题目传送门
1 /*
2 已知一向量为(x , y) 则将它旋转θ后的坐标为（x*cosθ- y * sinθ ， y*cosθ + x * sinθ）
3 应用到本题，x变为(xb - xa), y变为(yb - ya)相对A点的位置，即B绕着A点旋转60度至C点
4 注意：计算后加回A点的坐标才是相对于原点的坐标
5 详细解释：http://www.tuicool.com/articles/FnEZJb
6 */
7 #include <cstdio>
8 #include <cmath>
9 #include <algorithm>
10 #include <iostream>
11 #include <cstring>
12 using namespace std;
13
14 const double PI = acos (-1.0);
15
16 int main(void) //A Rescue The Princess
17 {
18 //freopen ("A.txt", "r", stdin);
19
20 int t;
21 while (scanf ("%d", &t) == 1)
22 {
23 while (t--)
24 {
25 double xa, ya, xb, yb, xd, yd;
26 scanf ("%lf%lf%lf%lf", &xa, &ya, &xb, &yb);
27 double xc = (xb - xa) * cos (PI/3.0) - (yb - ya) * sin (PI/3.0) + xa;
28 double yc = (yb - ya) * cos (PI/3.0) + (xb - xa) * sin (PI/3.0) + ya;
29 printf ("(%.2f,%.2f)
", xc, yc);
30 }
31 }
32
33 return 0;
34 }
1 /*
2 分各种情况讨论，这是我比赛写的，最后因为没有去绝对值而WA几次，
3 与上面的比较来看，可见好的思维和数学素养是多么重要:)
4 */
5 #include <cstdio>
6 #include <cmath>
7 #include <algorithm>
8 #include <iostream>
9 #include <cstring>
10 using namespace std;
11
12 int main(void) //A Rescue The Princess
13 {
14 //freopen ("A.txt", "r", stdin);
15
16 int t;
17 while (scanf ("%d", &t) == 1)
18 {
19 while (t--)
20 {
21 double xa, ya, xb, yb, xc, yc, xd, yd;
22 scanf ("%lf%lf%lf%lf", &xa, &ya, &xb, &yb);
23 double ab = sqrt ((xa-xb) * (xa-xb) + (ya-yb) * (ya-yb));
24 double cd = sqrt (3.0) / 2 * ab;
25 xd = (xa + xb) / 2; yd = (ya + yb) / 2;
26 if (ya == yb)
27 {
28 xc = (xa + xb) / 2;
29 if (xa < ab)
30 {
31 yc = ya + cd;
32 }
33 else
34 {
35 yc = ya - cd;
36 }
37 }
38 else if (xa == xb)
39 {
40 yc = (ya + yb) / 2;
41 if (ya > yb)
42 {
43 xc = xa + cd;
44 }
45 else
46 {
47 xc = xa - cd;
48 }
49 }
50 else
51 {
52 double k = (ya - yb) / (xa - xb);
53 k = -1.0 / k;
54 double q = atan (k);
55 //printf ("%.2f %.2f %.2f %.2f %.2f %.2f
", ab, cd, xd, yd, k, q);
56 if (k > 0)
57 {
58 if (xa < xb)
59 {
60 xc = xd + abs (cd * cos (q));
61 yc = yd + abs (cd * sin (q));
62 }
63 else
64 {
65 xc = xd - abs (cd * cos (q));
66 yc = yd - abs (cd * sin (q));
67 }
68 }
69 else
70 {
71 if (xa < xb)
72 {
73 xc = xd - abs (cd * cos (q));
74 yc = yd + abs (cd * sin (q));
75 }
76 else
77 {
78 xc = xd + abs (cd * cos (q));
79 yc = yd - abs (cd * sin (q));
80 }
81 }
82 }
83
84 printf ("(%.2f,%.2f)
", xc, yc);
85 }
86 }
87
88 return 0;
89 }