I - Jam's math problem

I - Jam’s math problem

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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Status

Description

Jam has a math problem. He just learned factorization.

He is trying to factorize ax^2+bx+c into the form of 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 T, means there are T(1 \leq T \leq 100 ) cases

Each case has one line,the line has 3 numbers a,b,c (1 \leq a,b,c \leq 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 x2+6∗x+5 into (x+1)(x+5)

题意：求二元一次方程根的个数。

思路：用图像法：当x=0时，f(x)=ax^2+bx+c=c>0,又a>0,可以简单画出图像，要满足题意需图像与x轴有交点，根据根的判别式判断，题中还有一要求，就是p,q,k,m都是整数，根据求根公式可得sqrt(b*b-4*a*c)一定要为整数。（数学公式不好打呀）

失误：第一次没有用数学方法，超时；第二次没考虑到根式必须开尽。看来数学题还是都有规律，都是特殊情况的，应该推出公式。

代码如下：

正确代码：
#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;

int main()
{
__int64 t,a,b,c,s,ans;
cin>>t;

while(t--)
{
cin>>a>>b>>c;

s=b*b-4*a*c;//有根
ans=sqrt(s);//根式开的尽

if(s>=0&&ans*ans==s)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}

超时代码：

#include<cstdio>
#include<iostream>
using namespace std;

int main()
{
__int64 t,p,q,a,b,c,flag,k,m;
cin>>t;

while(t--)
{
cin>>a>>b>>c;

flag=0;
for(p=1;p<=a;++p)
{
if(a%p!=0)
continue;

for(k=1;k<=c;++k)
{
if(c%k!=0)
continue;

q=a/p;
m=c/k;
if(q*k+m*p==b)
{
flag=1;
break;
}
}

if(flag==1)
break;
}

if(flag)
{
cout<<"YES"<<endl;
}

else{
cout<<"NO"<<endl;
}
}
return 0;
}