HUST 1541 Student’s question

1541 - Student’s question

时间限制：1秒 内存限制：128兆

696 次提交 134 次通过

题目描述

    YYis a student. He is tired of calculating the quadratic equation. He wants you to help him to get the result of the quadratic equation. The quadratic equation’ format is as follows: ax^2+bx+c=0.

输入

       The first line contains a single positive integer N, indicating the number of datasets. The next N lines are n datasets. Every line contains three integers indicating integer numbers a,b,c (a≠0).

输出

       For every dataset you should output the result in a single line. If there are two same results, you should just output once. If there are two different results, you should output them separated by a space. Be sure the later is larger than the former. Output the result to 2 decimal places. If there is no solution, output “NO”.

样例输入

3
1 2 1
1 2 3
1 -9 6

样例输出

-1.00
NO
0.73 8.27

题目链接：http://acm.hust.edu.cn/problem/show/1541

分析：此题坑点很多！比赛中看到有人WA了11次都没过！原因在于要取double型而非int型，取double，直接AC，虽然弱弱也WA了3次才想到这一点！

大意就是要求解方程的根，常规做法就是先判断△=b*b-4*a*c的值，小于0无解，输出NO；等于0，一解；大于0，两解！但要注意的是两个解要从小到大进行有序输出！

而如何求解x1,x2，根据韦达定理得：x1＋x2=－b/a，x1*x2=c/a去求解x1-x2=sqrt((x1+x2)*(x1+x2)-4*x1*x2)，然后就可以求出x1,x2啦！还要记得保留两位小数哟！

下面给出AC代码：

#include <bits/stdc++.h>
using namespace std;
int main()
{
    double n,a,b,c;
    while(cin>>n)
    {
        while(n--)
        {
            cin>>a>>b>>c;
            if(a==0)break;
            else
            {
            if(b*b-4*a*c>=0)
            {
                double t1=(-b)/a;
                double t2=c/a;
                double t3=sqrt(t1*t1-4*t2);
                double x1=(t1+t3)/2;
                double x2=(t1-t3)/2;
                if(x1==x2) cout<<fixed<<setprecision(2)<<x1<<endl;
                else if(x1<x2)
                cout<<fixed<<setprecision(2)<<x1<<" "<<x2<<endl;
                else if(x1>x2)
                    cout<<fixed<<setprecision(2)<<x2<<" "<<x1<<endl;
            }
            else cout<<"NO"<<endl;
            }
        }
    }
    return 0;
}