sgu 120

sgu120

题意:给出正n边形的两个点(x1,y1),(x2,y2),求正n边形所有点坐标

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首先要知道

向量(x,y)逆时针旋转a(rad)得到(x′,y′)
有这样一个关系>
x′=x∗cos(a)−y∗sin(a)
y′=x∗sin(a)+y∗cos(a)
证明：
记x轴旋转到向量(x,y)的角为b(rad) ，(x,y)长度为L

有>
x=L∗cos(b)
y=L∗sin(b)

x′=L∗cos(a+b)
y′=L∗sin(a+b)

而>
cos(a+b)=cos(a)∗cos(b)−sin(a)∗sin(b)
sin(a+b)=sin(a)∗cos(b)+cos(a)∗sin(b)

推得>
x′=x∗cos(a)−y∗sin(a)
y′=x∗sin(a)+y∗cos(a)

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解法：
套公式，列方程，解出正n边形中心O点坐标
然后用向量解出其他每个点的坐标

因为是保留6位小数，所以还要特判 -0.000000 的情况

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计算几何就是。。。唉不说了，说起来满满都是泪

#include<stdio.h>#include<stdlib.h>#include<math.h>#define eps 0.0000001struct node{double x,y;int s;}n1,n2,f,f1,ans;int n;double p,q;double pi;double a,b;int main(){    int i;    #ifndef ONLINE_JUDGE    freopen("sgu120.in","r",stdin);    freopen("sgu120.out","w",stdout);     #endif    scanf("%d%d%d",&n,&n1.s,&n2.s);    scanf("%lf%lf",&n1.x,&n1.y);    scanf("%lf%lf",&n2.x,&n2.y);    pi=acos(-1);    a=((double)n2.s-n1.s)/n*pi*2;    p=n1.x-n2.x*cos(a)+n2.y*sin(a);    q=n1.y-n2.x*sin(a)-n2.y*cos(a);    f.x=(p*((double)1-cos(a))-q*sin(a))/((double)1-cos(a))/2;    f.y=(p*sin(a)+q*((double)1-cos(a)))/((double)1-cos(a))/2;    b=((double)n1.s-1)/n*pi*2;    f1.x=(n1.x-f.x)*cos(b)-(n1.y-f.y)*sin(b);    f1.y=(n1.x-f.x)*sin(b)+(n1.y-f.y)*cos(b);    for(i=1;i<=n;i++)    {        b=((double)1-i)/n*pi*2;        ans.x=f1.x*cos(b)-f1.y*sin(b)+f.x;        ans.y=f1.x*sin(b)+f1.y*cos(b)+f.y;        if(fabs(ans.x)<eps)            printf("0.000000 ");        else            printf("%.6lf ",ans.x);        if(fabs(ans.y)<eps)            printf("0.000000\n");        else            printf("%.6lf\n",ans.y);    }    #ifndef ONLINE_JUDGE    fclose(stdin);    fclose(stdout);    #endif    return 0;}

话说这编辑器真好玩^_^