The 8th Shandong Province ACM College Students Programming Contest F question quadratic equation logical judgment + math question

Title:

Give you a proposition: "For any x, if$a⋅x^2+b⋅x+c=0$, then x is an integer." For any x, if$a⋅x^2+b⋅x+c=0$, then x is an integer.
Let you judge the proposition is true or false, output YES, NO;

analyze:

Judge all possibilities:

a	b	c	describe
0	0	0	x is any real number, eg.x = 0.1 makes the equation 0, but x is not an integer
0	0	！0	For any x, the equation is not 0, so the proposition is true
0	！0	！0	bx + c = 0; x = c/b, if c/b is an integer, x is an integer, otherwise not
！0	-	-	Unary quadratic equation, find the root formula, determine whether x1, x2 are integers

Code:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#define debug cout<<"**********"<<endl;
#define ll long long
#define yes cout<<"YES"<<endl;
#define no cout<<"NO"<<endl;
using namespace std;
const int maxn = 10000;
const int mod = 1e9+7;
int main()
{
    std::ios::sync_with_stdio(false);
    int T;
    cin>>T;
    double a,b,c;
    while(T--)
    {
        cin>>a>>b>>c;
        if(a == 0)
        {
            if(b == 0)
            {
                if(c == 0)
                {
                    no;
                }
                else
                {
                    yes;
                }
            }
            else
            {
                if((int)c%(int)b == 0)
                {
                    yes;
                }
                else
                {
                    no;
                }
            }
        }
        else
        {

            double k = b*b-4*a*c;
            if(k < 0)
            {
                yes;
            }
            else
            {
                double d = sqrt(k);
                double x1 = (-b+d)/(2*a);
                double x2 = (-b-d)/(2*a);
                int xx1 = x1;
                int xx2 = x2;
                if(x1 - xx1 ==0 && x2 - xx2 == 0)
                {
                    yes;
                }
                else
                {
                    no;
                }
            }
        }
    }
    return 0;
}