I thought I solved this problem but the program output "0". I don't see any problem. Thank you for helping.
Question :
n! means n × (n − 1) × ... × 3 × 2 × 1
For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800, and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
Find the sum of the digits in the number 100!
package projecteuler;
public class problem20 {
public static void main(String[] args)
{
int sayi=0;
int carpim=1;
for(int i=100;i>=1;i--)
{
carpim*=i;
}
String carp=""+carpim;
int[] dizi = new int[carp.length()];
String[] dizis=new String[carp.length()];
for(int i=0;i<carp.length();i++)
{
dizis[i]=carp.substring(i);
}
for(int i=0;i<carp.length();i++)
{
dizi[i]=Integer.parseInt(dizis[i]);
sayi+=dizi[i];
}
System.out.println(sayi);
}
}
100!
is 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000
, and that exceeds the valid range of an int
(by rather a lot). Try using a BigInteger
. To get you started,
BigInteger carpim = BigInteger.ONE;
for (int i = 100; i >= 1; i--) {
carpim = carpim.multiply(BigInteger.valueOf(i));
}
System.out.println(carpim);
The output of which is the number mentioned before.