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
Hint
Source
“浪潮杯”山东省第八届ACM大学生程序设计竞赛（感谢青岛科技大学）
解题感悟：这道题用到了离散数学中的知识，也就是蕴含式，前真后假为假，其余都是真。
#include<iostream>
#include<cmath>
using namespace std;
int main()
{int t,a,b,c,tmp;cin>>t;while(t--){cin>>a>>b>>c;tmp=b*b-4*a*c;if(a==0){if(b==0)if(c==0) //解可以取任意值，显然不满足条件cout<<"NO"<<endl;elsecout<<"YES"<<endl; //前件不成立else{if(c%b==0) //解为整数cout<<"YES"<<endl;else //解不是整数cout<<"NO"<<endl;}}else if(tmp<0) //无解cout<<"YES"<<endl;else{double res=sqrt(1.0*tmp);if((-b+res)/(2*a)-(int)((-b+res)/(2*a))==0&&(-b-res)/(2*a)-(int)((-b-res)/(2*a))==0) //两个解都是整数cout<<"YES"<<endl;elsecout<<"NO"<<endl;}}return 0;
}