hdu5615(基础)

Jam's math problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 881    Accepted Submission(s): 423

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

//hdu5615题目大意:对于一个方程ax^2+bx+c=0;问是否可以通过分解a、c的因子.将方程化为(px+m)(qx+k)=0的形式.//且(m、k、p、q大于零,a=p*q,c=m*k)//解题思路:首先由于要求p,q,m,k都大于零,所以可以可以从1~sqrt(a)和1~sqrt(c)枚举a和c的因子。//注：(a、c范围较大，所以就像判断素数的原理一样,只要枚举前i=sqrt(a)个数,后面的也已经通过a/i得到了)//看是否有满足 (p*m+q*k)==b或(p*k+q*m)==b;有则存在解，输出YES,否则输出NO. #include<stdio.h>#include<math.h>int main(){    int t,a,b,c,u,n;    int p,q,m,k,flag;scanf("%d",&t);while(t--){  flag=0;  scanf("%d%d%d",&a,&b,&c);  n=sqrt(a)+1;  u=sqrt(c)+1;  for(int i=1;i<=n;i++)    //枚举a的前sqrt(a)项因子   {   if(flag==1)  break;   //已经找到解，跳出循环。    if(a%i==0)   {     p=i;q=a/i;       for(int j=1;j<=u;j++)   //枚举c的前sqrt(c)项因子     {     if(c%j==0)     {       m=j;k=c/j;        if((p*k+q*m)==b||(q*k+p*m)==b)   //满足条件，flag=1,跳出循环。         {        flag=1;break;        }     }   }     }  }   if(flag==1) printf("YES\n");  else        printf("NO\n");}return 0;}