HDOJ 5615-Jam's math problem【数学】

心已赠人 2022-08-20 04:24 6阅读 0赞

# Jam's math problem #

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 196 Accepted Submission(s): 102

Problem Description

Jam has a math problem. He just learned factorization.  
He is trying to factorize  ax![Image 1][]2![Image 1][]\+bx\+c![Image 1][] into the form of  pqx![Image 1][]2![Image 1][]\+(qk\+mp)x\+km=(px\+k)(qx\+m)![Image 1][].  
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![Image 1][], means there are  T(1≤T≤100)![Image 1][] cases  
  
Each case has one line,the line has  3![Image 1][] numbers  a,b,c(1≤a,b,c≤100000000)![Image 1][]

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 $(x+1)(x+5)$

Source

[BestCoder Round \#70 ][BestCoder Round _70]

解题思路：

题目大意就是给出二元一次函数式的a，b，c问该式子能否被因式分解，初中学的十字交叉法，暴力就行了。

#include<stdio.h>
    #include<string.h>
    #include<cmath>
    #include<algorithm>
    #define LL __int64
    using namespace std;
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--)
        {
            bool wc=false;
            LL a,b,c;
            scanf("%I64d%I64d%I64d",&a,&b,&c);
            if((b*b-4*a*c)<0)
            {
                printf("NO\n");
            } 
            else
            {
                LL i,j;
                for(i=1;i<=sqrt(a);i++)
                {
                    for(j=1;j<=sqrt(c);j++)
                    {
                        if(a%i==0&&c%j==0)
                        {
                            LL xx,yy,uu,ii;
                            xx=i,yy=a/i,uu=j,ii=c/j;
                            if((xx*ii+yy*uu)==b||(xx*uu+yy*ii)==b)
                            {
                                wc=true;
                                break;
                            }
                        }
                    }
                    if(wc==true)
                    break;
                }
                if(wc==true)
                printf("YES\n");
                else
                printf("NO\n");
            }
        }
        return 0;
    }

[Image 1]: 
[BestCoder Round _70]: http://acm.hdu.edu.cn/search.php?field=problem&key=BestCoder+Round+%2370&source=1&searchmode=source