细节

With given integers a,b,c, you are asked to judge whether the following statement is true: "For any x, if a x² +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


题意：
给你三个数,a,b,c问这三个数构成的一元二次方程有没有非整数根，（如果有根，则一定整数，如果命题成立，输出yes 否则输出no），没有根的情况为yes。

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define minn 1e-8
using namespace std;

int main()
{
    int t,a,b,c,ans;
    double ans1;
    cin>>t;
    while(t--)
    {
        cin>>a>>b>>c;
        if(a==0)
        {
            if(b!=0)
            {
                if(c%b!=0)
                    printf("NO
");
                else
                    printf("YES
");
            }
            else
            {
                if(c==0)
                    printf("NO
");
                else
                    printf("YES
");
            }
        }
        else
        {
            double der=b*b-4*a*c;
            if(der<0)
            {
                printf("YES
");

            }
            else
            {
                double dd=sqrt(der);
                int dd1=int(sqrt(der));
                if(abs(dd1-dd)<minn&&((-b+dd1)%(2*a)==0&&(-b-dd1)%(2*a)==0))
                    printf("YES
");
                else
                    printf("NO
");
            }

        }

    } return 0;
}