数学 A Rescue The Princess

 

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)\n", 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\n", 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)\n", xc, yc); 85  } 86  } 87 88 return 0; 89 }