给定n个整数的数组，直线的斜率即m和直线的截距即c，计算点i≠j的点的有序对(i，j)的数量，使得点(A i ，A j )满足以给定的斜率和截距形成的线。
注意：直线的方程式为y = mx + c，其中m为直线的斜率，c为截距。
例子 ：
Input : m = 1, c = 1, arr[] = [ 1, 2, 3, 4, 2 ]
Output : 5 ordered points
Explanation : The equation of the line with given slope and intercept is : y = x + 1. The Number of pairs (i, j), for which (arri, arrj) satisfies the above equation of the line are : (1, 2), (1, 5), (2, 3), (3, 4), (5, 3).
Input : m = 2, c = 1, arr[] = [ 1, 2, 3, 4, 2, 5 ]
Output : 3 ordered points
方法1(蛮力)：
生成所有可能的对(i，j)，并检查特定的有序对(i，j)是否满足(arr i ，arr j )满足线y = mx + c的给定方程，且i≠j。如果该点有效(如果满足上述条件，则该点有效)，增加存储有效点总数的计数器。
C++
// CPP code to count the number of ordered
// pairs satisfying Line Equation
#include
using namespace std;
/* Checks if (i, j) is valid, a point (i, j)
is valid if point (arr[i], arr[j])
satisfies the equation y = mx + c And
i is not equal to j*/
bool isValid(int arr[], int i, int j,
int m, int c)
{
// check if i equals to j
if (i == j)
return false;
// Equation LHS = y, and RHS = mx + c
int lhs = arr[j];
int rhs = m * arr[i] + c;
return (lhs == rhs);
}
/* Returns the number of ordered pairs
(i, j) for which point (arr[i], arr[j])
satisfies the equation of the line
y = mx + c */
int findOrderedPoints(int arr[], int n,
int m, int c)
{
int counter = 0;
// for every possible (i, j) check
// if (a[i], a[j]) satisfies the
// equation y = mx + c
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
// (firstIndex, secondIndex)
// is same as (i, j)
int firstIndex = i, secondIndex = j;
// check if (firstIndex,
// secondIndex) is a valid point
if (isValid(arr, firstIndex, secondIndex, m, c))
counter++;
}
}
return counter;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
// equation of line is y = mx + c
int m = 1, c = 1;
cout << findOrderedPoints(arr, n, m, c);
return 0;
} Java
// Java code to find number of ordered
// points satisfying line equation
import java.io.*;
public class GFG {
// Checks if (i, j) is valid,
// a point (i, j) is valid if
// point (arr[i], arr[j])
// satisfies the equation
// y = mx + c And
// i is not equal to j
static boolean isValid(int []arr, int i,
int j, int m, int c)
{
// check if i equals to j
if (i == j)
return false;
// Equation LHS = y,
// and RHS = mx + c
int lhs = arr[j];
int rhs = m * arr[i] + c;
return (lhs == rhs);
}
/* Returns the number of ordered pairs
(i, j) for which point (arr[i], arr[j])
satisfies the equation of the line
y = mx + c */
static int findOrderedPoints(int []arr,
int n, int m, int c)
{
int counter = 0;
// for every possible (i, j) check
// if (a[i], a[j]) satisfies the
// equation y = mx + c
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
// (firstIndex, secondIndex)
// is same as (i, j)
int firstIndex = i,
secondIndex = j;
// check if (firstIndex,
// secondIndex) is a
// valid point
if (isValid(arr, firstIndex,
secondIndex, m, c))
counter++;
}
}
return counter;
}
// Driver Code
public static void main(String args[])
{
int []arr = { 1, 2, 3, 4, 2 };
int n = arr.length;
// equation of line is y = mx + c
int m = 1, c = 1;
System.out.print(
findOrderedPoints(arr, n, m, c));
}
}
// This code is contributed by
// Manish Shaw (manishshaw1)Python3
# Python code to count the number of ordered
# pairs satisfying Line Equation
# Checks if (i, j) is valid, a point (i, j)
# is valid if point (arr[i], arr[j])
# satisfies the equation y = mx + c And
# i is not equal to j
def isValid(arr, i, j, m, c) :
# check if i equals to j
if (i == j) :
return False
# Equation LHS = y, and RHS = mx + c
lhs = arr[j];
rhs = m * arr[i] + c
return (lhs == rhs)
