HDU-5615(Jam's math problem)(方程求解)

HDU-5615-Jam's math problem（方程求解）

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 870    Accepted Submission(s): 415

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

Hint

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

 

感覺這道題有點坑，可能是我英語不太好吧，  p,q,m,k are positive numbers。 p,q,m,k 不應該是正數嗎？可是按正數處理，樣例都不滿足。最後看了別人的題解才知道，當作整數處理。

一元二次方程的解：x1=(-b-sqrt(b*b-4*a*c))/(2*a)、x2=(-b+sqrt(b*b-4*a*c))/(2*a)，

剛開始面對這道題也不知道要如何求解，最後看了同學的題解恍然大悟，而且方法很簡單。

x1=(-1)*(k/p),x2=(-1)*(m/q).

即：(-1)*(k/p)=(-b-sqrt(b*b-4*a*c))/(2*a)、(-1)*(m/q)=(-b+sqrt(b*b-4*a*c))/(2*a)

，因k、p、m、q是整數，(2*a)、-b也是整數，只要sqrt(b*b-4*a*c)爲整數就滿足，輸出YES，否則輸出NO.

 

My  solution:

/*2016.3.15*/

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
int main()
{
	long long a,b,c,q,p,k,m,n,h;
	double j;
	scanf("%I64d",&n);
	while(n--)
	{
		scanf("%I64d%I64d%I64d",&a,&b,&c);
		j=b*b-4*a*c;
		//m=b*b-4*a*c;
		if(j>=0)
		{
			j=sqrt(j);//q=sqrt(m) 
			h=(int)j;//強制類型轉換 
			if(j-h)//if(q*q!=m) 
			printf("NO\n");
			else
			printf("YES\n");
		}
		else
		printf("NO\n");
	}	
	return 0;
 }