给出直角三角坐标平面上三角形其中两个顶点的坐标，求第三个顶点的坐标，要求保留小数点后两位小数
[编程题]正三角形的顶点位置
- 热度指数：1334 时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 32M，其他语言64M
- 算法知识视频讲解
输入描述:
有多组测试数据，输入的第一行是整数T(1≤T≤200)表示随后测试数据的组数。
每组测试数据占一行，由4个带两位小数由一个空格隔开的实数构成，表示已知的两个顶点的横纵坐标。
输出描述:
对应每组测试数据，输出对应的第三个顶点（两组解），如果两组解的横坐标不相等，则先输出横坐标较小的顶点，否则输出纵坐标较小的顶点，每组输出占一行，输出保留两位小数
示例1
输入
3 12.00 3.00 12.00 9.00 12.00 3.00 24.00 3.00 1.00 2.00 3.00 4.00
输出
6.80 6.00 17.20 6.00 18.00 -7.39 18.00 13.39 0.27 4.73 3.73 1.27
假设已知的两个点为p1和p2，第三个要求的点为p3。显然，向量p1->p2逆时针和顺时针旋转60°就可以得到向量p1->p3的坐标，而p1的坐标已知，进一步我们可以求得p3的两个可行解。
对于旋转角度Θ，有旋转矩阵为[[cosΘ,-sinΘ],[sinΘ,cosΘ]]，用行向量右乘旋转矩阵就可以得到旋转后的向量坐标。
import java.lang.String;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
class Point {
private double x, y;
public Point(double x, double y) {
this.x = x;
this.y = y;
}
public double getX() { return x; }
public double getY() { return y; }
}
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int T = Integer.parseInt(br.readLine().trim());
while(T-- > 0) {
String[] params = br.readLine().trim().split(" ");
Point point1 = new Point(Double.parseDouble(params[0]), Double.parseDouble(params[1]));
Point point2 = new Point(Double.parseDouble(params[2]), Double.parseDouble(params[3]));
Point[] points = getPoints(point1, point2);
System.out.printf("%.2f %.2f %.2f %.2f\n",
points[0].getX(), points[0].getY(), points[1].getX(), points[1].getY());
}
}
private static Point[] getPoints(Point p1, Point p2) {
Point[] points = new Point[2];
// 向量p1->p2的坐标
Point v = new Point(p2.getX() - p1.getX(), p2.getY() - p1.getY());
// 将向量分别逆时针和顺时针旋转60°就可以得到向量p1->p3的坐标
double[][] rotate1 = new double[2][2];
rotate1[0][0] = Math.cos(Math.toRadians(60));
rotate1[0][1] = -Math.sin(Math.toRadians(60));
rotate1[1][0] = Math.sin(Math.toRadians(60));
rotate1[1][1] = Math.cos(Math.toRadians(60));
Point v1 = new Point(v.getX()*rotate1[0][0] + v.getY()*rotate1[1][0],
v.getX()*rotate1[0][1] + v.getY()*rotate1[1][1]);
double[][] rotate2 = new double[2][2];
rotate2[0][0] = Math.cos(Math.toRadians(-60));
rotate2[0][1] = -Math.sin(Math.toRadians(-60));
rotate2[1][0] = Math.sin(Math.toRadians(-60));
rotate2[1][1] = Math.cos(Math.toRadians(-60));
Point v2 = new Point(v.getX()*rotate2[0][0] + v.getY()*rotate2[1][0],
v.getX()*rotate2[0][1] + v.getY()*rotate2[1][1]);
// 而p1坐标已知，进一步可以求得p3的坐标
Point p3_1 = new Point(v1.getX() + p1.getX(), v1.getY() + p1.getY());
Point p3_2 = new Point(v2.getX() + p1.getX(), v2.getY() + p1.getY());
// 按照题中要求的顺序排列两个点
if(p3_1.getX() != p3_2.getX()){
if(p3_1.getX() < p3_2.getX()){
points[0] = p3_1;
points[1] = p3_2;
}else{
points[0] = p3_2;
points[1] = p3_1;
}
}else{
if(p3_1.getY() < p3_2.getY()){
points[0] = p3_1;
points[1] = p3_2;
