周练D - Jam's math problem

最新推荐文章于 2022-11-28 22:48:06 发布

原创最新推荐文章于 2022-11-28 22:48:06 发布 · 364 阅读

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#Jams math problem #Jam has a math probl

本博客探讨了一个关于数学表达式因式分解的问题，通过实例解释如何判断给定的表达式是否能被正整数p、q、m、k因式分解，并提供了两种解决方法，帮助理解数学表达式的分解过程。

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D - Jam's math problem

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

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

into the form of

.
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

, means there are

cases

Each case has one line,the line has

numbers

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 $(x+1)(x+5)$

这个题我刚开始做的是用一个二维数组，把mpqk的所有情况全部列举出来，但是不知道为啥，一直wa，后来看别人代码都很简单，说是根号下德尔塔一定是整数，找了几个大神，也没解释出所以然，后来又看到另一种方法，跟我第一次的差不多，试着敲了一遍，就过了，真神奇！

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
__int64 p[10000010][2],q[10000010][2];
int main()
{
	__int64 i,j,m,a,b,c,z,x;
	scanf("%I64d",&m);
	while(m--)
	{
		scanf("%I64d%I64d%I64d",&a,&b,&c);
		int flag=0;
		for(i=1;i*i<=a;i++)
		{
			if(a%i==0)
			{
				z=a/i;
				for(j=1;j*j<=c;j++)
				{
					if(c%j==0)
					{
						x=c/j;
						if(i*j+z*x==b||i*x+j*z==b)
						{
							flag=1;
							break;
						}
					}
				}

			}
				if(flag)
				break;
		}
		if(flag)
		printf("YES\n");
		else
		printf("NO\n");
	}
	return 0;
}