Thank you guys, i love u all :) On Sep 25, 1:30 pm, kujta1 <[email protected]> wrote: > public class Client { > public static void main(String[] args) { > BigInteger a[]= new BigInteger[1010]; > a[0]=BigInteger.ONE; > a[1]=BigInteger.ONE; > for(int i=2;i<1010;i++) > { > String f=""; > f=f+i; > a[i]=a[i-1].multiply(BigInteger.valueOf(i)); > > } > > //System.out.println(a[5]); > // so in the array a are stored all values of > factorial > > } > > } > > a[0]=1; > a[1]=1; > On Sep 25, 4:16 pm, सारंग <[email protected]> wrote: > > > > > > > > > Hello Agaly, > > Use mathematics and its Formula that is n! (n Factorial); > > double product = 0.0; > > Object O = new Object(); // U can use java.lang.Float as direct object data > > type of float data type. > > O.factorial(1000); > > > double Factorial(double number){ > > if(number ==0 || number == 1) > > return 1; > > else > > return(number * (Factorial (number-1)) > > > } > > > * Please let me know anyone is having more efficient way solving this > > problem. > > > Thanks and Regards > > Sarang Gandhi > > > On Sun, Sep 25, 2011 at 6:19 AM, Agaly Rahmanov > > <[email protected]>wrote: > > > > Can you answer to this question? > > > > 1. Write Java variable declaration > > > > Declare a variable to store a product number (1-1000) > > > > __________________________________________________ > > > > -- > > > To post to this group, send email to > > > [email protected] > > > To unsubscribe from this group, send email to > > > [email protected] > > > For more options, visit this group at > > >http://groups.google.com/group/javaprogrammingwithpassion?hl=en
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