LightOJ 1305-Area of a Parallelogram【几何】

我不是女神ヾ 2022-08-21 13:39 3阅读 0赞

1305 - Area of a Parallelogram

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   <td style="color:rgb(121,114,104); background:rgb(245,250,250)">Time Limit:&nbsp;<span style="color:rgb(180,95,4)">1 second(s)</span></td> 
   <td style="color:rgb(121,114,104); background:rgb(245,250,250)">Memory Limit:&nbsp;<span style="color:rgb(180,95,4)">32 MB</span></td> 
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A parallelogram is a quadrilateral with two pairs of parallel sides. See the picture below:

![a.png][]

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 **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, 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.

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   <td> <h1 style="color:green">Sample Input</h1> </td> 
   <td> <h1 style="color:green">Output for Sample Input</h1> </td> 
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   <td> <p>3</p> <p>0 0 10 0 10 10</p> <p>0 0 10 0 10 -20</p> <p>-12 -10 21 21 1 40</p> </td> 
   <td> <p>Case 1: 0 10 100</p> <p>Case 2: 0 -20 200</p> <p>Case 3: -32 9 1247</p> </td> 
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--------------------

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;
    }

[a.png]: /images/20220724/c5690eb06fe14b64aa5cf6683ea1b893.png