HDU 5615 Jam's math problem <数学>

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最新推荐文章于 2024-07-21 14:30:00 发布

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本文介绍了一个具体的数学问题——将特定形式的二次多项式进行因式分解的方法。文章详细阐述了如何通过寻找合适的正整数因子来判断是否能将给定的二次多项式分解为两个一次多项式的乘积。

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Jam's math problem

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

Jam has a math problem. He just learned factorization.
He is trying to factorize

into the form of

.
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

, means there are

cases

Each case has one line,the line has

numbers

Output

You should output the "YES" or "NO".

Sample Input

Sample Output

    
     YES
NO

Hint

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

Source

BestCoder Round #70

代码：

#include<cstdio>
#define LL long long
void solve()
{
     LL a,b,c;
     scanf("%lld%lld%lld",&a,&b,&c);
     LL i,j;
     int ga=0,gb=0;
     int a1[35000],a2[35000],b1[35000],b2[35000];
     for (i=1;i*i<=a;i++)
        if (a%i==0)
        {
            a1[ga]=i;
            a2[ga++]=a/i;
        }
    for (i=1;i*i<=c;i++)
    {
        if (c%i==0)
        {
            b1[gb]=i;
            b2[gb++]=c/i;
        }
    }


    LL aa,bb;

    for (i=0;i<ga;i++)
    for (j=0;j<gb;j++)
    {
        aa=a1[i]*b1[j];
        bb=a2[i]*b2[j];
        if (aa+bb==b||aa-bb==b||bb-aa==b||-aa-bb==c)
        {
            printf("YES\n");
            return;
        }
        aa=a1[i]*b2[j];
        bb=a2[i]*b1[j];
        if (aa+bb==b||aa-bb==b||bb-aa==b||-aa-bb==c)
        {
            printf("YES\n");
            return;
        }
    }
    printf("NO\n");
    return ;
}
int main()
{
    int t;scanf("%d",&t);
    while (t--)
    {
        solve();
    }
    return 0;
}