zoukankan      html  css  js  c++  java
  • 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, N1¹N2separated 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;
    }
    

      

  • 相关阅读:
    mybatis 配置之<typeAliases>别名配置元素设置
    <q> 与 <blockquote> 的区别
    line-height属性
    ol/ul/dl的区别
    xampp修改mysql 启动脚本
    linux 设置查看文本行数
    vim跳出括号的方法
    vim中不同模式的帮助信息的查找
    ubuntu下安装自动补全YouCompleteMe
    ubuntu下将CapsLock改为Ctrl键
  • 原文地址:https://www.cnblogs.com/xuesu/p/4005598.html
Copyright © 2011-2022 走看看