UVa 10250 The other two trees: computational geometry

Source: Internet
Author: User
Tags integer numbers time limit

10250-the other two Trees

Time limit:3.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=101&page=show_ problem&problem=1191

Have a quadrilateral shaped land whose opposite fences are of equal of length. You are have four neighbors whose lands are exactly adjacent to your four, which fences you means a have common and all of them. For example if your have a fence of length d in one side, this fence of length d is also the fence of the adjacent neighbor On that side. The adjacent neighbors have no fence in common among themselves and their lands don ' t also. The main difference between their land and your are that their, lands all square are. All your neighbors have a is at the center of their lands. Given the Cartesian coordinates of trees of two opposite neighbors, you'll have to find the Cartesian coordinates the Other two trees.


The input file contains several lines of input. Each line contains four floating point or integer numbers x1, y1, x2, Y2, where (x1, y1), (x2, y2) are the coordinates of The trees of two opposite neighbors. The Input is terminated by the end of file.


For each line of input produce one line of output which contains "impossible." Without the quotes, if you cannot Determine the coordinates of the other two trees. Otherwise, print four floating point numbers separated through a single spaces with ten digits after the decimal point ax1, Ay1, AX2, Ay2, where (Ax1, ay1) and (AX2, ay2) are the coordinates of the other two trees. The output is checked with special judge and so don ' t worry about the ordering of the points or small precision Errors. The sample output would make it clear.

Sample Input

10 0-10 0

10 0-10 0

10 0-10 0

Sample Output

0.0000000000 10.0000000000 0.0000000000-10.0000000000

0.0000000000 10.0000000000-0.0000000000-10.0000000000

0.0000000000-10.0000000000 0.0000000000 10.0000000000

Relative proximity of the median distance d1,d2 satisfies d1=d2 (proofs slightly)

Complete code:

#include <cstdio>  
int main ()  
    double x1, x2, y1, y2;  
    while (~SCANF ("%lf%lf%lf%lf", &x1, &y1, &x2, &y2))  
        printf ("%f%f%f%f\n", (x1 + x2 + y2-y1)/2, (x 1-x2 + y1 + y2)/2, (x1 + x2 + y1-y2)/2, (x2-x1 + y1 + y2)/2);  
    return 0;  

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