}else{
points[0] = p3_2;
points[1] = p3_1;
}
}
return points;
}
}
发表于 2021-02-24 17:52:37
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#include<iostream>
#include<iomanip>
#include<vector>
#include<cmath>
using namespace std;
int main()
{
int n;
while(cin>>n)
{
vector<vector<double>> vec;
for(int i=0;i<n;i++)
{
vector<double> v(4,0);
cin>>v[0]>>v[1]>>v[2]>>v[3];
vec.push_back(v);
}
for(int i=0;i<n;i++)
{
double mid_x=(vec[i][0]+vec[i][2])/2.0;
double mid_y=(vec[i][1]+vec[i][3])/2.0;
double l=sqrt((vec[i][0]-vec[i][2])*(vec[i][0]-vec[i][2])+(vec[i][1]-vec[i][3])*(vec[i][1]-vec[i][3]));
double x1 = mid_x - sqrt(3.0) / 2.0*l*sin(atan((vec[i][1] - vec[i][3]) / (vec[i][2] - vec[i][0])));
double y1 = mid_y - sqrt(3.0) / 2.0*l*cos(atan((vec[i][1] - vec[i][3]) / (vec[i][2] - vec[i][0])));
double x2 = mid_x + sqrt(3.0) / 2.0*l*sin(atan((vec[i][1] - vec[i][3]) / (vec[i][2] - vec[i][0])));
double y2 = mid_y + sqrt(3.0) / 2.0*l*cos(atan((vec[i][1] - vec[i][3]) / (vec[i][2] - vec[i][0])));
if(x1<x2 ||(x1==x2&&y1<y2))
cout<<fixed<<setprecision(2)<<x1<<" "<<y1<<" "<<x2<<" "<<y2<<endl;
else
cout<<fixed<<setprecision(2)<<x2<<" "<<y2<<" "<<x1<<" "<<y1<<endl;
}
}
}
发表于 2020-08-12 15:00:51
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import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
/**
* @author wylu
*/
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while (t-- != 0) {
String[] strs = br.readLine().split(" ");
double x1 = Double.parseDouble(strs[0]), y1 = Double.parseDouble(strs[1]);
double x2 = Double.parseDouble(strs[2]), y2 = Double.parseDouble(strs[3]);
double dx = x2 - x1, dy = y2 - y1;
double edgeLen = Math.sqrt(dx * dx + dy * dy);
double height = Math.sqrt(3) / 2.0 * edgeLen;
double resX1, resY1, resX2, resY2;
if (dx == 0) {
resX1 = x1 - height;
resX2 = x1 + height;
resY1 = (y1 + y2) / 2.0;
resY2 = resY1;
} else if (dy == 0) {
resX1 = (x1 + x2) / 2.0;
resX2 = resX1;
resY1 = y1 - height;
resY2 = y1 + height;
} else {
double midX = (x1 + x2) / 2.0, midY = (y1 + y2) / 2.0;
double sine = Math.abs(dy / edgeLen), k = dy / dx;
double ox = height * sine, oy = ox / k;
resX1 = midX - ox;
resY1 = midY + oy;
resX2 = midX + ox;
resY2 = midY - oy;
}
System.out.printf("%.2f %.2f %.2f %.2f\n", resX1, resY1, resX2, resY2);
}
}
}
发表于 2019-03-16 10:17:00
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import java.util.*;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0){
double x1 = sc.nextDouble(), y1 = sc.nextDouble(),
x2 = sc.nextDouble(), y2 = sc.nextDouble();
double dx = x2 - x1, dy = y2 - y1;
double radians = Math.toRadians(60.0);
double x3 = x1 + Math.cos(radians) * dx + Math.sin(radians) * dy
