HDU 4998 (点的旋转) Rotate
为了寻找等效旋转操作，我们任选两个点P0和Q0，分别绕这n个点旋转一定的角度后最终得到Pn和Qn
然后已知：P0和Pn共圆，Q0和Qn共圆。所以要找的等效旋转点就是这两个线段的垂直平分线交点O。
等效的角度的计算，可以利用已知的等腰三角形(这里有两个)△P0PnR，做一条垂线(三线合一的性质)，再利用反三角函数计算半角，再乘二
还有一种特殊情况就是，如果答案比平角要大，我们计算的角度就不对了。
此时可以让P0逆时针旋转90°得到一个P1，然后将P1和Pn的坐标分别代入直线P0R的方程，如果异号，说明P1和Pn在直线两侧；同号说明同侧。
也就是异侧的话，就要将所求角度用2π减去它。
1 //#define LOCAL 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <cmath> 6 using namespace std; 7 8 void Rote(double& a1, double& b1, double a2, double b2, double p) 9 { 10 double tempx = a1, tempy = b1; 11 tempx = cos(p)*(a1 - a2) + sin(p)*(b2 - b1) + a2; 12 tempy = sin(p)*(a1 - a2) + cos(p)*(b1 - b2) + b2; 13 a1 = tempx; 14 b1 = tempy; 15 } 16 17 double dis(double a1, double b1, double a2, double b2) 18 { 19 return sqrt((a1-a2)*(a1-a2) + (b1-b2)*(b1-b2)); 20 } 21 22 int main(void) 23 { 24 #ifdef LOCAL 25 freopen("Bin.txt", "r", stdin); 26 #endif 27 28 int T; 29 scanf("%d", &T); 30 while(T--) 31 { 32 int n; 33 scanf("%d", &n); 34 double P0x = 0.123, P0y = 0.312, Q0x = 1.589, Q0y = 1.455; 35 double Pnx = P0x, Pny = P0y, Qnx = Q0x, Qny = Q0y; 36 for(int i = 0; i < n; ++i) 37 { 38 double P1x, P1y, alpha; 39 scanf("%lf%lf%lf", &P1x, &P1y, &alpha); 40 Rote(Pnx, Pny, P1x, P1y, alpha); 41 Rote(Qnx, Qny, P1x, P1y, alpha); 42 } 43 double A1 = P0x - Pnx, B1 = P0y - Pny, C1 = (P0x*P0x + P0y*P0y - Pnx*Pnx - Pny*Pny) / 2; 44 double A2 = Q0x - Qnx, B2 = Q0y - Qny, C2 = (Q0x*Q0x + Q0y*Q0y - Qnx*Qnx - Qny*Qny) / 2; 45 double ansx = (C1*B2 - C2*B1) / (A1*B2 - A2*B1); 46 double ansy = (C1*A2 - C2*A1) / (B1*A2 - B2*A1); 47 48 double zx = (P0x + Pnx) / 2, zy = (P0y + Pny) / 2; 49 double ansa = acos(dis(zx,zy, ansx, ansy) / dis(P0x, P0y, ansx, ansy)) * 2; 50 51 double P1x = ansx - (P0y - ansy), P1y = ansy + (P0x - ansx); 52 if(((P0y-ansy)*(P1x-ansx)-(P0x-ansx)*(P1y-ansy)) * ((P0y-ansy)*(Pnx-ansx)-(P0x-ansx)*(Pny-ansy)) < 0) 53 ansa = 3.1415926536 * 2 - ansa; 54 55 printf("%.10lf %.10lf %.10lf\n", ansx, ansy, ansa); 56 } 57 return 0; 58 }
还有一种复数的方法，<complex>里面已经有了现成的复数数据类型和函数了，用起来很方便。
这是上交一队的代码，膜拜了
1 #include <cstdlib> 2 #include <cctype> 3 #include <cstring> 4 #include <cstdio> 5 #include <cmath> 6 #include <algorithm> 7 #include <vector> 8 #include <string> 9 #include <iostream> 10 #include <sstream> 11 #include <map> 12 #include <set> 13 #include <queue> 14 #include <stack> 15 #include <fstream> 16 #include <numeric> 17 #include <iomanip> 18 #include <bitset> 19 #include <list> 20 #include <stdexcept> 21 #include <functional> 22 #include <utility> 23 #include <ctime> 24 #include <cassert> 25 #include <complex> 26 using namespace std; 27 #define rep(i,a,n) for (int i=a;i<n;i++) 28 #define per(i,a,n) for (int i=n-1;i>=a;i--) 29 #define pb push_back 30 #define mp make_pair 31 #define all(x) (x).begin(),(x).end() 32 #define fi first 33 #define se second 34 #define SZ(x) ((int)(x).size()) 35 #define ACCU accumulate 36 #define TWO(x) (1<<(x)) 37 #define TWOL(x) (1ll<<(x)) 38 #define clr(a) memset(a,0,sizeof(a)) 39 #define POSIN(x,y) (0<=(x)&&(x)<n&&0<=(y)&&(y)<m) 40 #define PRINTC(x) cout<<"Case #"<<++__<<": "<<x<<endl 41 #define POP(x) (__builtin_popcount(x)) 42 #define POPL(x) (__builtin_popcountll(x)) 43 typedef vector<int> VI; 44 typedef vector<string> VS; 45 typedef vector<double> VD; 46 typedef long long ll; 47 typedef long double LD; 48 typedef pair<int,int> PII; 49 typedef pair<ll,ll> PLL; 50 typedef vector<ll> VL; 51 typedef vector<PII> VPII; 52 typedef complex<double> CD; 53 const int inf=0x20202020; 54 const ll mod=1000000007; 55 const double eps=1e-9; 56 const double pi=3.1415926535897932384626; 57 const int DX[]={1,0,-1,0},DY[]={0,1,0,-1}; 58 ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} 59 ll powmod(ll a,ll b,ll mod) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} 60 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;} 61 // head 62 63 int _,n; 64 CD k,b,p; 65 double x,y,c; 66 int main() { 67 for (scanf("%d",&_);_;_--) { 68 k=CD(1,0);b=CD(0,0); 69 scanf("%d",&n); 70 rep(i,0,n) { 71 scanf("%lf%lf%lf",&x,&y,&c); 72 p=CD(x,y); 73 k=k*CD(cos(c),sin(c)); 74 b=(b-p)*CD(cos(c),sin(c))+p; 75 } 76 double the=arg(k); 77 while (the<0) the+=2*pi; 78 while (the>=2*pi) the-=2*pi; 79 p=b/(CD(1,0)-k); 80 printf("%.10f %.10f %.10f\n",real(p),imag(p),the); 81 } 82 }
