C++ Chudnovsky Formula for Pi
문제
I am trying to make a C++ application calculate pi for me. I have tried to implement the Chudnovsky formula with no luck.
Here is my code:
#include <iostream>
#include <cmath>
long fac(long num) {
if (num == 1)
return 1;
return fac(num - 1) * num;
}
int main() {
using namespace std;
double pi;
for (long k = 0; k < 10; k++) {
pi += (pow(-1, k) * fac(6 * k) * (13591409 + (545140134 * k))) / (fac(3 * k) * pow(fac(k), 3) * pow(640320, 3 * k + 3/2));
}
pi *= 12;
cout << 1 / pi << endl;
system("pause");
return 0;
}
The goal of this was to have the program output 10 iterations of the Chudnovsky formula. Instead, I got this:
call of overloaded `pow(int, long int&)' is ambiguous
해결책
You never initialize pi
, so your code has undefined behaviour.
Your fac
function does not correctly handle 0
(fac(0)
should be 1
).
3/2
evaluates to 1
(because it uses integer division, which truncates), and this makes your formula evaluate to the completely wrong answer.
다른 팁
You almost certainly want to do all the math on doubles, to avoid lot of time-killing conversions. You probably also want to use an iterative implementation of fac
instead of a recursive one (not that recursion is going to be a big problem, but this is a prime example of when recursion should really be avoided because it gains you nothing). Of course you also need to initialize pi
as others have already pointed out.
#include <iostream>
#include <iomanip>
#include <cmath>
double fac(double num) {
double result = 1.0;
for (double i=2.0; i<num; i++)
result *= i;
return result;
}
int main() {
using namespace std;
double pi=0.0;
for (double k = 0.0; k < 10.0; k++) {
pi += (pow(-1.0,k) * fac(6.0 * k) * (13591409.0 + (545140134.0 * k)))
/ (fac(3.0 * k) * pow(fac(k), 3.0) * pow(640320.0, 3.0 * k + 3.0/2.0));
}
pi *= 12.0;
cout << setprecision(15) << 1.0 / pi << endl;
return 0;
}
pow(-1, k) is ineffective as it is a direct translation from math formula to the code.
Use this instead:
(k%2==1?-1.0:1.0)*fac(...
Edit:
Also your fac code is far from optimal too.