| 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 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 |
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 |
【题解】题中没有限定第四个点的位置吗？？？？？不是这样的，细心地人不难发现1+3-2就是第四个点，然后可以根据三角形的面积公式进行计算。
公式一：S = abs((x2-x1)*(y3-y1)-(x3-x1)*(y2-y1)) / 2;
公式二：S = ah÷2 三角形的面积=底×高÷2；
公式三：S = ab×sin×1/2 三角形任意两边之积×这两边的夹角的正弦值÷2；
公式四：2、海伦公式:S = √[p(p-a)(p-b)(p-c) ]其中p=1/2(a+b+c)；
公式五：s=1/2的周长*内切圆半径；
在这一题中建议采用第一种方式，相对而言来说最简单！！！！！！
AC代码：
1 #include<cstdio> 2 #include<cmath> 3 #include<algorithm> 4 using namespace std; 5 struct node{ 6 double x, y; 7 }s[5]; 8 int main() 9 { 10 int t, Case = 1; 11 scanf("%d", &t); 12 while(t--) 13 { 14 for(int i = 1; i <= 3; i++) 15 scanf("%lf %lf", &s[i].x, &s[i].y); 16 s[4].x = s[1].x + s[3].x - s[2].x; 17 s[4].y = s[1].y + s[3].y - s[2].y; 18 double sum = abs((s[2].x-s[1].x)*(s[3].y-s[1].y) - (s[3].x-s[1].x)*(s[2].y-s[1].y)); 19 printf("Case %d: %.lf %.lf %.lf ", Case++, s[4].x, s[4].y, sum); 20 } 21 return 0; 22 }