一道实验题 优化1000,5000,10000,三个数的阶乘计算，显示优化前后三个数的计算时间进行对比。

优化措施除了“biginteger”方法字符串方法外还可以用什么方法呢。简要说明优化措施

源代码如下：

class Huge

{

private int[] digits;

public Huge(int nDigits)

{

digits = new int[nDigits];

}

private Huge add(Huge n2)

{

Huge result = new Huge(digits.length);

int carry = 0;

for (int k = 0; k < digits.length; k++)

{

int sum = carry + digits[k] + n2.digits[k];

result.digits[k] = sum % 10;

carry = sum / 10;

}

return result;

}

private Huge multiplyDigit(int digit)

{

Huge result = new Huge(digits.length);

int carry = 0;

for (int k = 0; k < digits.length; k++)

{

int prod = carry + digit * digits[k];

result.digits[k] = prod % 10;

carry = prod / 10;

}

return result;

}

private Huge multiply(int number)

{

int weight = 0;

Huge result = new Huge(digits.length);

while (number > 0)

{

int d = number % 10;

number /= 10;

Huge n = multiplyDigit(d);

n.shift(weight++);

result = result.add(n);

}

return result;

}

private void output()

{

int k = digits.length - 1;

while (digits[k--] == 0) ;

for (k = k + 1 ; k >= 0; k--)

System.out.printf("%d", digits[k]);

System.out.println();

}

private void shift(int weight)

{

for (int k = digits.length - 1; k >= 0; k--)

digits[k] = (k >= weight)? digits[k - weight] : 0;

}

public static void main(String[] args)

{

int N = Integer.parseInt(args[0]);

// calculate num-of-digits required for storing factorial(n)

int nbits = 0;

for (int k = 1; k <= N; k++)

nbits += (int)Math.ceil(Math.log10(k));

Huge result = new Huge(nbits);

result.digits[0] = 1;

for (int k = 2; k <= Integer.parseInt(args[0]); k++)

result = result.multiply(k);

result.output();

}

}