h>
 #include<conio.h>
 void  main()
  {
   long int n;
   int i,su[51],sm[51],sum[51],x,y,k,w,q[51],a[51];
   clrscr();
   for(i=0;i<=50;i++)
     {
       su[i]=0;
       sm[i]=0;
       a[i]=0;
       q[i]=0;
       sum[i]=0;
     }
   for(n=2;n<=1000000;n++)
     {
       if(n==2)
          a[0]=1;
       for(i=0;i<=50;i++)
         {
           q[i]=a[i]/n;
           if(q[i]==0)
              a[i+1]+=a[i]*10;
           else
              a[i+1]+=(a[i]%n)*10;
           a[i]=q[i];
         }
       for(i=50;i>=0;i--)
         {
           if(n%2==0)
            {
              su[i]+=a[i];
              x=su[i]%10;
              y=su[i]/10;
              if(y!=0)
               {
                 su[i]=x;
                 su[i-1]+=y;
               }
           }
          else if(n%2!=0)
            {
              sm[i]+=a[i];
              w=sm[i]%10;
              k=sm[i]/10;
              if(k!=0)
               {
                 sm[i]=w;
                 sm[i-1]+=k;
               }
            }
      }
    }
   for(i=50;i>=0;i--)
    {
      if(su[i]>=sm[i])
        sum[i]=su[i]-sm[i];
      else if(su[i]<sm[i])
        {
           sum[i]=10+su[i]-sm[i];
           su[i-1]=su[i-1]-1;
        }
    }
   printf("0.");
   for(i=1;i<=50;i++)
   printf("%d",sum[i]);
   getch();
  }