#include<bits/stdc++.h> #define de(x) cout<<#x<<"="<<x<<endl; #define dd(x) cout<<#x<<"="<<x<<" "; #define rep(i,a,b) for(int i=a;i<(b);++i) #define repd(i,a,b) for(int i=a;i>=(b);--i) #define repp(i,a,b,t) for(int i=a;i<(b);i+=t) #define ll long long #define mt(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define pii pair<int,int> #define pdd pair<double,double> #define pdi pair<double,int> #define mp(u,v) make_pair(u,v) #define sz(a) (int)a.size() #define ull unsigned long long #define ll long long #define pb push_back #define PI acos(-1.0) #define qc std::ios::sync_with_stdio(false) #define db double #define all(a) a.begin(),a.end() const int mod = 1e9+7; const int maxn = 2e2+5; const double eps = 1e-6; using namespace std; bool eq(const db &a, const db &b) { return fabs(a - b) < eps; } bool ls(const db &a, const db &b) { return a + eps < b; } bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); } ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); }; ll lcm(ll a,ll b) { return a/gcd(a,b)*b; } ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } struct Point{ db x,y; }ans[maxn],p1,p2,o; Point get_mid(Point p1,Point p2){ Point ret; ret.x = (p1.x + p2.x) / 2; ret.y = (p1.y + p2.y) / 2; return ret; } Point get_o(Point p1,Point p2,int n1,int n2,int n){//正多边形中点,外接圆圆心 Point mid = get_mid(p1,p2),ret; db ang = PI / 2.0 - 1.0 * (n2 - n1) * PI / n; // 向量的思想 ret.x = mid.x - (p1.y - mid.y) * tan(ang); ret.y = mid.y + (p1.x - mid.x) * tan(ang); return ret; } db f(db x){//会出现-0的情况 return fabs(x)<eps?0:x; } int main(){ int n,n1,n2; scanf("%d%d%d",&n,&n1,&n2); scanf("%lf%lf%lf%lf",&p1.x,&p1.y,&p2.x,&p2.y); if(n1 > n2) swap(n1,n2), swap(p1,p2); n1--; n2--; o = get_o(p1,p2,n1,n2,n); db ang = 2.0 * PI / n; ans[n1] = p1; // dd(o.x)de(o.y) for(int i = n1+1; ;++i){ if(i%n==n1) break; // x0,y0绕rx0,ry0逆时针旋转a度 // x0= (x - rx0)*cos(a) - (y - ry0)*sin(a) + rx0 ; // y0= (x - rx0)*sin(a) + (y - ry0)*cos(a) + ry0 ; ans[i%n].x = o.x + (ans[(i-1+n)%n].x - o.x)*cos(-ang) - (ans[(i-1+n)%n].y - o.y)*sin(-ang); ans[i%n].y = o.y + (ans[(i-1+n)%n].x - o.x)*sin(-ang) + (ans[(i-1+n)%n].y - o.y)*cos(-ang); } rep(i,0,n) printf("%.8f %.8f\n",f(ans[i].x),f(ans[i].y)); return 0; }
SGU 120
题意：给你正n边形上的两个点，让你求出正n边形上的所有点，1 - n是顺时针
收获:求正n边形中心或者说是外接圆的圆心，向量旋转的公式[x*cosA-y*sinA x*sinA+y*cosA]（向量逆时针旋转A）