Counting trailing zeros of numbers resulted from f

问题:

I'm trying to count trailing zeros of numbers that are resulted from factorials (meaning that the numbers get quite large). Following code takes a number, compute the factorial of the number, and count the trailing zeros. However, when the number is about as large as 25!, numZeros don't work.

public static void main(String[] args) {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    double fact;
    int answer;

    try {
        int number = Integer.parseInt(br.readLine());
        fact = factorial(number);
        answer = numZeros(fact);
    }
    catch (NumberFormatException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }
}

public static double factorial (int num) {
    double total = 1;
    for (int i = 1; i <= num; i++) {
        total *= i;
    }
    return total;
}   

public static int numZeros (double num) {
    int count = 0;
    int last = 0;   

    while (last == 0) {
        last = (int) (num % 10);
        num = num / 10;
        count++;
    }

    return count-1;
}

I am not worrying about the efficiency of this code, and I know that there are multiple ways to make the efficiency of this code BETTER. What I'm trying to figure out is why the counting trailing zeros of numbers that are greater than 25! is not working.

Any ideas?

回答1:

Your task is not to compute the factorial but the number of zeroes. A good solution uses the formula from http://en.wikipedia.org/wiki/Trailing_zeros (which you can try to prove)

def zeroes(n):
    i = 1
    result = 0
    while n >= i:
        i *= 5
        result += n/i  # (taking floor, just like Python or Java does)
    return result

Hope you can translate this to Java. This simply computes [n / 5] + [n / 25] + [n / 125] + [n / 625] + ... and stops when the divisor gets larger than n.

DON'T use BigIntegers. This is a bozosort. Such solutions require seconds of time for large numbers.

回答2:

You only really need to know how many 2s and 5s there are in the product. If you're counting trailing zeroes, then you're actually counting "How many times does ten divide this number?". if you represent n! as q*(2^a)*(5^b) where q is not divisible by 2 or 5. Then just taking the minimum of a and b in the second expression will give you how many times 10 divides the number. Actually doing the multiplication is overkill.

Edit: Counting the twos is also overkill, so you only really need the fives.

And for some python, I think this should work:

def countFives(n):
    fives = 0   
    m = 5
    while m <= n:
        fives = fives + (n/m)
        m = m*5
    return fives

回答3:

The double type has limited precision, so if the numbers you are working with get too big the double will be only an approximation. To work around this you can use something like BigInteger to make it work for arbitrarily large integers.

回答4:

You can use a DecimalFormat to format big numbers. If you format your number this way you get the number in scientific notation then every number will be like 1.4567E7 this will make your work much easier. Because the number after the E - the number of characters behind the . are the number of trailing zeros I think.

I don't know if this is the exact pattern needed. You can see how to form the patterns here

DecimalFormat formater = new DecimalFormat("0.###E0");

回答5:

My 2 cents: avoid to work with double since they are error-prone. A better datatype in this case is BigInteger, and here there is a small method that will help you:

public class CountTrailingZeroes {

    public int countTrailingZeroes(double number) {
        return countTrailingZeroes(String.format("%.0f", number));
    }

    public int countTrailingZeroes(String number) {
        int c = 0;
        int i = number.length() - 1;

        while (number.charAt(i) == '0') {
            i--;
            c++;
        }

        return c;

    }

    @Test
    public void $128() {
        assertEquals(0, countTrailingZeroes("128"));
    }

    @Test
    public void $120() {
        assertEquals(1, countTrailingZeroes("120"));
    }

    @Test
    public void $1200() {
        assertEquals(2, countTrailingZeroes("1200"));
    }

    @Test
    public void $12000() {
        assertEquals(3, countTrailingZeroes("12000"));
    }

    @Test
    public void $120000() {
        assertEquals(4, countTrailingZeroes("120000"));
    }

    @Test
    public void $102350000() {
        assertEquals(4, countTrailingZeroes("102350000"));
    }

    @Test
    public void $1023500000() {
        assertEquals(5, countTrailingZeroes(1023500000.0));
    }
}

回答6:

This is how I made it, but with bigger > 25 factorial the long capacity is not enough and should be used the class Biginteger, with witch I am not familiar yet:)

public static void main(String[] args) {
    // TODO Auto-generated method stub
    Scanner in = new Scanner(System.in);
    System.out.print("Please enter a number : ");
    long number = in.nextLong();
    long numFactorial = 1;

    for(long i = 1; i <= number; i++) {
        numFactorial *= i;
    }
    long result = 0;
    int divider = 5;
    for( divider =5; (numFactorial % divider) == 0; divider*=5) {
         result += 1;
    }

    System.out.println("Factorial of n is: " + numFactorial);
    System.out.println("The number contains " + result + " zeroes at its end.");

    in.close();

 }

}

回答7:

The best with logarithmic time complexity is the following:

public int trailingZeroes(int n) {
    if (n < 0)
        return -1;

    int count = 0;
    for (long i = 5; n / i >= 1; i *= 5) {
        count += n / i;
    }

    return count;
}

shamelessly copied from http://www.programcreek.com/2014/04/leetcode-factorial-trailing-zeroes-java/

回答8:

I had the same issue to solve in Javascript, and I solved it like:

var number = 1000010000;
var str = (number + '').split(''); //convert to string
var i = str.length - 1; // start from the right side of the array
var count = 0; //var where to leave result
for (;i>0 && str[i] === '0';i--){
    count++;
}
console.log(count) // console shows 4

This solution gives you the number of trailing zeros.

var number = 1000010000;
var str = (number + '').split(''); //convert to string
var i = str.length - 1; // start from the right side of the	array
var count = 0; //var where to leave result
for (;i>0 && str[i] === '0';i--){
	count++;
}
console.log(count)

回答9:

Java's doubles max out at a bit over 9 * 10 ^ 18 where as 25! is 1.5 * 10 ^ 25. If you want to be able to have factorials that high you might want to use BigInteger (similar to BigDecimal but doesn't do decimals).

回答10:

I wrote this up real quick, I think it solves your problem accurately. I used the BigInteger class to avoid that cast from double to integer, which could be causing you problems. I tested it on several large numbers over 25, such as 101, which accurately returned 24 zeros.

The idea behind the method is that if you take 25! then the first calculation is 25 * 24 = 600, so you can knock two zeros off immediately and then do 6 * 23 = 138. So it calculates the factorial removing zeros as it goes.

public static int count(int number) {
    final BigInteger zero = new BigInteger("0");
    final BigInteger ten = new BigInteger("10");
    int zeroCount = 0;
    BigInteger mult = new BigInteger("1");
    while (number > 0) {
        mult = mult.multiply(new BigInteger(Integer.toString(number)));
        while (mult.mod(ten).compareTo(zero) == 0){
            mult = mult.divide(ten);
            zeroCount += 1;
        }
        number -= 1;
    }
    return zeroCount;
}

Since you said you don't care about run time at all (not that my first was particularly efficient, just slightly more so) this one just does the factorial and then counts the zeros, so it's cenceptually simpler:

public static BigInteger factorial(int number) {
    BigInteger ans = new BigInteger("1");
    while (number > 0) {
        ans = ans.multiply(new BigInteger(Integer.toString(number)));
        number -= 1;
    }
    return ans;
}

public static int countZeros(int number) {
    final BigInteger zero = new BigInteger("0");
    final BigInteger ten = new BigInteger("10");
    BigInteger fact = factorial(number);
    int zeroCount = 0;
    while (fact.mod(ten).compareTo(zero) == 0){
        fact = fact.divide(ten);
        zeroCount += 1;
    }
}