,y3 = y1 - Math.sin(radians) * dx + Math.cos(radians) * dy;
radians = Math.toRadians(-60.0);
double x4 = x1 + Math.cos(radians) * dx + Math.sin(radians) * dy
,y4 = y1 - Math.sin(radians) * dx + Math.cos(radians) * dy;
if(x3 > x4 || (x3 == x4 && y3 > y4)){
double tmp = x3;
x3 = x4;
x4 = tmp;
tmp = y3;
y3 = y4;
y4 = tmp;
}
System.out.println(String.format("%.2f",x3) + " " + String.format("%.2f",y3)
+ " " + String.format("%.2f",x4) + " " + String.format("%.2f",y4));
}
}
}
发表于 2022-03-21 22:19:30
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对于数值计算和用到公式的地方，python比java要方便很多，适合在时间紧迫的面试时候用
from math import sqrt
for i in range(int(input())):points = list(map(float, input().split()))
value = [points[2] - points[0], points[3] - points[1]]
side1 = [points[0] + value[0] / 2 - value[1] * sqrt(3) / 2, points[1] + value[1] / 2 + value[0] * sqrt(3) / 2] #c和d一左一右
side2 = [points[0] + value[0] / 2 + value[1] * sqrt(3) / 2, points[1] + value[1] / 2 - value[0] * sqrt(3) / 2] #c和d一左一右
if value[1] > 0:
result = (side1[0], side1[1], side2[0], side2[1])
elif value[1] < 0:
result = (side2[0], side2[1], side1[0], side1[1])
else:
result = (side1[0], min(side1[1], side2[1]), side2[0], max(side1[1], side2[1]))
print("%.2f %.2f %.2f %.2f" % result)
发表于 2021-03-31 16:39:40
回复(0)
#include<iostream>
#include<vector>
#include<math.h>
#include<algorithm>
#include<iomanip>
using namespace std;
int main()
{
int n;
while (cin >> n)
{
vector<vector<double>> p(n, vector<double>(4, 0));
for (int i = 0; i < n; i++)
{
cin >> p[i][0]; cin >> p[i][1]; cin >> p[i][2]; cin >> p[i][3];
}
vector<pair<double, double>> temp;
vector<vector<pair<double, double>>> res;
double a1, b1; double a2, b2;
double x, y;//中点
double dx, dy;//从中点开始的坐标增量
double l;//两点距离
for (int i = 0; i < n; i++)
{
if (p[i][3] - p[i][1] == 0 || p[i][2] - p[i][0] == 0)//平行于x轴或者y轴
{
if (p[i][3] - p[i][1] == 0)
{
pair<double, double> t;
t.first = (p[i][2] + p[i][0]) / 2;
t.second = p[i][3] + abs(p[i][2] - p[i][0]) * sqrt(3) / 2;
temp.push_back(t);
t.first = (p[i][2] + p[i][0]) / 2;
t.second = p[i][3] - abs(p[i][2] - p[i][0]) * sqrt(3) / 2;
temp.push_back(t);
sort(temp.begin(), temp.end());
res.push_back(temp);
temp.clear();
}
else
{
pair<double, double> t;
t.first = p[i][2] + abs(p[i][1] - p[i][3]) * sqrt(3) / 2;
t.second = (p[i][1] + p[i][3]) / 2;
temp.push_back(t);
t.first = p[i][2] - abs(p[i][1] - p[i][3]) * sqrt(3) / 2;
t.second = (p[i][1] + p[i][3]) / 2;
temp.push_back(t);
sort(temp.begin(), temp.end());
res.push_back(temp);
temp.clear();
}
}
else
{
a1 = (p[i][3] - p[i][1]) / (p[i][2] - p[i][0]);
a2 = -1 / a1;
x = (p[i][2] + p[i][0]) / 2; y = (p[i][3] + p[i][1]) / 2;
//cout << x << " " << y << endl;
b2 = y - a2 * x;
