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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);
    }
}   
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1 Answer

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.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
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