解方程组ax^2+bx+c=0

cjliux

于 2015-03-21 21:10:27 发布

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分类专栏：简单数学题 C/C++ 文章标签： c++ 解方程

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这段代码展示了如何用C++解决二次方程ax^2 + bx + c = 0的情况，包括实数解、双实数解和复数解。程序首先检查a是否为零，然后计算判别式并根据其值输出相应的解。

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#include<iostream>
#include<cmath>
using namespace std;

int main(void)
{
	double a,b,c;
	while(cin >> a >> b >> c)
	{
		if(fabs(a)<1e-15) {
			cout << '(' <<a << ')' << "*x^2" << '+' << '(' << b << ')' << "*x"
				<< '+' << '(' << c << ')' << " = 0" 
				<<" is not a quadratic equation." << endl;
			if(fabs(b)<1e-15 && fabs(c)<1e-15) {
				cout << "Sln: x = 0" << endl;
			} else if(fabs(b)<1e-15) {
				cout << "No Sln." << endl;
			} else {	//b != 0.
				cout << "Sln: " << -c/b << endl;
			}
		} else {
			cout << '(' <<a << ')' << "*x^2" << '+' << '(' << b << ')' << "*x"
				<< '+' << '(' << c << ')' << " = 0" 
				<< " is a quadratic equation." << endl;
			double delt = b*b-4*a*c;
			double p = -b/(2*a), q = sqrt(fabs(delt))/(2*a);
			if(fabs(delt)<1e-15) {
				cout << "Sln: x1 = x2 = " << -b/(2*a) << endl;
			} else if(delt > 1e-15) {
				//double p = -b/(2*a), q = sqrt(delt)/(2*a);
				cout << "Sln: x1 = " << p+q <<", x2 = " << p-q << '.' << endl;
			} else {
				cout << "Sln: x1 = " << p << '+' << q << 'i' << ','
						<< " x2 = " << p << '-' << q << "i." << endl;
			}
		}
	}
	return 0;
}