HDU 5615 Jam's math problem

Problem Description

Jam has a math problem. He just learned factorization. He is trying to factorize ax^2+bx+cax​2​​+bx+c into the form of pqx^2+(qk+mp)x+km=(px+k)(qx+m)pqx​2​​+(qk+mp)x+km=(px+k)(qx+m). He could only solve the problem in which p,q,m,k are positive numbers. Please help him determine whether the expression could be factorized with p,q,m,k being postive.

Input

The first line is a number $T$, means there are $T(1\le T\le 100)$ cases

Each case has one line,the line has $3$ numbers $a,b,c(1\le a,b,c\le 100000000)$

Output

You should output the "YES" or "NO".

Sample Input

2
1 6 5
1 6 4

Sample Output

YES NO 
    
    
    
    

     
     
     
     

       Hint 
     
The first case turn 
     
     
     
     x^2+6*x+5x​2​​+6∗x+5 into 
     
     
     
     (x+1)(x+5)
     
     
     
     $(x+1)(x+5$
     
     
     
     )
    
    
    
    
    
    
    
    

     
     
     
     直接上的暴力大法，然而其实只要判断delta是不是平方数就好了
    
    
    
    
    
    
    
    

     
     
     
     
     
     
     
     
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
#include<queue>
#include<iostream>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 10;

int T, n, m, t1, t2, a, b, c;
struct point
{
    int x, y;
    point(){}
    point(int x, int y) :x(x), y(y){}
}p[maxn], q[maxn];

int main()
{
    scanf("%d", &T);
    while (scanf("%d%d%d", &a, &b, &c) != EOF, T--)
    {
        t1 = t2 = 0;
        for (int i = 1; i*i <= a; i++)
            if (a%i == 0) p[t1++] = point(i, a / i);
        for (int i = 1; i*i <= c; i++)
            if (c%i == 0) q[t2++] = point(i, c / i);
        bool flag = false;
        for (int i = 0; i < t1&&!flag; i++)
        {
            for (int j = 0; j < t2; j++)
            if (p[i].x*q[j].x + p[i].y*q[j].y == b || p[i].y*q[j].x + p[i].x*q[j].y == b)
            {
                flag = true; break;
            }
        }
        if (flag) printf("YES\n"); else printf("NO\n");
    }
    return 0;
}