C++14(g++5.4) 解法, 执行用时: 14ms, 内存消耗: 504K, 提交时间: 2020-07-18 12:56:36

#include<bits/stdc++.h>
using namespace std;
const double eps=1e-2;
int sgn(double x)
{
	if(x<-eps)return -1;
	if(x>eps)return 1;
	return 0;
}
struct Point
{
	double x,y;
	void read(){scanf("%lf%lf",&x,&y);}
	Point operator -(const Point &t)const{return {x-t.x,y-t.y};}
	double operator *(const Point &t)const{return x*t.y-y*t.x;}
	double norm(){return x*x+y*y;}
}p[30],v[30];
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		int x=0,y;
		for(int i=1;i<=20;i++)p[i].read();
		for(int i=1;i<=20;i++)v[i]=p[i%20+1]-p[i];
		for(int i=1;i<=20;i++)if(fabs(v[i].norm()-36)<eps)x=i;
		if(fabs(v[x%20+1].norm()-1)<eps)y=x%20+1;
		else y=(x+18)%20+1;
		printf("%s\n",sgn(v[x]*v[y])<0?"right":"left");
	}
}

C++(g++ 7.5.0) 解法, 执行用时: 18ms, 内存消耗: 716K, 提交时间: 2023-05-01 17:10:15

#include<bits/stdc++.h>
using namespace std;
struct po
{
	double x,y;
} a[30];
double dis(po x,po y) //边长
{
	return sqrt((x.x-y.x)*(x.x-y.x)+(x.y-y.y)*(x.y-y.y));
}
double cj(po x,po y,po z) //叉积
{
	return (y.x-x.x)*(z.y-x.y)-(y.y-x.y)*(z.x-y.x);
}
int main()
{
	int t,i;
	bool flag;
	const double e=1e-5;
	scanf("%d",&t);
	while(t--)
	{
		flag=0;
		for(i=0;i<20;i++)
			scanf("%lf%lf",&a[i].x,&a[i].y);
		for(i=0;i<20;i++)
			if(fabs(dis(a[i],a[(i+1)%20])-9)<e)
			{
				po x=a[i],y=a[(i+1)%20],z=a[(i+2)%20];
				if(cj(x,y,z)>0 && fabs(dis(y,z)-6)<e || cj(x,y,z)<0 && fabs(dis(y,z)-8)<e) flag=1;
				break;
			}
				
		if(flag) printf("left\n");
		else printf("right\n");
	}
    return 0;
}

Python3 解法, 执行用时: 168ms, 内存消耗: 4816K, 提交时间: 2023-05-01 14:21:00

def dis(a,b):
    return (a[0]-b[0])**2+(a[1]-b[1])**2

for _ in range(int(input())):
    l=[list(map(float,input().split())) for _ in range(20)]
    x=y=0
    for z in l:
        x+=z[0]
        y+=z[1]
    
    x/=20
    y/=20

    l+=l
    for i in range(10,30):
        d=dis(l[i],l[i+1])
        if 80<d:
            a=(l[i][0]+l[i+1][0])/2
            b=(l[i][1]+l[i+1][1])/2
            
            if dis(l[i-1],l[i])>dis(l[i+1],l[i+2]):
                c,d=l[i-1]
            else:
                c,d=l[i+2]
                        
            x-=a
            y-=b
            c-=a
            d-=b
            
            print('right' if x*d<y*c else 'left')

C++11(clang++ 3.9) 解法, 执行用时: 14ms, 内存消耗: 376K, 提交时间: 2020-07-18 13:02:16

#include <cmath>
#include <cstdio>
#include <algorithm>

double x[25],y[25];
int T,n;

double dis(int i,int j){return std::sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));}

bool same(double a,double b){return std::fabs(a-b)<=1e-5;}

double cross(double x,double y,double _x,double _y){return x*_y-y*_x;}

int main(){
	scanf("%d",&T);
	while(T--){
		n=20;
		for(int i=1;i<=n;i++)scanf("%lf%lf",&x[i],&y[i]);
		for(int i=1;i<=n;i++){
			int j=i%n+1,k=j%n+1;
			if(same(dis(i,j),9)){
				bool ans=same(dis(j,k),8)^(cross(x[i]-x[j],y[i]-y[j],x[k]-x[j],y[k]-y[j])<0);
				printf("%s\n",ans?"left":"right");
				break;
			}
		}
	}
}