LOJ 1305 - Area of a Parallelogram（数学j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 Band (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

 题意：求平行四边形的第四个定点以及平行四边形的面积

#include<cstdio>#include<cstdlib>struct Node{   int x;   int y;}node[10];int main(){    int t,kcase = 1;    scanf("%d",&t);    while(t--)    {        for(int i = 0 ; i < 3 ; i++)            scanf("%d%d",&node[i].x,&node[i].y);        node[3].x = node[2].x + node[0].x - node[1].x;        node[3].y = node[2].y + node[0].y - node[1].y;        int area = abs((node[2].y - node[1].y) * (node[0].x - node[1].x) - (node[0].y - node[1].y) * (node[2].x - node[1].x));        printf("Case %d: %d %d %d\n",kcase++,node[3].x,node[3].y,area);    }    return 0;}