【lightoj1305】 Area of a Parallelogram

喜欢ヅ旅行 2022-07-17 01:41 33阅读 0赞

B - B 圆周率用acos(-1.0) 使用longlong

**Time Limit:**1000MS **Memory Limit:**32768KB **64bit IO Format:**%lld & %llu

Submit [Status][] [Practice][] [LightOJ 1305][] [![12x12_uDebug.png][]uDebug][12x12_uDebug.png_uDebug]

Description

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

![04d453f1e094f0c5ffec3488ef1438df][]

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.

Sample Input

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

要考虑斜率不存在的情况，就是分母为0的时候，坑的我，哎，不说了。另外面积输出要四舍五入，虽然他说输入三个整数，但不能直接（int）类型转换，否则出现误差。

#include<cstdio>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    int main() {
    	int T,p=0;
    	scanf("%d",&T);
    	while(T--) {
    		int x1,x2,x3,y1,y2,y3;
    		scanf("%d%d%d%d%d%d",&x1,&y1,&x2,&y2,&x3,&y3);
    		int x4,y4;
    		x4=x1+x3-x2;
    		y4=y1+y3-y2;
    		double s,k,l,d;
    		if(x1==x2) {
    			s=(double)abs(y2-y1)*(double)abs(x4-x1);
    		} else {
    			k=(double)(y2-y1)/(double)(x2-x1);
    			l=fabs((double)(k*x4-y4-k*x1+y1))/(double)sqrt(k*k+1.0);
    			d=(double)sqrt((double)(x2-x1)*(x2-x1)+(double)(y2-y1)*(y2-y1));
    			s=l*d;
    		}
    		printf("Case %d: %d %d %.lf\n",++p,x4,y4,s);
    	}
    	return 0;
    }

[Status]: http://acm.hust.edu.cn/vjudge/contest/130068#status//B/0
[Practice]: http://acm.hust.edu.cn/vjudge/problem/27014
[LightOJ 1305]: http://acm.hust.edu.cn/vjudge/problem/27014/origin
[12x12_uDebug.png]: http://acm.hust.edu.cn/vjudge/images/12x12_uDebug.png
[12x12_uDebug.png_uDebug]: https://www.udebug.com/LOJ/1305
[04d453f1e094f0c5ffec3488ef1438df]: http://7xjob4.com1.z0.glb.clouddn.com/04d453f1e094f0c5ffec3488ef1438df