BestCoder Round #70 Jam's math problem（hdu 5615）

Problem Description

Jam has a math problem. He just learned factorization. He is trying to factorize ax^2+bx+cax?2??+bx+c into the form of pqx^2+(qk+mp)x+km=(px+k)(qx+m)pqx?2??+(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 TT, means there are T(1 \leq T \leq 100 )T(1≤T≤100) cases

Each case has one line,the line has 33 numbers a,b,c (1 \leq a,b,c \leq 100000000)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

Copy

YES
NO

Hint

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

题意：给你一个一元二次方程的三个系数a ,b,c问你是否能用十字相乘的方法分解这个式子

题解：直接暴力枚举，当然要优化下枚举的方法，不然会超时滴，优化：因为最大数据是一亿，一亿可以分解为一万乘一万，因为这样我们分解的两个数一定不可能超过一万，先将c的所有因子存在数组中，然后在计算出a的所有因子，一个一个试即可

#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD doublea
#define MAX 10100
#define mod 10007
using namespace std;
int ans[MAX];
int main()
{
    int n,m,j,i,t,k;
    int a,b,c,Min1,Min2;
	scanf("%d",&t);
	while(t--)
	{
	    scanf("%d%d%d",&a,&b,&c);
	    Min1=min(a,10000);
	    Min2=min(c,10000);
	    k=1;n=m=1;
	    for(i=1;i<=Min2;i++)
	    {
	    	n=c/i;
	    	if(n*i==c)
	    		ans[k++]=i;
	    }
	    int flag=0;
	    for(i=1;i<=Min1;i++)
	    {
	    	m=a/i;
	    	if(i*m==a)
	    	{
	    		for(j=1;j<k;j++)
	    		{
	    			if((i*ans[j]+m*(c/ans[j])==b)||(m*ans[j]+i*(c/ans[j])==b))
	    			{
	    				flag=1;
	    				break;
	    			}
	    		}
	    	}
	    	if(flag)
	    	    break;
	    }
	    if(flag) printf("YES\n");
	    else printf("NO\n");
	}
	return 0;
}

BestCoder Round #70 Jam's math problem（hdu 5615）