Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem E

The Other Two Trees

Input: standard input

Output: standard output

Time Limit: 2 seconds

You have a quadrilateral shaped land whose opposite fences are of equal length. You have four neighbors whose lands are exactly adjacent to your four fences, that means you have a common fence with all of them. For example if you 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 also don’t intersect. The main difference between their land and your land is that their lands are all square shaped. All your neighbors have a tree at the center of their lands. Given the Cartesian coordinates of trees of two opposite neighbors, you will have to find the Cartesian coordinates of the other two trees.

Input

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. Input is terminated by end of file.

Output

For each line of input produce one line of output which contains the line “Impossible.” without the quotes, if you cannot determine the coordinates of the other two trees. Otherwise, print four floating point numbers separated by a single space 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 will be checked with special judge program, so don’t worry about the ordering of the points or small precision errors. The sample output will 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

(World Final Warm-up Contest, Problem Setter: Shahriar Manzoor)

題目描述：

有一個正方形，四個頂點放樹，給對角的兩棵樹座標，問另外兩棵樹的座標。

題目解法：

從對角兩棵的中心點，走半個法向量過去即可。

#include <stdio.h>

int main() {
    double x1, x2, y1, y2;
    while(scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2) == 4) {
        double cx, cy;
        double vx, vy;
        double vvx, vvy;
        cx = (x1+x2)/2;
        cy = (y1+y2)/2;
        vx = x2-x1;
        vy = y2-y1;
        vvx = -vy/2;
        vvy = vx/2;
        printf("%.10lf %.10lf %.10lf %.10lf\n", cx+vvx, cy+vvy, cx-vvx, cy-vvy);
    }
    return 0;
}