LightOJ 1305-Area of a Parallelogram【几何】

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
3 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

 

PROBLEM SETTER: JANE ALAM JAN

解题思路：

输入时逆时针输入点的坐标，输出最后一个点。

第四点的寻找方法：x1+x3-x2，y1+y3-y2.

<pre name="code" class="cpp">#include<stdio.h>
#include<cmath>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
	int x,y;
}d[3];
int main()
{
	int t;
	scanf("%d",&t);
	int tt=0;
	while(t--)
	{
		scanf("%d%d%d%d%d%d",&d[0].x,&d[0].y,&d[1].x,&d[1].y,&d[2].x,&d[2].y);
		printf("Case %d: ",++tt);
		printf("%d %d ",d[0].x+d[2].x-d[1].x,d[0].y+d[2].y-d[1].y);
		double a,b,c;
		a=sqrt((d[0].x-d[1].x)*(d[0].x-d[1].x)+(d[0].y-d[1].y)*(d[0].y-d[1].y));
		b=sqrt((d[2].x-d[1].x)*(d[2].x-d[1].x)+(d[2].y-d[1].y)*(d[2].y-d[1].y));
		c=sqrt((d[0].x-d[2].x)*(d[0].x-d[2].x)+(d[0].y-d[2].y)*(d[0].y-d[2].y));
		double p=(a+b+c)/2; 
		double S=sqrt(p*(p-a)*(p-b)*(p-c));
		printf("%.0lf\n",S*2);
	}
	return 0;
}