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)△P₀P_nR，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P_n的坐标分别代入直线P₀R的方程，如果异号，说明P₁和P_n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。

 //#define LOCAL
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 using namespace std;

 void Rote(double& a1, double& b1, double a2, double b2, double p)
 {
     double tempx = a1, tempy = b1;
     tempx = cos(p)*(a1 - a2) + sin(p)*(b2 - b1) + a2;
     tempy = sin(p)*(a1 - a2) + cos(p)*(b1 - b2) + b2;
     a1 = tempx;
     b1 = tempy;
 }

 double dis(double a1, double b1, double a2, double b2)
 {
     return sqrt((a1-a2)*(a1-a2) + (b1-b2)*(b1-b2));
 }

 int main(void)
 {
     #ifdef LOCAL
         freopen("Bin.txt", "r", stdin);
     #endif

     int T;
     scanf("%d", &T);
     while(T--)
     {
         int n;
         scanf("%d", &n);
         double P0x = 0.123, P0y = 0.312, Q0x = 1.589, Q0y = 1.455;
         double Pnx = P0x, Pny = P0y, Qnx = Q0x, Qny = Q0y;
         for(int i = ; i < n; ++i)
         {
             double P1x, P1y, alpha;
             scanf("%lf%lf%lf", &P1x, &P1y, &alpha);
             Rote(Pnx, Pny, P1x, P1y, alpha);
             Rote(Qnx, Qny, P1x, P1y, alpha);
         }
         double A1 = P0x - Pnx, B1 = P0y - Pny, C1 = (P0x*P0x + P0y*P0y - Pnx*Pnx - Pny*Pny) / ;
         double A2 = Q0x - Qnx, B2 = Q0y - Qny, C2 = (Q0x*Q0x + Q0y*Q0y - Qnx*Qnx - Qny*Qny) / ;
         double ansx = (C1*B2 - C2*B1) / (A1*B2 - A2*B1);
         double ansy = (C1*A2 - C2*A1) / (B1*A2 - B2*A1);

         double zx = (P0x + Pnx) / , zy = (P0y + Pny) / ;
         double ansa = acos(dis(zx,zy, ansx, ansy) / dis(P0x, P0y, ansx, ansy)) * ;

         double P1x = ansx - (P0y - ansy), P1y = ansy + (P0x - ansx);
         if(((P0y-ansy)*(P1x-ansx)-(P0x-ansx)*(P1y-ansy)) * ((P0y-ansy)*(Pnx-ansx)-(P0x-ansx)*(Pny-ansy)) < )
             ansa = 3.1415926536 *  - ansa;

         printf("%.10lf %.10lf %.10lf\n", ansx, ansy, ansa);
     }
     return ;
 }

代码君

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

 #include <cstdlib>
 #include <cctype>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <iostream>
 #include <sstream>
 #include <map>
 #include <set>
 #include <queue>
 #include <stack>
 #include <fstream>
 #include <numeric>
 #include <iomanip>
 #include <bitset>
 #include <list>
 #include <stdexcept>
 #include <functional>
 #include <utility>
 #include <ctime>
 #include <cassert>
 #include <complex>
 using namespace std;
 #define rep(i,a,n) for (int i=a;i<n;i++)
 #define per(i,a,n) for (int i=n-1;i>=a;i--)
 #define pb push_back
 #define mp make_pair
 #define all(x) (x).begin(),(x).end()
 #define fi first
 #define se second
 #define SZ(x) ((int)(x).size())
 #define ACCU accumulate
 #define TWO(x) (1<<(x))
 #define TWOL(x) (1ll<<(x))
 #define clr(a) memset(a,0,sizeof(a))
 #define POSIN(x,y) (0<=(x)&&(x)<n&&0<=(y)&&(y)<m)
 #define PRINTC(x) cout<<"Case #"<<++__<<": "<<x<<endl
 #define POP(x) (__builtin_popcount(x))
 #define POPL(x) (__builtin_popcountll(x))
 typedef vector<int> VI;
 typedef vector<string> VS;
 typedef vector<double> VD;
 typedef long long ll;
 typedef long double LD;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> PLL;
 typedef vector<ll> VL;
 typedef vector<PII> VPII;
 typedef complex<double> CD;
 const int inf=0x20202020;
 const ll mod=;
 const double eps=1e-;
 const double pi=3.1415926535897932384626;
 const int DX[]={,,-,},DY[]={,,,-};
 ll powmod(ll a,ll b) {ll res=;a%=mod;for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
 ll powmod(ll a,ll b,ll mod) {ll res=;a%=mod;for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
 // head

 int _,n;
 CD k,b,p;
 double x,y,c;
 int main() {
     for (scanf("%d",&_);_;_--) {
         k=CD(,);b=CD(,);
         scanf("%d",&n);
         rep(i,,n) {
             scanf("%lf%lf%lf",&x,&y,&c);
             p=CD(x,y);
             k=k*CD(cos(c),sin(c));
             b=(b-p)*CD(cos(c),sin(c))+p;
         }
         double the=arg(k);
         while (the<) the+=*pi;
         while (the>=*pi) the-=*pi;
         p=b/(CD(,)-k);
         printf("%.10f %.10f %.10f\n",real(p),imag(p),the);
     }
 }

外来的代码君