递归--n阶勒让德多项式的值

#include <iostream>

int main()
{
	double pfunc(double x,int n);

	double xparam;
	int n;
	std::cout<<"input the parameter x and nubmer n: ";
	std::cin>>xparam>>n;
	
	std::cout<<"the result of the function is: "<<pfunc(xparam,n)<<std::endl;
	return 0;
}

double pfunc(double x,int n)
{
	double p;
	if (n==0) p=1;
	else if (n==1) p=x;
	else
	{
		p=((2*n-1)*x*pfunc(x,n-1)-(n-1)*pfunc(x,n-2))/n;
	}

	return p;
}#include <iostream>

int main()
{
	double pfunc