求直线通过给定点的方程
给定一个包含平面内 N 个坐标点的数组 arr ，任务是检查坐标点是否位于一条直线上。如果它们位于一条直线上，则打印是以及该直线的方程，否则打印否。
示例:
输入: arr[] = {{1，1}、{2，2}、{3，3}} 输出: 是 1x- 1y=0
输入: arr[] = {{0，1}，{2，0}} 输出: 是 2y+x-2 = 0
输入: arr[] = {{1，5}，{2，2}，{4，6}，{3，5}} 输出:否
方法:想法是找到可以使用数组中给定的任意一对点形成的直线的方程，如果所有其他点都满足使用这对点形成的直线的方程，那么所有这些点一起形成一条直线。所以，如果所有的点都满足直线方程，那么打印是后跟直线方程，否则打印否。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if a straight line
// can be formed using N points
void isStraightLinePossibleEq(
vector<pair<int, int> > arr,
int n)
{
// First pair of point (x0, y0)
int x0 = arr[0].first;
int y0 = arr[0].second;
// Second pair of point (x1, y1)
int x1 = arr[1].first;
int y1 = arr[1].second;
int dx = x1 - x0,
dy = y1 - y0,
c = dy * x0 - dx * y0;
// Loop to iterate over the points
// and check whether they satisfy
// the equation or not.
for (int i = 2; i < n; i++) {
int x = arr[i].first, y = arr[i].second;
if ((dx * y) - (dy * x) != c) {
cout << "No";
return;
}
}
cout << "Yes" << endl;
cout << dy << "x-" << dx
<< "y=" << c << "\n";
}
// Driver Code
int main()
{
// Array of points
vector<pair<int, int> > arr
= { { 0, 0 }, { 1, 1 }, { 3, 3 }, { 2, 2 } };
int N = 2;
// Function Call
isStraightLinePossibleEq(arr, 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.*;
public class GFG
{
// Function to check if a straight line
// can be formed using N points
static void isStraightLinePossibleEq(int arr[][], int n)
{
// First pair of point (x0, y0)
int x0 = arr[0][0];
int y0 = arr[0][1];
// Second pair of point (x1, y1)
int x1 = arr[1][0];
int y1 = arr[1][1];
int dx = x1 - x0,
dy = y1 - y0,
c = dy * x0 - dx * y0;
// Loop to iterate over the points
// and check whether they satisfy
// the equation or not.
for (int i = 2; i < n; i++) {
int x = arr[i][0], y = arr[i][1];
if ((dx * y) - (dy * x) != c) {
System.out.print("No");
return;
}
}
System.out.print("Yes" + "\n");
System.out.print(dy + "x-" + dx
+ "y=" + c + "\n");
}
// Driver Code
public static void main(String args[])
{
// Array of points
int arr[][] = {{ 0, 0 }, { 1, 1 }, { 3, 3 }, { 2, 2 }};
int N = 2;
// Function Call
isStraightLinePossibleEq(arr, N);
}
}
// This code is contributed by Samim Hossain Mondal.
Python 3
# Python code for the above approach
# Function to check if a straight line
# can be formed using N points
def isStraightLinePossibleEq(arr, n):
# First pair of point (x0, y0)
x0 = arr[0][0]
y0 = arr[0][1]
# Second pair of point (x1, y1)
x1 = arr[1][0]
y1 = arr[1][1]
dx = x1 - x0
dy = y1 - y0
c = dy * x0 - dx * y0
# Loop to iterate over the points
# and check whether they satisfy
# the equation or not.
for i in range(2, n):
x = arr[i][0], y = arr[i][1]
if (dx * y) - (dy * x) != c:
print("No")
return
print("Yes")
print(str(dy)+ "x" +"-" + str(dx)+ "y"+ "="+ str(c))
# Driver Code
# Array of points
arr = [[0, 0], [1, 1], [3, 3], [2, 2]]
N = 2
# Function Call
isStraightLinePossibleEq(arr, N)
# This code is contributed by Potta Lokesh
C
// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG
{
// Function to check if a straight line
// can be formed using N points
static void isStraightLinePossibleEq(int [,]arr, int n)
{
// First pair of point (x0, y0)
int x0 = arr[0, 0];
int y0 = arr[0, 1];
// Second pair of point (x1, y1)
int x1 = arr[1, 0];
int y1 = arr[1, 1];
int dx = x1 - x0,
dy = y1 - y0,
c = dy * x0 - dx * y0;
// Loop to iterate over the points
// and check whether they satisfy
// the equation or not.
for (int i = 2; i < n; i++) {
int x = arr[i, 0], y = arr[i, 1];
if ((dx * y) - (dy * x) != c) {
Console.Write("No");
return;
}
}
Console.Write("Yes" + "\n");
Console.Write(dy + "x-" + dx
+ "y=" + c + "\n");
}
// Driver Code
public static void Main()
{
// Array of points
int[,] arr = new int[4, 2] {{ 0, 0 }, { 1, 1 }, { 3, 3 }, { 2, 2 }};
int N = 2;
// Function Call
isStraightLinePossibleEq(arr, N);
}
}
// This code is contributed by Samim Hossain Mondal.
java 描述语言
<script>
// JavaScript program for the above approach
// Function to check if a straight line
// can be formed using N points
const isStraightLinePossibleEq = (arr, n) => {
// First pair of point (x0, y0)
let x0 = arr[0][0];
let y0 = arr[0][1];
// Second pair of point (x1, y1)
let x1 = arr[1][0];
let y1 = arr[1][1];
let dx = x1 - x0,
dy = y1 - y0,
c = dy * x0 - dx * y0;
// Loop to iterate over the points
// and check whether they satisfy
// the equation or not.
for (let i = 2; i < n; i++) {
let x = arr[i][0], y = arr[i][1];
if ((dx * y) - (dy * x) != c) {
cout << "No";
return;
}
}
document.write("Yes<br/>");
document.write(`${dy}x-${dx}y=${c}<br/>`);
}
// Driver Code
// Array of points
let arr = [[0, 0], [1, 1], [3, 3], [2, 2]];
let N = 2;
// Function Call
isStraightLinePossibleEq(arr, N);
// This code is contributed by rakeshsahni
</script>
Output
Yes
1x-1y=0
时间复杂度:O(N) T3】辅助空间: O(1)
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