HDOJ 5615 Jam's math problem

最新推荐文章于 2018-07-30 19:51:08 发布

纯真zwj

最新推荐文章于 2018-07-30 19:51:08 发布

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本文链接：https://blog.youkuaiyun.com/zwj1452267376/article/details/50613031

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本文介绍了一个数学问题，即如何将形式为ax²+bx+c的多项式因式分解为(px+k)(qx+m)的形式，并给出了具体的解决方法及示例代码。

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Jam's math problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 182 Accepted Submission(s): 92

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

Sample Output

  
  
   
   YES
NO


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

题意： ax^2+bx+c ==> qpx^2+(qk+mp)x+km == (px+k)*(qx+m) 给出a，b，c；问是否存在正整数p，q，

m，k。

题解：枚举p，q，k，m啊，再判断（qk+mp）||（qm+pk）是否等于b就行了。啊啊啊啊啊啊！！！！！！这么简单我都没想到，没救了。。。。

代码如下：

#include<cstdio>
#include<cstring>
#include<cmath>
int main()
{
	int t,i,j,a,b,c,q,p,k,m,flag;
	scanf("%d",&t);
	while(t--)
	{
		flag=0;
		scanf("%d%d%d",&a,&b,&c);
		for(i=1;i<=sqrt(a*1.0);++i)
		{
			if(a%i==0)
			{
				p=i;
				q=a/p;
				for(j=1;j<=sqrt(c*1.0);++j)
				{
					if(c%j==0)
					{
						k=j;
						m=c/k;
						if(k*q+m*p==b||k*p+m*q==b)
						{
							flag=1;
							break;
						}
					}
				}
			}
			if(flag)
				break;
		}
		if(flag)
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}