C prog-Fact of 1,00,000

[Pragadheesh]

Its a question asked in my C test..
Factorial of One Lakh(1,00,000)
The problem is ,wat the data type can hold the result..?? is there any other way?
plz help me frenz..

[mediator]

Even the scientific calculator of FC5 gives "error" on that! I tried "long" in my factorial program, but it computes well to 12! .

[kalpik]

Hehe.. Just use string pointers, and you will be limited only by the ammount of memory your system has!

[aadipa]

Try to create something like BigInt in java.

Refer to this code

enum Sign { NON_NEGATIVE, NEGATIVE};

typedef enum Sign SIGN;

struct BigInt {
   char number[100];
   SIGN sign;
   } ONE  = {"1",NON_NEGATIVE},
     ZERO = {"0",NON_NEGATIVE};

typedef struct BigInt BigInteger;

char signChar(BigInteger b) {
   return  (b.sign==NEGATIVE) ? '-' : ' ';
   }

BigInteger getBigInt(void) {

   BigInteger b;

   scanf("%s",b.number);
   if(b.number[0]=='-') {
      xstrcpy(b.number, &b.number[1]);
      b.sign=NEGATIVE;
      }
   else {
      b.sign=NON_NEGATIVE;
      }
   return b;
   }


BigInteger addBigInt(BigInteger b1, BigInteger b2) {
   BigInteger result;

   int l1,l2,l;
   int i;
   int i1,i2,carry;

   if(b1.sign != b2.sign) {
      b2.sign = (b2.sign+1)%2;
      return subBigInt(b1,b2);
      }

   l1 = xstrlen(b1.number);
   l2 = xstrlen(b2.number);

   l = l1<l2?l2:l1;

   l1--;l2--;

   result.number[l]=NULL;
   carry=0;

   for(i=0; i<l; i++, l1--, l2--) {
      i1 = (l1>=0 ? ( b1.number[l1]) - '0' : 0);
      i2 = (l2>=0 ? ( b2.number[l2]) - '0' : 0);
      result.number[i] = ( i1 + i2 + carry)%10 + '0';
      carry = ( i1 + i2 + carry)/10;
      }

   if(carry != 0) {
      result.number[i++] = carry + '0';
      }
   result.number[i] = NULL;
   xstrcpy(xstrrev(result.number),result.number);

   result.sign = b1.sign;
   return result;
   }

BigInteger subBigInt(BigInteger b1, BigInteger b2) {
   BigInteger result;

   int l1,l2,l;
   int i;
   int i1,i2,carry;

   if(b1.sign != b2.sign) {
      b2.sign = (b2.sign+1)%2;
      return addBigInt(b1,b2);
      }

   l1 = xstrlen(b1.number);
   l2 = xstrlen(b2.number);

   l = l1<l2?l2:l1;

   l1--;l2--;

   result.number[l]=NULL;
   result.sign = b1.sign;
   carry=0;

   for(i=l-1; i>=0; i--, l1--, l2--) {
      i1 = (l1>=0 ? ( b1.number[l1]) - '0' : 0);
      i2 = (l2>=0 ? ( b2.number[l2]) - '0' : 0);
      result.number[i] = ( i1 - i2 + carry + 10)%10 + '0';
      carry = ( i1 - i2 + carry)<0?-1:0;
      }

   if(carry!=0) {
      for(i=l-1; i>=0; i--) {
	 result.number[i] = '9' - result.number[i] + '0';
	 }
      if(result.sign == NON_NEGATIVE)
	 result = addBigInt(result, ONE);
      else
	 result = subBigInt(result, ONE);

      result.sign = (result.sign+1)%2;
      }
   return result;
   }

BigInteger mulBigInt(BigInteger b1, BigInteger b2) {

   int l1,l2;
   int i;

   BigInteger temp,result;

   temp = ZERO;
   result = ZERO;

   l1 = xstrlen(b1.number);
   l2 = xstrlen(b2.number);

   for(i=0; i<l2; i++) {
      int c,j,num,carry=0;

      for(c=0,j=0;c<i;c++,j++) {
	 temp.number[j] = '0';
	 }

      for(c=0; c<l1 || carry!=0; c++,j++) {
	 num = (b2.number[l2-i-1] - '0')
	     * (c<l1 ? b1.number[l1-c-1] - '0' : 0)
	     + carry;
	 temp.number[j] = num%10 +'0';
	 carry = num/10;
	 }

      temp.number[j]=NULL;
      xstrcpy(temp.number, xstrrev(temp.number) );
      result = addBigInt(temp, result);
      }

   if(b1.sign != b2.sign) {
      result.sign = NEGATIVE;
      }

   return result;
   }

printf("\nEnter first number : ");
b1 = getBigInt();
.....
printf("\n   %075s%c", result.number, signChar(result));

[mediator]

How?? I'm not so good at pointers in C !

[Pragadheesh]

kalpik plz help... how to use string pointers.. even 'm not all dat good in pointers..