Jam's math problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 990 Accepted Submission(s): 474

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

21 6 51 6 4

Sample Output

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

Source

/*思路：题目意思是判定 ax2+bx+c能不能分解成 pqx2+(qk+mp)x+km=(px+k)(qx+m)，其中参数都是正整数  即十字相乘， 刚看到这个题目第一感觉就是把复杂度降下来，枚举p,k就行，两重循环，刚开始误以为p枚举1到sqrt(a)就行 ，这样是错误的，因为p,q代表不同角色，可能是大小反过来，所以枚举的 时候i若能被a整除，则另一个p是a/i这样才对。参考网上思路1： 为了降低复杂度，将2个循环先分开处理一下，将不能整除的先去掉，这样就行 参考网上思路2：对于形如ax2+bx+c的多项式，在判定它能否使用十字分解法分解因式时，可以使用Δ=b2-4ac进行判定。当Δ为完全平方数时，可以在整数范围对该多项式进行十字相乘。*//*解法1*//*#include 
    
     #include 
     
      using namespace std;const int N=2*10000+5;int nump[N],numk[N];int lenp,lenk;int main(){		int t,a,b,c,p,q,k,m;	int n1,n2;	cin>>t;	while(t--){		cin>>a>>b>>c;		n1=sqrt(a);		n2=sqrt(c);		bool ok=0;		lenp=0;		lenk=0;	  for(p=1;p<=n1;p++){	     if(a%p==0){	    	nump[lenp++]=p;	    		nump[lenp++]=a/p; 		 }	  }	  for(k=1;k<=n2;k++){	     if(c%k==0){	    	numk[lenk++]=k;	    		numk[lenk++]=c/k; 		 }	  }	  for(int i=0;i
      
       #include 
       
        using namespace std;int main(){		long long a,b,c; 	int t;	cin>>t;	while(t--){		bool ok=0;		cin>>a>>b>>c;		long long tmp=sqrt(b*b-4*a*c);		if(tmp*tmp==b*b-4*a*c)	            ok=1;		if(ok)		 cout<<"YES"<