hdoj Jam's math problem 5615 （数学十字相乘法）暴力

Jam's math problem

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

Total Submission(s): 734    Accepted Submission(s): 346


[align=left]Problem Description[/align]
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.
 

[align=left]Input[/align]
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)
 

[align=left]Output[/align]
You should output the "YES" or "NO".
 

[align=left]Sample Input[/align]

2 1 6 5 1 6 4

 

[align=left]Sample Output[/align]

YES NO

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

Jam有道数学题想向你请教一下，他刚刚学会因式分解比如说,x^2+6x+5=(x+1)(x+5)x​2​​+6x+5=(x+1)(x+5)
就好像形如 ax^2+bx+cax​2​​+bx+c => pqx^2+(qk+mp)x+km=(px+k)(qx+m)pqx​2​​+(qk+mp)x+km=(px+k)(qx+m)
但是他很蠢，他只会做p,q,m,kp,q,m,k为正整数的题目
请你帮助他，问可不可以分解

输入描述

第一行TT，表示T(1 \leq T \leq 100 )T(1≤T≤100)组数据。
接下来TT组数据：
每组数据一行，一个三个整数a,b,ca,b,c，一组数据一行 (1 \leq a,b,c \leq 100000000)(1≤a,b,c≤100000000)

输出描述

对于每组数据，输出"YES"或者"NO".

输入样例

2
1 6 5
1 6 4

输出样例

YES
NO

Hint

第一组数据可以分成(x+1)(x+5)=x^2+6*x+5(x+1)(x+5)=x​2​​+6∗x+5

//a数组存放aa的因子，c数组存放cc的因子。//直接暴力枚举所有情况。。。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define INF 0x3f3f3f3f
#define ll long long
#define N 10010
using namespace std;
struct zz
{
int x;
int y;
}a
,c
;
int main()
{
int t,aa,bb,cc;
int i,j;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&aa,&bb,&cc);
int k=0;
for(i=1;i<=sqrt(aa);i++)
{
if(aa%i==0)
{
a[k].x=i;
a[k++].y=aa/i;
}
}
int kk=0;
for(i=1;i<=sqrt(cc);i++)
{
if(cc%i==0)
{
c[kk].x=i;
c[kk++].y=cc/i;
}
}
int flag=0;
for(i=0;i<k;i++)
{
for(j=0;j<kk;j++)
{
if((a[i].x*c[j].x+a[i].y*c[j].y==bb)||(a[i].x*c[j].y+a[i].y*c[j].x==bb))
{
flag=1;
break;
}
}
}
if(flag)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}