杭电5615 Jam's math problem
2016-03-09 20:26
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Jam's math problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 821 Accepted Submission(s): 390
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 $(x+1)(x+5)$
Source
BestCoder Round #70
果不其然的A了,来来来,让我先装一波,抱紧我,我要起飞了,什么?你不懂一元二次方程的解?什么,你没有听说过十字交叉法?让赵大师教教你,对于a*x*x+b*x+c这个一元二次方程,如果解存在那么解的公式为
一元二次方程的求根公式导出过程如下:
(为了配方,两边各加
)
(化简得)。
那么对于方程(p*x-k)*(q*x-m)它的解为k/p和m/q,那么问题就来了,如果两个方程的解一样,是不是就可以说明两个方程一样?因为题目要求p,q,m,k,为整数,首先a,b,c为整数,那么2*a肯定为整数,-b也是整数,只要根号德尔塔为整数就满足了,不过得先判断德尔塔大于0 ,小于0就是没有解,而对于(p*x-k)*(q*x-m)肯定有解,所以德尔塔必定大于0,所以可得到代码:
#include<stdio.h> #include<string.h> #include<algorithm> #include<math.h> using namespace std; __int64 i,j,k,l,m,n,x,y,z; int main() { scanf("%I64d",&l); while(l--) { scanf("%I64d%I64d%I64d",&x,&y,&z); m=y*y-4*x*z; n=sqrt(m); if(m<0) { printf("NO\n"); continue; } if(n*n==m)//德尔塔开方后为整数 printf("YES\n"); else printf("NO\n"); } }
k
ax2+bx+c
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