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

 1 #include <iostream>
 2 #include <math.h>
 3
 4 using namespace std;
 5
 6 double _abs(double a)
 7 {
 8     if (a < 0)
 9         return -a;
10     return a;
11 }
12
13 int main()
14 {
15     ios::sync_with_stdio(false);
16     int t;
17     cin >> t;
18     while (t--)
19     {
20         double a, b, c;
21         cin >> a >> b >> c;
22         if (!a && !b)
23         {
24             if (!c)
25                 cout << "NO" << endl;
26             else
27                 cout << "YES" << endl;
28         }
29         else if (!a && b)
30         {
31             if (_abs(c/b - (int)(c/b)) > 0.0000001)
32                 cout << "NO" << endl;
33             else
34                 cout << "YES" << endl;
35         }
36         else
37         {
38             if ((b * b - 4 * a * c) >= 0)
39             {
40                 double x1 = (-b + sqrt(b*b - 4 * a*c)) / (2 * a);
41                 double x2 = (-b - sqrt(b*b - 4 * a*c)) / (2 * a);
42                 //cout << x1 << ends << x2 << endl;
43                 if (_abs(x1 - (int)x1) > 0.0000001 || _abs(x2 - (int)x2) > 0.0000001)
44                     cout << "NO" << endl;
45                 else
46                     cout << "YES" << endl;
47             }
48             else
49                 cout << "YES" << endl;
50         }
51     }
52
53     return 0;
54 }

原文地址：https://www.cnblogs.com/qq965921539/p/9360727.html