l = sqrt(pow(p[i][3] - p[i][1], 2) + pow(p[i][2] - p[i][0], 2));
dx = sqrt(pow(l * sqrt(3) / 2,2) / (1 + pow(a2, 2)));
dy = abs(a2) * dx;
//cout << dx << " " << dy << endl;
pair<double, double> t;
if (a2 > 0)
{
t.first = x + dx;
t.second = y + dy;
temp.push_back(t);
t.first = x - dx;
t.second = y - dy;
temp.push_back(t);
sort(temp.begin(), temp.end());
res.push_back(temp);
temp.clear();
}
else
{
t.first = x - dx;
t.second = y + dy;
temp.push_back(t);
t.first = x + dx;
t.second = y - dy;
temp.push_back(t);
sort(temp.begin(), temp.end());
res.push_back(temp);
temp.clear();
}
}
}
for (int i = 0; i < res.size(); i++)
{
cout << fixed;
cout << setprecision(2) << res[i][0].first << " " << res[i][0].second << " " << res[i][1].first << " " << res[i][1].second << endl;
}
}
return 0;
}
发表于 2020-11-21 14:58:43
回复(0)
#include<iostream>
#include<iomanip>
#include<cmath>
int main()
{
int T;
double b;
while(std::cin>>T)
{
for(int i=0;i<T;i++)
{
double b[4]={0.0};
for(int j=0;j<4;j++)std::cin>>b[j];
if(b[0]==b[2])std::cout<<std::fixed<<std::setprecision(2)<<b[0]-abs(b[3]-b[1])*sqrt(3)/2<<" "<<(b[1]+b[3])/2<<" "<<b[0]+abs(b[3]-b[1])*sqrt(3)/2<<" "<<(b[1]+b[3])/2<<std::endl;
else if(b[1]==b[3])std::cout<<std::fixed<<std::setprecision(2)<<(b[0]+b[2])/2<<" "<<b[1]-abs(b[2]-b[0])*sqrt(3)/2<<" "<<(b[0]+b[2])/2<<" "<<b[1]+abs(b[2]-b[0])*sqrt(3)/2<<std::endl;
else
{
double k1=(b[3]-b[1])/(b[2]-b[0]);
double k2=-1/k1;
double x0=(b[0]+b[2])/2;
double y0=(b[1]+b[3])/2;
double d=(y0-b[1])*(y0-b[1])+(x0-b[0])*(x0-b[0]);
std::cout<<std::fixed<<std::setprecision(2)<<x0-sqrt(3*d/(k2*k2+1))<<" "<<y0-k2*sqrt(3*d/(k2*k2+1))<<" "<<x0+sqrt(3*d/(k2*k2+1))<<" "<<y0+k2*sqrt(3*d/(k2*k2+1))<<std::endl;
}
}
}
return 0;
#include<iomanip>
#include<cmath>
int main()
{
int T;
double b;
while(std::cin>>T)
{
for(int i=0;i<T;i++)
{
double b[4]={0.0};
for(int j=0;j<4;j++)std::cin>>b[j];
if(b[0]==b[2])std::cout<<std::fixed<<std::setprecision(2)<<b[0]-abs(b[3]-b[1])*sqrt(3)/2<<" "<<(b[1]+b[3])/2<<" "<<b[0]+abs(b[3]-b[1])*sqrt(3)/2<<" "<<(b[1]+b[3])/2<<std::endl;
else if(b[1]==b[3])std::cout<<std::fixed<<std::setprecision(2)<<(b[0]+b[2])/2<<" "<<b[1]-abs(b[2]-b[0])*sqrt(3)/2<<" "<<(b[0]+b[2])/2<<" "<<b[1]+abs(b[2]-b[0])*sqrt(3)/2<<std::endl;
else
{
double k1=(b[3]-b[1])/(b[2]-b[0]);
double k2=-1/k1;
double x0=(b[0]+b[2])/2;
double y0=(b[1]+b[3])/2;
double d=(y0-b[1])*(y0-b[1])+(x0-b[0])*(x0-b[0]);
std::cout<<std::fixed<<std::setprecision(2)<<x0-sqrt(3*d/(k2*k2+1))<<" "<<y0-k2*sqrt(3*d/(k2*k2+1))<<" "<<x0+sqrt(3*d/(k2*k2+1))<<" "<<y0+k2*sqrt(3*d/(k2*k2+1))<<std::endl;
}
}
}
return 0;
}
考一个数学专业的
发表于 2020-09-14 20:01:34
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复数
from math import sqrt
