HDOJ 5615 Jam's math problem (数学题枚举)

Jam's math problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 298    Accepted Submission(s): 160

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

   
   
   
   

    
    
    
    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：查看能不能分解那个二元一次方程 思路：暴力枚举就好了，开始没想到这样能过就胡乱写了一个。。。。 ac代码： 
     
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
int main()
{
    int t;
    ll a,b,c,i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d%I64d%I64d",&a,&b,&c);
        int bz=0;
        for(i=1;i*i<=a;i++)
        {
            if(a%i)
            continue;
            for(j=1;j*j<=c;j++)
            {
                if(c%j)
                continue;
                if(i*j+(a/i)*(c/j)==b||i*(c/j)+j*(a/i)==b)
                {
                    bz=1;
                    break;
                }
            }
            if(bz)
            break;
        }
        if(bz)
        printf("YES\n");
        else
        printf("NO\n");
    }
    return 0;
}