我将如何更清晰地编写此基本C ++程序，提高运行效率或返回更精确的数字？

Question

It is supposed to evaluate e^pi - pi.

#include <iostream>
#include <cmath>
#include <iomanip>

using namespace std;
long double Pie();
long double Factorial(double n);
long double E();

int main()
{
    long double answer = pow(E(),Pie()) - Pie();
    cout << setprecision(20);
    cout << answer;

    return 0;
}
long double Pie()
{
    long double a = 1;
    long double b = (1 / sqrtl(2));
    long double t = (1.0 / 4.0);
    long double p = 1;

    long double aPlaceholder;
    for (int i = 1; i < 5; i++)
    {
        aPlaceholder = a;
        a = (a + b) / 2;
        b = sqrtl(aPlaceholder * b);
        t = t - p * (aPlaceholder - a) * (aPlaceholder - a);
        p = 2 * p;
    }
    long double nicePie;
    nicePie = (a + b) * (a + b) / (4 * t);
    return nicePie;
}

long double E()
{
    long double e = 0;
    for(double i = 0; i < 20; i++)
    e += 1.0 / Factorial(i);
    return e;
}

long double Factorial(double n)
{
    if(n == 0)
    return 1;
    int i = n - 1;
    while (i > 0)
    {
      n *= i;
      i--;
    }
    return n;
}

The scenario is that I want to evaluate e, raise it to the power of pi, and then subtract pi from the result and then print the answer to the screen. Another aspect to the scenario is that this is a basic C++ program.

Answer 1

cmath provides π and e as predefined constants as M_PI and M_E accurate within the precision of double , but it's not mandatory by C++ standard.

You can just do double pi = acos(-1);

Answer 2

• <cmath> provides double long exp(double long) : http://www.cplusplus.com/reference/clibrary/cmath/exp/ as well as the Pi constant M_PI in double precision.
• Boost provides Pi in long double precision :
  const long double pi = boost::math::constants::pi<long double>();

That being said there is some calculus in your code that you do multiple times while it is not needed:

• Pie is called twice.
• In E() , Factorial is called at every iteration while you could multiply the previous result with i .

.

long double E()
{
  long double e = 0;
  long double fact_i = 1;
  for(double i = 1; i < 20; i++)
  {
    fact_i *= i;
    e += 1.0 / fact_i;
  }
  return e;
}