for i in range(int(input())):
a = list(map(float, input().split()))
b = [a[2] - a[0], a[3] - a[1]]
c = [a[0] + b[0] / 2 - b[1] * sqrt(3) / 2, a[1] + b[1] / 2 + b[0] * sqrt(3) / 2]
d = [a[0] + b[0] / 2 + b[1] * sqrt(3) / 2, a[1] + b[1] / 2 - b[0] * sqrt(3) / 2]
if b[1] > 0: o = (c[0], c[1], d[0], d[1])
elif b[1] < 0: o = (d[0], d[1], c[0], c[1])
else: o = (c[0], min(c[1], d[1]), d[0], max(c[1], d[1]))
print("%.2f %.2f %.2f %.2f" % o)
发表于 2020-07-15 22:11:13
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import math n=int(input()) arr=[] for i in range(n): arr.append([float(x) for x in input().split()]) c,midx,midy,l,px1,px2,py1,py2=[],[],[],[],[],[],[],[] for i in range(n): c.append(math.sqrt((arr[i][0]-arr[i][2])**2+(arr[i][1]-arr[i][3])**2)) midx.append(abs((arr[i][0]+arr[i][2])/2)) midy.append(abs((arr[i][1]+arr[i][3])/2)) l.append((math.sqrt(3)/2)*c[i]) px1.append(midx[i]-abs(arr[i][1]-arr[i][3])/c[i]*l[i]) px2.append(midx[i]+abs(arr[i][1]-arr[i][3])/c[i]*l[i]) temp1=arr[i][3]-arr[i][1] temp2=arr[i][2] - arr[i][0] if temp1!=0 and temp2!=0: py1.append( midy[i] + abs(arr[i][0] - arr[i][2]) / c[i] * l[i] * ((arr[i][3] - arr[i][1])*(arr[i][2] - arr[i][0]))/abs(((arr[i][3] - arr[i][1])*(arr[i][2] - arr[i][0])))) py2.append( midy[i] - abs(arr[i][0] - arr[i][2]) / c[i] * l[i] * ((arr[i][3] - arr[i][1])*(arr[i][2] - arr[i][0]))/abs(((arr[i][3] - arr[i][1])*(arr[i][2] - arr[i][0])))) else: py1.append(midy[i] + abs(arr[i][0] - arr[i][2]) / c[i] * l[i]) py2.append(midy[i] - abs(arr[i][0] - arr[i][2]) / c[i] * l[i]) for i in range(n): if px1[i]>px2[i]: print('%.2f %.2f %.2f %.2f'%(px2[i],py2[i],px1[i],py1[i])) elif px1[i]<px2[i]: print('%.2f %.2f %.2f %.2f'%(px1[i],py1[i],px2[i],py2[i])) else: if py1[i]>py2[i]: print('%.2f %.2f %.2f %.2f'%(px2[i],py2[i],px1[i],py1[i])) else: print('%.2f %.2f %.2f %.2f'%(px1[i],py1[i],px2[i],py2[i]))
发表于 2019-06-19 00:12:55
回复(0)
纯数学题
#include<iostream>
#include<algorithm>
#include<iomanip>
using namespace std;
int main()
{
int n;
cin>>n;
double g3=sqrt(3);
for(int i=0;i<n;i++)
{
double x1,y1,x2,y2,x0,y0,d,x,y,xx,yy,k,hx,hy;
cin>>x1>>y1>>x2>>y2;
if(x2==x1)
{
d=abs(y2-y1);
y=min(y1,y2)+d/2;
yy=y;
x=x1+d*g3/2;
xx=x1-d*g3/2;
}
else if(y2==y1)
{
d=abs(x2-x1);
x=min(x1,x2)+d/2;
xx=x;
y=y1+d*g3/2;
yy=y1-d*g3/2;
}
else
{
d=sqrt(pow((y2-y1),2)+pow((x2-x1),2));
k=(x1-x2)/(y2-y1);
x0=(x1+x2)/2;y0=(y1+y2)/2;
hx=g3*d/2/sqrt(1+pow(k,2));
hy=abs(k*hx);
if(k<0)
{
xx=x0-hx;
yy=y0+hy;
x=x0+hx;
y=y0-hy;
}
else
{
xx=x0-hx;
yy=y0-hy;
x=x0+hx;
y=y0+hy;
}
}
cout<<setiosflags(ios::fixed)<<setprecision(2);
cout<<xx<<" "<<yy<<" "<<x<<" "<<y<<endl;
}
return 0;
}
发表于 2019-04-28 10:55:57
回复(4)
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