Jam's math problem<hdoj5615>_题目描述 jam刚学会因式分解.对于ax2+bx+c,他会尝试分解为pqx2 + (qk+mp)x -优快云博客

本文介绍了一种通过编程方式解决特定形式多项式是否可以分解为两个一次因式的乘积的问题。具体而言，对于形式为 ax² + bx + c 的多项式，探讨了其是否能被分解为 (px + k)(qx + m) 的形式，其中 p, q, m, k 均为正整数。

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Problem Description

Jam has a math problem. He just learned factorization.
He is trying to factorize

ax2+bx+c into the form of

pqx2+(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≤T≤100) cases

Each case has one line,the line has

3 numbers

a,b,c(1≤a,b,c≤100000000)

Output

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

Sample Input

2
1 6 5
1 6 4

Sample Output

YES
NO
求a*x^2+b*x+c能否分解因式  暴力循环查找
pqx2+(qk+mp)x+km=(px+k)(qx+m)找出符合的p q n m
#include<cstdio>
int main()
{
	int t;
	scanf("%d",&t);	
	long long a,b,c;
	while(t--)
	{
		scanf("%lld%lld%lld",&a,&b,&c);
		int flog=0;
		for(long long i=1;i*i<=a;i++)
		{
			if(a%i!=0)
			continue;
			for(long long j=1;j*j<=c;j++)
			{
					if(c%j!=0)
			        continue;
				if(b==((i*j)+(a/i*c/j))||b==(i*(c/j)+j*(a/i)))
				{
							flog=1;
							break;
				}
		
			}
			if(flog==1)
			break;
		}
		if(flog==0)
		{
				printf("NO\n");
		}
		else
		{
				printf("YES\n");	
		}

	}
	return 0;
 }