杭电hdu 5615 Jam's math problem【胡搞过】

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

  
  

   
   2
1 6 5
1 6 4
  
  

 

Sample Output

  
  

   
   YES
NO


   
   

    
    

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

因为题干说：

ax​2​​+bx+c => pqx2+(qk+mp)x+km=(px+k)(qx+m)pqx^2+(qk+mp)x+km=(px+k)(qx+m)pqx​2​​+(qk+mp)x+km=(px+k)(qx+m)

如果有这样的转换，ｐ，ｋ，ｑ，ｍ四个数要正整数，由十字相乘相关理论我们能知道，如果要有p，k，q，m，那么首先要有解，所以b*b-4*a*c要>0，然而因为p，k，q，m是正整数，所以代表x1,x2都是有理数，有理数是什么鬼呢？就是解不带根号，我们知道有求根公式，其中2*a，-b都保证是自然数了，如果根号下b*b-4*a*c也保证是有理数我们就就能保证解是自然数，那么如何保证根号下b*b-4*a*c是有理数呢？那么b*b-4*a*c就是平方数啦~

然后我们根据推论，就能得到AC代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
using namespace std;
#define ll long long int
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ll a,b,c;
        scanf("%I64d%I64d%I64d",&a,&b,&c);
        double d=b*b-4*a*c;
        if(d>1e-10&&int(sqrt(d))==sqrt(d))
        {
            printf("YES\n");
        }
        else
        {
            printf("NO\n");
        }
    }
}