SGU 120 Arhipelago

題目實質：

要你先求出此正多邊形的中心，再不斷的瞬時針旋轉求答案。

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當你逆時針旋轉角，則向量(x, y)會變成下圖：



透過上式，可以解二元一次方程式求出中心

再來因為他是要瞬時針旋轉，記得要改成負的

然後Pi值要精準，因為 x, y 都很大（Pi = 3.1415926535897932384626）

//By momo
#include<cstdio>
#include<cmath>
#include<algorithm>
#define N 200
#define EPS 1e-9
#define Pi (3.1415926535897932384626)
using namespace std;

int n, a, b;
double ansx[N], ansy[N];
void find(double a1, double b1, double c1, double a2, double b2, double c2, double& x, double& y){
    x = (b2*c1-b1*c2) / (a1*b2-a2*b1);
    y = (a2*c1-a1*c2) / (a2*b1-a1*b2);
}

int main(){
   
    double x, y, x1, y1, x2, y2;
    scanf("%d%d%d", &n, &a, &b);
    scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
    if(a > b) b += n;
   
    double theta = 2.0 * Pi * (b-a) / n;
    double phi   = 2.0 * Pi *  (-1) / n;
    double sn  = sin(theta), cs  = cos(theta);
    double snp = sin(  phi), csp = cos(  phi);
    find(cs-1, sn, x1*cs+y1*sn-x2, -sn, cs-1, y1*cs-x1*sn-y2, x, y);
   
    double xi = x1-x, yi = y1-y;
    for(int i = 0; i < n; i++){
        int pos = (a+i-1) % n;
        ansx[pos] = xi+x; ansy[pos] = yi+y;
        double nx = xi * csp - yi * snp;
        double ny = yi * csp + xi * snp;
        xi = nx, yi = ny;
    }
    for(int i = 0; i < n; i++) printf("%.6lf %.6lf\n", ansx[i]+EPS, ansy[i]+EPS);

}