HDOJ 5615-Jam's math problem【数学】

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 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

21 6 51 6 4

 

Sample Output

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

 

Source

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 __int64using 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;}