HDU-5615-Jam's math problem（方程求解）

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 870    Accepted Submission(s): 415

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 $(x+1)(x+5)$
 
感觉这道题有点坑，可能是我英语不太好吧，  p,q,m,k are positive numbers。    p,q,m,k 不应该是正数吗？可是按正数处理，样例都不满足。最后看了别人的题解才知道，当作整数处理。
一元二次方程的解：x1=(-b-sqrt(b*b-4*a*c))/(2*a)、x2=(-b+sqrt(b*b-4*a*c))/(2*a)，
刚开始面对这道题也不知道要如何求解，最后看了同学的题解恍然大悟，而且方法很简单。
x1=(-1)*(k/p),x2=(-1)*(m/q).
即：(-1)*(k/p)=(-b-sqrt(b*b-4*a*c))/(2*a)、(-1)*(m/q)=(-b+sqrt(b*b-4*a*c))/(2*a)
，因k、p、m、q是整数，(2*a)、-b也是整数，只要sqrt(b*b-4*a*c)为整数就满足，输出YES，否则输出NO.

 

My  solution:

/*2016.3.15*/

#include<stdio.h>#include<math.h>#include<algorithm>using namespace std;int main(){long long a,b,c,q,p,k,m,n,h;double j;scanf("%I64d",&n);while(n--){scanf("%I64d%I64d%I64d",&a,&b,&c);j=b*b-4*a*c;//m=b*b-4*a*c;if(j>=0){j=sqrt(j);//q=sqrt(m) h=(int)j;//强制类型转换 if(j-h)//if(q*q!=m) printf("NO\n");elseprintf("YES\n");}elseprintf("NO\n");}return 0; }