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

21 6 51 6 4

Sample Output

YESNO
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是不是完全平方数就好了。
#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;}