CodeForces - 630N Forecast （解一元二次方程组）

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本文介绍了一种解决一元二次方程的方法，并通过具体的代码实现了方程ax² + bx + c = 0的求解过程。文章强调了输出根时应先输出较大的根。

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CodeForces - 630N

Forecast

Time Limit: 500MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

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 ofax² + 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

Source

Experimental Educational Round: VolBIT Formulas Blitz

//题意：

给出一元二次方程组，a*x*x+b*x+c=0，给出a,b,c，求解此方程组。

注意：输出时大的一个解首先输出。。。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define INF 0x3f3f3f3f
#define ll long long
#define N 10010
#define M 1000000007
using namespace std;
int main()
{
	int a,b,c;
	while(scanf("%d%d%d",&a,&b,&c)!=EOF)
	{
		double p=(double)sqrt(b*b-4*a*c);
		double x1=(-1.0*b+p)/(2*a);
		double x2=(-1.0*b-p)/(2*a);
		if(x1<x2)
			printf("%.7lf\n%.7lf\n",x2,x1);
		else
			printf("%.7lf\n%.7lf\n",x1,x2);
	}
	return 0;
}