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：acm.hust.edu.cn/problem/sho…

  • 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：

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 int main()
     4 {
     5     double n,a,b,c;
     6     while(cin>>n)
     7     {
     8         while(n--)
     9         {
    10             cin>>a>>b>>c;
    11             if(a==0)break;
    12             else
    13             {
    14             if(b*b-4*a*c>=0)
    15             {
    16                 double t1=(-b)/a;
    17                 double t2=c/a;
    18                 double t3=sqrt(t1*t1-4*t2);
    19                 double x1=(t1+t3)/2;
    20                 double x2=(t1-t3)/2;
    21                 if(x1==x2) cout<<fixed<<setprecision(2)<<x1<<endl;
    22                 else if(x1<x2)
    23                 cout<<fixed<<setprecision(2)<<x1<<" "<<x2<<endl;
    24                 else if(x1>x2)
    25                     cout<<fixed<<setprecision(2)<<x2<<" "<<x1<<endl;
    26             }
    27             else cout<<"NO"<<endl;
    28             }
    29         }
    30     }
    31     return 0;
    32 }