使用对数恒等式的大数阶乘

给定非常大的数字N ，任务是使用Log查找数字的阶乘。

Factorial of a non-negative integer is the multiplication of all integers smaller than or equal to N.

前面我们已经讨论了一个简单的程序来查找本文中的阶乘。在这里，我们将讨论找到大数阶乘的有效方法。

例子：

Input: N = 100
Output: 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

Input: N = 50
Output: 30414093201713378043612608166064768844377641568960512000000000000

方法：最常见的迭代版本在预期的O(N)时间中运行。但是随着数字变大，假设乘法需要固定的时间将是错误的。天真的方法需要O(K * M)的时间进行乘法运算，其中K是乘数的长度，M是被乘数的长度。因此，该想法是使用对数属性：

我们知道$N! = \prod_{i=1}^{N} i$和$ln (ab) = ln(a) + ln(b)$

所以：
$ln (N!) = ln(}\prod_{i=1}^{N} i) = \sum_{i=1}^{N} ln(i)$

另一个属性是
$e^{ln(N!)} = N!$
用ln(N！)的值代替。

下面是上述方法的实现：

C++

// C++ program to compute the
// factorial of big numbers
  
#include 
using namespace std;
  
// Maximum number of digits
// in output
#define MAX 1000
  
// Function to find the factorial
// of large number and return
// them in string format
string factorial(long long n)
{
    if (n > MAX) {
        cout << " Integer Overflow"
             << endl;
        return "";
    }
  
    long long counter;
    long double sum = 0;
  
    // Base case
    if (n == 0)
        return "1";
  
    // Calculate the sum of
    // logarithmic values
  
    for (counter = 1; counter <= n;
         counter++) {
        sum = sum + log(counter);
    }
  
    // Number becomes too big to hold in
    // unsigned long integers.
    // Hence converted to string
    // Answer is sometimes under
    // estimated due to floating point
    // operations so round() is used
    string result
        = to_string(round(exp(sum)));
  
    return result;
}
  
// Driver code
int main()
{
    clock_t tStart = clock();
    string str;
    str = factorial(100);
    cout << "The factorial is: "
         << str << endl;
  
    // Calculates the time taken
    // by the algorithm to execute
    cout << "Time taken: " << setprecision(10)
         << ((double)(clock() - tStart)
             / CLOCKS_PER_SEC)
         << " s" << endl;
}

输出：

The factorial is: 93326215443944231979346762015249956831505959550546075483971433508015162170687116519232751238036777284091181469944786448222582618323317549251483571058789842944.000000
Time taken: 0.000114 s

时间复杂度： O(N) ，其中N是给定的数字。