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

In one of his notebooks, Euclid gave a complex procedure for solving the following problem. With computers, perhaps there is an easier way.

In a 2D plane, consider a line segment AB, another point C which is not collinear with AB, and a triangle DEF. The goal is to find points G and H such that:

• H is on the ray AC (it may be closer to A than C or further away, but angle CAB is the same as angle HAB)
• ABGH is a parallelogram (AB is parallel to GH, AH is parallel to BG)
• The area of parallelogram ABGH is the same as the area of triangle DEF

Input 

Input consists of multiple datasets. Each dataset will consist of twelve real numbers, with no more than 3 decimal places each, on a single line. Those numbers will represent the x and y coordinates of points A through F , as follows:

x_A y_A x_B y_B x_C y_C x_D y_D x_E y_E x_F y_F

Points A, B and C are guaranteed to not be collinear. Likewise, D, E and F are also guaranteed to be non-collinear. Every number is guaranteed to be in the range from -1000.0...1000.0 inclusive.

End of the input will be a line with twelve zero values (0.0).

Output 

For each input set, print a single line with four floating point numbers. These represent points G and H, like this:

x_G y_G x_H y_H

Print all values to a precision of 3 decimal places (rounded, NOT truncated). Print a single space between numbers.

Sample Input 

0 0 5 0 0 5 3 2 7 2 0 4 
1.3 2.6 12.1 4.5 8.1 13.7 2.2 0.1 9.8 6.6 1.9 6.7 
0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

Sample Output 

5.000 0.800 0.000 0.800 
13.756 7.204 2.956 5.304

---

很簡單的推導。

#include <stdio.h>
#include <math.h>
struct Pt {
    double x, y;
    void scan() {
        scanf("%lf %lf", &x, &y);
    }
};
#define eps 1e-8
double cross(Pt o, Pt a, Pt b) {
    return (a.x-o.x)*(b.y-o.y)-(a.y-o.y)*(b.x-o.x);
}
double dist(Pt a, Pt b) {
    return sqrt(pow(a.x-b.x,2)+pow(a.y-b.y,2));
}
int main() {
    Pt in[6];
    int i, j, k;
    while(1) {
        int eof = 1;
        for(i = 0; i < 6; i++) {
            in[i].scan();
            if(fabs(in[i].x) > eps || fabs(in[i].y) > eps)
                eof = 0;
        }
        if(eof) break;
        double areaDEF = fabs(cross(in[3], in[4], in[5])/2);
        double areaABC = fabs(cross(in[0], in[1], in[2])/2);
        double distAB = dist(in[0], in[1]);
        double distAC = dist(in[0], in[2]);
        double sintheta = areaABC*2/distAB/distAC;
        double distAH = areaDEF/sintheta/distAB;
        Pt H, G;
        H.x = in[0].x + (in[2].x-in[0].x)*distAH/distAC;
        H.y = in[0].y + (in[2].y-in[0].y)*distAH/distAC;
        G.x = H.x + (in[1].x-in[0].x);
        G.y = H.y + (in[1].y-in[0].y);
        printf("%.3lf %.3lf %.3lf %.3lf\n", G.x, G.y, H.x, H.y);
    }
    return 0;
}
/*
0 0 5 0 0 5 0 0 0 1 1 0
0 0 5 0 0 5 3 2 7 2 0 4
1.3 2.6 12.1 4.5 8.1 13.7 2.2 0.1 9.8 6.6 1.9 6.7
0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
*/