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
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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
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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 - 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==)break;
else
{
if(b*b-*a*c>=)
{
double t1=(-b)/a;
double t2=c/a;
double t3=sqrt(t1*t1-*t2);
double x1=(t1+t3)/;
double x2=(t1-t3)/;
if(x1==x2) cout<<fixed<<setprecision()<<x1<<endl;
else if(x1<x2)
cout<<fixed<<setprecision()<<x1<<" "<<x2<<endl;
else if(x1>x2)
cout<<fixed<<setprecision()<<x2<<" "<<x1<<endl;
}
else cout<<"NO"<<endl;
}
}
}
return ;
}