lightoj-1305 - Area of a Parallelogram【数学】【向量法求平行四边形面积】

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题目链接：点击打开链接

1305 - Area of a Parallelogram

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Time Limit: 1 second(s)

Memory Limit: 32 MB

A parallelogram is a quadrilateral with two pairs of parallel sides. See the picture below:

Fig: a parallelogram

Now you are given the co ordinates of A, B and C, you have to find the coordinates of D and the area of the parallelogram. The orientation of ABCD should be same as in the picture.

Input

Input starts with an integer T (≤ 1000), denoting the number of test cases.

Each case starts with a line containing six integers A_x, A_y, B_x, B_y, C_x, C_y where (A_x, A_y) denotes the coordinate of A, (B_x, B_y) denotes the coordinate of B and (C_x, C_y) denotes the coordinate of C. Value of any coordinate lies in the range [-1000, 1000]. And you can assume that A, B and C will not be collinear.

Output

For each case, print the case number and three integers where the first two should be the coordinate of D and the third one should be the area of the parallelogram.

Sample Input

Output for Sample Input

0 0 10 0 10 10

0 0 10 0 10 -20

-12 -10 21 21 1 40

Case 1: 0 10 100

Case 2: 0 -20 200

Case 3: -32 9 1247

题解：点到直线的距离公式啊，有木有

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
double ax,ay,bx,by,cx,cy;
int main()
{int t,text=0;scanf("%d",&t);while(t--){scanf("%lf %lf %lf %lf %lf %lf",&ax,&ay,&bx,&by,&cx,&cy);double midx=(ax+cx)/2;double midy=(ay+cy)/2;double dx=midx*2-bx;double dy=midy*2-by;
//		double base=sqrt((ax-bx)*(ax-bx)+(ay-by)*(ay-by));double height=(ax-bx)*dy-(ax-bx)*ay+(by-ay)*dx+(ay-by)*ax;double ans=abs(height);printf("Case %d: %.lf %.lf %.lf\n",++text,dx,dy,ans);}return 0;
}

涨姿势了：平行四边形的面积可以用两相邻向量求，公式就是 x1 * y2 - x2 * y1

这里有个详细的链接：点击打开链接

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