hello.cpp

这是一个利用VC++编写的求解极值的多元的二次方程程序

#include <iostream.h>
#include "quad.h"

void PrintResults(quadratic &gook);
void Intersection(quadratic &mutha, quadratic &phuqa);
int main( void )
{
A:	char choice;
	double a, b, c, d, e, f;
//	double domain;
	cout << "*** QUADRATIC EQUATION CALCULATOR ***" << endl;
	cout << "           ax^2 + bx + c" << endl;
	cout << "Please enter the \'a\' variable now: ";
	cin >> a;
	cout << "Please enter the \'b\' variable now: ";
	cin >> b;
	cout << "Please enter the \'c\' variable now: ";
	cin >> c;
	quadratic shihdog(a, b, c);
	cout << "           " << shihdog.get_a() << "x^2 + " << shihdog.get_b() << "x + " << shihdog.get_c() << endl;
	PrintResults(shihdog);
//	Break for Calculation of Range
//	cout << endl << "Please enter a Domain variable (any x integer): ";
//	cin >> domain;
	cout << endl << "--You will now enter in a second equation in same format--" << endl;
	cout << "Please enter another \'a\' variable now: ";
	cin >> d;
	cout << "Please enter another \'b\' variable now: ";
	cin >> e;
	cout << "Please enter another \'c\' variable now: ";
	cin >> f;
	quadratic shihdawg(d, e, f);
	cout << "           " << shihdawg.get_a() << "x^2 + " << shihdawg.get_b() << "x + " << shihdawg.get_c() << endl;
	PrintResults(shihdawg);
	Intersection(shihdog, shihdawg);
	cout << "Would you like to enter new data (y/n)? ";
Z:	cin >> choice;
	switch (choice)
	{
		case 'y':
			goto A;
			break;
		case 'n':
			cout << "***GOODBYE***";
			getchar();
			break;
		default:
			goto Z;
			break;
	}
}

void PrintResults(quadratic &gook)
{
//	Break For Xint
	cout << endl << "X-Intercepts: ";
	if (gook.numxintercept() == 0)
		cout << "NONE" << endl;
	else if (gook.numxintercept() == 1)
		cout << "x=" << gook.xint1() << endl;
	else
		cout << "x=" << gook.xint1() << ", x=" << gook.xint2() << endl;
//	Break For Vertex
	cout << "      Vertex: (" << gook.xvertex() << "," << gook.yvertex() << ")" << endl;
//	Break for Min or Max BS
	cout << " f(x) status: " << gook.min_or_max() << endl;
}

void Intersection(quadratic &mutha, quadratic &phuqa)
{
	cout << endl;
	if (!mutha.hasintercepts(phuqa))
		cout << "NO POINTS OF INTERSECTION" << endl;
	else
		cout << "Intersection Points: (" << mutha.xintercept1(phuqa) << "," << mutha.yintercept1(phuqa) << ") (" << mutha.xintercept2(phuqa) << "," << mutha.yintercept2(phuqa) << ")" << endl;
}