快速切题 sgu120. Archipelago 计算几何

120. Archipelago

time limit per test: 0.25 sec. 
memory limit per test: 4096 KB

 

Archipelago Ber-Islands consists of N islands that are vertices of equiangular and equilateral N-gon. Islands are clockwise numerated. Coordinates of island N₁ are (x₁, y₁), and island N₂ – (x₂, y₂). Your task is to find coordinates of all N islands.

 

Input

In the first line of input there are N, N₁ and N₂ (3£ N£ 150, 1£ N₁,N₂£N, N₁¹N₂) separated by spaces. On the next two lines of input there are coordinates of island N₁ and N₂ (one pair per line) with accuracy 4digits after decimal point. Each coordinate is more than -2000000 and less than 2000000.

 

Output

Write N lines with coordinates for every island. Write coordinates in order of island numeration. Write answer with 6 digits after decimal point.

 

Sample Input

4 1 3

1.0000 0.0000

1.0000 2.0000

Sample Output

1.000000 0.000000

0.000000 1.000000

1.000000 2.000000

2.000000 1.000000

思路:首先旋转N1N2一定角度到圆心方向利用三角形求出圆心O,然后旋转oN1得到其他点

#include <cstdio>

#include <cmath>

#include <algorithm>

using namespace std;

const int maxn=150;

const double eps=1e-12;

const double pie=acos(-1);

double xo,yo;

double x[maxn],y[maxn];

int n,n1,n2;

typedef pair<double,double> P;

P rot(double x,double y,double angle){

    P p1;

    p1.first=y*sin(angle)+x*cos(angle);

    p1.second=y*cos(angle)-x*sin(angle);

    return p1;

}

int main(){

    scanf("%d%d%d",&n,&n1,&n2);n1--;n2--;

    scanf("%lf%lf%lf%lf",x+n1,y+n1,x+n2,y+n2);

    if(n1>n2)swap(n1,n2);

    P p1=P(x[n2]-x[n1],y[n2]-y[n1]);

    int gap=n2-n1;

    double angle=pie/n*gap;

    double angle2=pie/2-angle;if(angle<eps)angle+=pie;//取夹角不可能钝角或<0

    p1=rot(p1.first,p1.second,angle2);

    p1.first=-p1.first/2/cos(angle2);

    p1.second=-p1.second/2/cos(angle2);

    xo=-p1.first+x[n1],yo=-p1.second+y[n1];

    for(int i=(n1+1)%n;i!=n1;i=(i+1)%n){

        p1=rot(p1.first,p1.second,2*pie/n);

        x[i]=p1.first+xo;

        y[i]=p1.second+yo;

    }

    for(int i=0;i<n;i++){

        printf("%.6f %.6f\n",x[i],y[i]);

    }

    return 0;

}