hdoj--5615--Jam's math problem（数学）（交叉相乘）



Jam's math problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 861    Accepted Submission(s): 411

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

我记得以前应该是这么做的，找出二次项系数跟常数项个子所有因子，然后组合，看是否可以组合产生一次项系数，，，本来以为会超时的，但是水过了 - - ||

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;int a[10000+10],b[10000+10];int main(){int t;scanf("%d",&t);while(t--){memset(a,0,sizeof(a));memset(b,0,sizeof(b));int A,B,C;scanf("%d%d%d",&A,&B,&C);int cnt1,cnt2;cnt1=cnt2=0;for(int i=1;i<=sqrt(A);i++)if(A%i==0)a[cnt1++]=i;for(int i=1;i<=sqrt(C);i++)if(C%i==0)b[cnt2++]=i;int f=false;for(int i=0;i<cnt1;i++){int p=a[i];int q=A/a[i];for(int j=0;j<cnt2;j++){int k=b[j];int m=C/b[j];if(m*p+q*k==B){f=true;break;}if(p*k+q*m==B){f=true;break;}}if(f) break;}if(f) printf("YES\n");else printf("NO\n");}return 0;}