# Returns the number of ordered pairs
# (i, j) for which point (arr[i], arr[j])
# satisfies the equation of the line
# y = mx + c */
def findOrderedPoints(arr, n, m, c) :
counter = 0
# for every possible (i, j) check
# if (a[i], a[j]) satisfies the
# equation y = mx + c
for i in range(0, n) :
for j in range(0, n) :
# (firstIndex, secondIndex)
# is same as (i, j)
firstIndex = i
secondIndex = j
# check if (firstIndex,
# secondIndex) is a valid point
if (isValid(arr, firstIndex,
secondIndex, m, c)) :
counter = counter + 1
return counter
# Driver Code
arr = [ 1, 2, 3, 4, 2 ]
n = len(arr)
# equation of line is y = mx + c
m = 1
c = 1
print (findOrderedPoints(arr, n, m, c))
# This code is contributed by Manish Shaw
# (manishshaw1)C#
// C# code to find number of ordered
// points satisfying line equation
using System;
class GFG {
// Checks if (i, j) is valid,
// a point (i, j) is valid if
// point (arr[i], arr[j])
// satisfies the equation
// y = mx + c And
// i is not equal to j
static bool isValid(int []arr, int i,
int j, int m, int c)
{
// check if i equals to j
if (i == j)
return false;
// Equation LHS = y,
// and RHS = mx + c
int lhs = arr[j];
int rhs = m * arr[i] + c;
return (lhs == rhs);
}
/* Returns the number of ordered pairs
(i, j) for which point (arr[i], arr[j])
satisfies the equation of the line
y = mx + c */
static int findOrderedPoints(int []arr, int n,
int m, int c)
{
int counter = 0;
// for every possible (i, j) check
// if (a[i], a[j]) satisfies the
// equation y = mx + c
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
// (firstIndex, secondIndex)
// is same as (i, j)
int firstIndex = i, secondIndex = j;
// check if (firstIndex,
// secondIndex) is a valid point
if (isValid(arr, firstIndex, secondIndex, m, c))
counter++;
}
}
return counter;
}
// Driver Code
public static void Main()
{
int []arr = { 1, 2, 3, 4, 2 };
int n = arr.Length;
// equation of line is y = mx + c
int m = 1, c = 1;
Console.Write(findOrderedPoints(arr, n, m, c));
}
}
// This code is contributed by
// Manish Shaw (manishshaw1)PHP
输出 ：
5
时间复杂度： O(n 2 )
方法2(高效)：
给定一个点的轴坐标，对于每个x，都有一个唯一的y值，并且y的值不过是m * x + c。因此，对于数组arr的每个可能的x坐标，计算满足该行方程的y的唯一值在该数组中出现了多少次。将数组arr的所有整数的计数存储在映射中。现在，对于每个值arr i ，在答案中加上m * arr i + c的出现次数。对于给定的i，m * a [i] + c在数组中出现x次，然后将x添加到我们的计数器中以获得总有效点，但是需要检查a [i] = m * a [i] + c，那么很明显，因为这在数组中发生了x次，所以在第i个索引处出现了一次，而其余(x – 1)次出现是有效的y坐标，因此将(x – 1)加到我们的点数计数器上。
// CPP code to find number of ordered
// points satisfying line equation
#include
using namespace std;
/* Returns the number of ordered pairs
(i, j) for which point (arr[i], arr[j])
satisfies the equation of the line
y = mx + c */
int findOrderedPoints(int arr[], int n,
int m, int c)
{
int counter = 0;
// map stores the frequency
// of arr[i] for all i
unordered_map frequency;
for (int i = 0; i < n; i++)
frequency[arr[i]]++;
for (int i = 0; i < n; i++)
{
int xCoordinate = arr[i];
int yCoordinate = (m * arr[i] + c);
// if for a[i](xCoordinate),
// a yCoordinate exists in the map
// add the frequency of yCoordinate
// to the counter
if (frequency.find(yCoordinate) !=
frequency.end())
counter += frequency[yCoordinate];
// check if xCoordinate = yCoordinate,
// if this is true then since we only
// want (i, j) such that i != j, decrement
// the counter by one to avoid points
// of type (arr[i], arr[i])
if (xCoordinate == yCoordinate)
counter--;
}
return counter;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
int m = 1, c = 1;
cout << findOrderedPoints(arr, n, m, c);
return 0;
}
输出：
5
时间复杂度： O(n)
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