Codeforces--630N--Forecast（方程求解）

最新推荐文章于 2021-08-15 21:01:35 发布

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最新推荐文章于 2021-08-15 21:01:35 发布

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原文链接：http://www.cnblogs.com/playboy307/p/5273412.html

本文介绍了一种解决特定形式的一元二次方程的方法，该方程有两个实数根，通过计算公式得出乐观和悲观预测对应的解，并提供了一个C++实现示例。

N - Forecast

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Description

The Department of economic development of IT City created a model of city development till year 2100.

To prepare report about growth perspectives it is required to get growth estimates from the model.

To get the growth estimates it is required to solve a quadratic equation. Since the Department of economic development of IT City creates realistic models only, that quadratic equation has a solution, moreover there are exactly two different real roots.

The greater of these roots corresponds to the optimistic scenario, the smaller one corresponds to the pessimistic one. Help to get these estimates, first the optimistic, then the pessimistic one.

Input

The only line of the input contains three integers a, b, c ( - 1000 ≤ a, b, c ≤ 1000) — the coefficients of ax² + bx + c = 0 equation.

Output

In the first line output the greater of the equation roots, in the second line output the smaller one. Absolute or relative error should not be greater than 10^- 6.

Sample Input

Input

1 30 200

Output

-10.000000000000000
-20.000000000000000

给出了一个一元二次方程A*x^2+b*x+c=0求两个根，从小到大输出

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int main()
{
	double a,b,c;
	double ans1,ans2;
	cin>>a>>b>>c;
	ans1=(-1*b+sqrt(b*b-4*a*c))/(2*a);
	ans2=(-1*b-sqrt(b*b-4*a*c))/(2*a);
	if(ans1>ans2)
	printf("%.8lf\n",ans1),printf("%.8lf\n",ans2);
	else 
	printf("%.8lf\n",ans2),printf("%.8lf\n",ans1);
}

转载于:https://www.cnblogs.com/playboy307/p/5273412.html