第八届山东ACM省赛F题-quadratic equation

这个题困扰了我长达1年多，终于在今天下午用两个小时理清楚啦

要注意的有以下几点：

1.a=b=c=0时 因为x有无穷种答案，所以不对

2.注意精度问题

3.b^2-4ac<0时也算对

Problem Description

With given integers a,b,c, you are asked to judge whether the following statement is true: "For any x, if a⋅+b⋅x+c=0, then x is an integer."

Input

The first line contains only one integer T(1≤T≤2000), which indicates the number of test cases.
For each test case, there is only one line containing three integers a,b,c(−5≤a,b,c≤5).

Output

or each test case, output “YES” if the statement is true, or “NO” if not.

Sample Input

3
1 4 4
0 0 1
1 3 1

Sample Output

YES
YES
NO

 #include <iostream>
 #include <math.h>

 using namespace std;

 double _abs(double a)
 {
     if (a < )
         return -a;
     return a;
 }

 int main()
 {
     ios::sync_with_stdio(false);
     int t;
     cin >> t;
     while (t--)
     {
         double a, b, c;
         cin >> a >> b >> c;
         if (!a && !b)
         {
             if (!c)
                 cout << "NO" << endl;
             else
                 cout << "YES" << endl;
         }
         else if (!a && b)
         {
             if (_abs(c/b - (int)(c/b)) > 0.0000001)
                 cout << "NO" << endl;
             else
                 cout << "YES" << endl;
         }
         else
         {
             if ((b * b -  * a * c) >= )
             {
                 double x1 = (-b + sqrt(b*b -  * a*c)) / ( * a);
                 double x2 = (-b - sqrt(b*b -  * a*c)) / ( * a);
                 //cout << x1 << ends << x2 << endl;
                 if (_abs(x1 - (int)x1) > 0.0000001 || _abs(x2 - (int)x2) > 0.0000001)
                     cout << "NO" << endl;
                 else
                     cout << "YES" << endl;
             }
             else
                 cout << "YES" << endl;
         }
     }

     return ;
 }