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  • LightOJ 1305Area of a Parallelogram

    Description

    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 ofABCD 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 Ax, Ay, Bx, By, Cx, Cy where (Ax, Ay) denotes the coordinate of A(Bx, By) denotes the coordinate of B and (Cx, Cy) denotes the coordinate of C. Value of any coordinate lies in the range [-1000, 1000]. And you can assume that A, Band 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

    3

    0 0 10 0 10 10

    0 0 10 0 10 -20

    -12 -10 21 21 1 40

    Sample Output

    Case 1: 0 10 100

    Case 2: 0 -20 200

    Case 3: -32 9 1247

     每个点的相对位置是一定的  坐标好求   对角的坐标相加是相等的
    求面积 直接求三角型ABC面积*2  
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<queue>
    #include<algorithm>
    using namespace std;
    int main()
    {
    	int t;
    	scanf("%d",&t);
    	int ans=0;
    	while(t--)
        {
        	ans++;
         double x1,y1,x2,y2,x3,y3;
        	scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3);
           int	x4=(int)x1+x3-x2;
        	int y4=(int)y1+y3-y2;
            double c=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
            double a=sqrt((x2-x3)*(x2-x3)+(y2-y3)*(y2-y3));
            double b=sqrt((x1-x3)*(x1-x3)+(y1-y3)*(y1-y3));
    	   double s=abs(sqrt(1-(a*a+b*b-c*c)/(2*a*b)*(a*a+b*b-c*c)/(2*a*b))*a*b);
        	printf("Case %d: %d %d %.0lf
    ",ans,x4,y4,s);
    	}
    	return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/kingjordan/p/12027026.html
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