hdoj--5615--Jam's math problem（数学）（交叉相乘）_三角函数交叉相乘csdn-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_29963431/article/details/50898490

本文探讨了一个特定的数学问题，即如何将形式为ax²+bx+c的二次方程分解为pqx²+(qk+mp)x+km=(px+k)(qx+m)的形式，特别关注于寻找正整数p、q、m和k的情况下方程是否可分解。

Jam's math problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 861 Accepted Submission(s): 411

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

Source

BestCoder Round #70

我记得以前应该是这么做的，找出二次项系数跟常数项个子所有因子，然后组合，看是否可以组合产生一次项系数，，，本来以为会超时的，但是水过了 - - ||

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int a[10000+10],b[10000+10];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		int A,B,C;
		scanf("%d%d%d",&A,&B,&C);
		int cnt1,cnt2;
		cnt1=cnt2=0;
		for(int i=1;i<=sqrt(A);i++)
		if(A%i==0)
		a[cnt1++]=i;
		
		for(int i=1;i<=sqrt(C);i++)
		if(C%i==0)
		b[cnt2++]=i;
		
		int f=false;
		for(int i=0;i<cnt1;i++)
		{
			int p=a[i];
			int q=A/a[i];
			for(int j=0;j<cnt2;j++)
			{
				int k=b[j];
				int m=C/b[j];
				if(m*p+q*k==B)
				{
					f=true;
					break;
				}
				if(p*k+q*m==B)
				{
					f=true;
					break;
				}
			}
			if(f) break;
		}
		if(f) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}