zoukankan      html  css  js  c++  java
  • Codeforces--630N--Forecast(方程求解)

    N - Forecast

    Crawling in process... Crawling failed Time Limit:500MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

    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
    ",ans1),printf("%.8lf
    ",ans2);
    	else 
    	printf("%.8lf
    ",ans2),printf("%.8lf
    ",ans1);
    }

  • 相关阅读:
    li float后IE下有空格
    [转]输入框对齐问题
    footer贴在底部的布局
    css3.0参考手册
    Java变量的命名规范
    刷题01
    前端面试题
    Cadence学习封装制作(焊盘)
    Cadence学习文档后缀简介
    Cadence学习PCB设计(序)
  • 原文地址:https://www.cnblogs.com/playboy307/p/5273412.html
Copyright © 2011-2022 走看看