在一个圆内找到 N 个随机点
原文:https://www . geesforgeks . org/find-n-random-points-in-a-circle/
给定四个整数 N、R、X 和 Y ，使其代表一个以【X、Y】为圆心坐标的半径为 R 的圆。任务是在圆内或圆上随机找 N 点。 示例:
输入: R = 12，X = 3，Y = 3，N = 5 输出: (7.05，-3.36) (5.21，-7.49) (7.53，0.19) (-2.37，12.05) (1.45，11.80) 输入: R = 5，X = 1，Y = 1，N = 3 输出:
方法:要在圆内或圆上找到一个随机点，我们需要两个分量，一个角(θ)和距中心的距离(D)。之后现在，点(x i ，y i )可以表示为:
xi = X + D * cos(theta)
yi = Y + D * sin(theta)
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
#define PI 3.141592653589
// Return a random double between 0 & 1
double uniform()
{
return (double)rand() / RAND_MAX;
}
// Function to find the N random points on
// the given circle
vector<pair<double, double> > randPoint(
int r, int x, int y, int n)
{
// Result vector
vector<pair<double, double> > res;
for (int i = 0; i < n; i++) {
// Get Angle in radians
double theta = 2 * PI * uniform();
// Get length from center
double len = sqrt(uniform()) * r;
// Add point to results.
res.push_back({ x + len * cos(theta),
y + len * sin(theta) });
}
// Return the N points
return res;
}
// Function to display the content of
// the vector A
void printVector(
vector<pair<double, double> > A)
{
// Iterate over A
for (pair<double, double> P : A) {
// Print the N random points stored
printf("(%.2lf, %.2lf)\n",
P.first, P.second);
}
}
// Driver Code
int main()
{
// Given dimensions
int R = 12;
int X = 3;
int Y = 3;
int N = 5;
// Function Call
printVector(randPoint(R, X, Y, N));
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
class GFG{
static final double PI = 3.141592653589;
static class pair
{
double first, second;
public pair(double first,
double second)
{
super();
this.first = first;
this.second = second;
}
}
// Return a random double between 0 & 1
static double uniform(){return Math.random();}
// Function to find the N random points on
// the given circle
static Vector<pair> randPoint(int r, int x,
int y, int n)
{
// Result vector
Vector<pair> res = new Vector<pair>();
for(int i = 0; i < n; i++)
{
// Get Angle in radians
double theta = 2 * PI * uniform();
// Get length from center
double len = Math.sqrt(uniform()) * r;
// Add point to results.
res.add(new pair(x + len * Math.cos(theta),
y + len * Math.sin(theta)));
}
// Return the N points
return res;
}
// Function to display the content of
// the vector A
static void printVector(Vector<pair> A)
{
// Iterate over A
for(pair P : A)
{
// Print the N random points stored
System.out.printf("(%.2f, %.2f)\n",
P.first, P.second);
}
}
// Driver Code
public static void main(String[] args)
{
// Given dimensions
int R = 12;
int X = 3;
int Y = 3;
int N = 5;
// Function call
printVector(randPoint(R, X, Y, N));
}
}
// This code is contributed by Rajput-Ji
C
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
static readonly double PI = 3.141592653589;
class pair
{
public double first, second;
public pair(double first,
double second)
{
this.first = first;
this.second = second;
}
}
// Return a random double between 0 & 1
static double uniform()
{
return new Random().NextDouble();
}
// Function to find the N random points on
// the given circle
static List<pair> randPoint(int r, int x,
int y, int n)
{
// Result vector
List<pair> res = new List<pair>();
for(int i = 0; i < n; i++)
{
// Get Angle in radians
double theta = 2 * PI * uniform();
// Get length from center
double len = Math.Sqrt(uniform()) * r;
// Add point to results.
res.Add(new pair(x + len * Math.Cos(theta),
y + len * Math.Sin(theta)));
}
// Return the N points
return res;
}
// Function to display the content of
// the vector A
static void printList(List<pair> A)
{
// Iterate over A
foreach(pair P in A)
{
// Print the N random points stored
Console.Write("({0:F2}, {1:F2})\n",
P.first, P.second);
}
}
// Driver Code
public static void Main(String[] args)
{
// Given dimensions
int R = 12;
int X = 3;
int Y = 3;
int N = 5;
// Function call
printList(randPoint(R, X, Y, N));
}
}
// This code is contributed by 29AjayKumar
Output:
(7.05, -3.36)
(5.21, -7.49)
(7.53, 0.19)
(-2.37, 12.05)
(1.45, 11.80)
时间复杂度:O(N) T5】空间复杂度: O(N)
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