I - Jam's math problem_jam's store time limit: 1000ms, special time limit-优快云博客

本文探讨了I-Jam提出的数学问题，即如何将二次多项式ax^2+bx+c分解为pqx^2+(qk+mp)x+km的形式，并确保所有系数均为正整数。文章提供了一种有效的方法来解决此问题，并通过示例说明了解决方案的有效性。

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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 $x^2+6*x+5$ 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;
}