题目链接：http://acm.timus.ru/problem.aspx?space=1&num=1046
参考博客：http://hi.baidu.com/cloudygoose/item/21fee021a5db348d9d63d17b
参考资料（向量的旋转）：http://www.cnblogs.com/woodfish1988/archive/2007/09/10/888439.html
题目大意：就是已知n个点，n个角。点Mi可以与多边形Ai和Ai+1构成等腰三角形，顶角为ang[i]. 现在要你求出这个多边形的n的顶点。
算法思路：刚开始想几何性质，怎么也想不出来一个好的思路。没办法，网搜才知道要用解方程的方法。蛋疼的是没写过，别人的代码有是懂非懂的。慢慢琢磨才发现其实现方程求解的思路。先看看那个参考博客的思路吧。
我只是大致翻译一下他的思想： 我们知道： Ai+1 = Rotate(Ai-Mi,ang[i]) + Mi; (画画图就知道) 整理下就是：Ai+1 = P1' * Ai + P2';
而由A1递推来 ： Ai = P1 * A1 + P2; 我们编程时要不断更新这个P1和P2;
具体看代码：
代码：
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std; const double eps = 1e-;
const double PI = acos(-1.0);
const double INF = 1000000000000000.000; struct Point{
double x,y;
Point(double x=, double y=) : x(x),y(y){ } //构造函数
};
typedef Point Vector; struct Circle{
Point c;
double r;
Circle() {}
Circle(Point c,double r): c(c),r(r) {}
};
Vector operator + (Vector A , Vector B){return Vector(A.x+B.x,A.y+B.y);}
Vector operator - (Vector A , Vector B){return Vector(A.x-B.x,A.y-B.y);}
Vector operator * (double p,Vector A){return Vector(A.x*p,A.y*p);}
Vector operator / (Vector A , double p){return Vector(A.x/p,A.y/p);} bool operator < (const Point& a,const Point& b){
return a.x < b.x ||( a.x == b.x && a.y < b.y);
} int dcmp(double x){
if(fabs(x) < eps) return ;
else return x < ? - : ;
}
bool operator == (const Point& a, const Point& b){
return dcmp(a.x - b.x) == && dcmp(a.y - b.y) == ;
} ///向量（x,y）的极角用atan2(y,x);
inline double Dot(Vector A, Vector B){ return A.x*B.x + A.y*B.y; }
inline double Length(Vector A) { return sqrt(Dot(A,A)); }
inline double Angle(Vector A, Vector B) { return acos(Dot(A,B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B) { return A.x*B.y - A.y * B.x; } Vector vecunit(Vector v){ return v / Length(v);} //单位向量
double torad(double deg) { return deg/ * PI; }
Vector Rotate(Vector A, double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad)); } /*************************************分 割 线*****************************************/ int main()
{
//freopen("E:\\acm\\input.txt","r",stdin); const int maxn = ;
Point M[maxn],P1,P2; double ang[maxn];
int n; cin>>n;
for(int i=; i<=n; i++)
{
scanf("%lf %lf",&M[i].x,&M[i].y);
}
for(int i=; i<=n; i++)
{
scanf("%lf",&ang[i]);
ang[i] = torad(ang[i]);
} P1 = Point(,);
P2 = Point(,); for(int i=; i<=n; i++)
{
P1 = Rotate(P1,ang[i]);
P2 = Rotate(P2,ang[i]);
P2 = P2 + M[i] - Rotate(M[i],ang[i]);
} P1.x -= ;
P2.x = -P2.x;
P2.y = -P2.y; Point ans; //求ans时，把P1,P2看成复平面中的点，即P1表示为P1.x+P2.y*i; 然后ans = P2/P1,用虚数就可求出。
ans.x = (P1.x*P2.x+P1.y*P2.y)/(P1.x*P1.x+P1.y*P1.y);
ans.y = (-P1.y*P2.x+P1.x*P2.y)/(P1.x*P1.x+P1.y*P1.y); for(int i=; i<=n; i++)
{
printf("%.2lf %.2lf\n",ans.x,ans.y);
ans = M[i] + Rotate(ans-M[i],ang[i]);
}
}