//C6.cpp
//Adanved Euler's Methom

#include<iostream.h>

const N=20;

double f(double x,double y)
{
	return x*y*y*(-1);
}

//n為[0，5]之間欲求離散點個數，將會設為20
void EulerMethom(double x0,double y0,int n)
{
	double y1,h;
	int i;
	h=5/(double)n;
	for (i=0;i<n;i++)
	{
		y1=y0+h*f(x0+i*h,y0);
		y1=y0+h*(f(x0+(i+1)*h,y1)+f(x0+i*h,y0))/2;
		cout<<"The vaule at "<<(i+1)*h<<"is: "<<y1<<endl;
		y0=y1;
	}
}

void main()
{
	double x0=0,y0=2;   
	EulerMethom(x0,y0,N);
}

//N=500時，y(2.5)=0.275871
//N=20時，y(2.5)=0.282357，距精確值0.275862相當遠
//迭代次數對結果精確性影響非常大，N=500時，程序耗時2秒多
//且有相當的誤差，Euler法確實不敢恭維