计算对(i，j)的对数，以使arr [i] * arr [j] = arr [i] + arr [j]

给定长度为N的数组arr []，计算对数(i，j)，以使arr [i] * arr [j] = arr [i] + arr [j]和0 <= i 还假定数组的元素可以是任何正整数，包括零。

例子：

Input : arr[] = {2, 0, 3, 2, 0} 
Output : 2

Input : arr[] = {1, 2, 3, 4}
Output : 0

简单的解决方案：
我们可以生成数组的所有可能的对，并对满足给定条件的那些对进行计数。

下面是上述方法的实现：

CPP

// C++ program to count pairs (i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
  
#include 
using namespace std;
  
// Function to return the count of pairs(i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
long countPairs(int arr[], int n)
{
    long count = 0;
  
    for (int i = 0; i < n - 1; i++) {
        for (int j = i + 1; j < n; j++) {
  
            // Increment count if condition satisfy
            if (arr[i] * arr[j] == arr[i] + arr[j])
                count++;
        }
    }
  
    // Return count of pairs
    return count;
}
  
// Driver code
int main()
{
  
    int arr[] = { 2, 0, 3, 2, 0 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Get and print count of pairs
    cout << countPairs(arr, n);
  
    return 0;
}

Java

// Java program to count pairs (i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
  
class GFG {
    // Function to return the count of pairs(i, j)
    // such that arr[i] * arr[j] = arr[i] + arr[j]
    static long countPairs(int arr[], int n)
    {
        long count = 0;
  
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
  
                // Increment count if condition satisfy
                if (arr[i] * arr[j] == arr[i] + arr[j])
                    count++;
            }
        }
  
        // Return count of pairs
        return count;
    }
  
    // Driver code
    public static void main(String[] args)
    {
  
        int arr[] = { 2, 0, 3, 2, 0 };
        int n = arr.length;
  
        // Get and print count of pairs
        System.out.println(countPairs(arr, n));
    }
}

Python3

# Python3 program to count pairs (i, j) 
# such that arr[i] * arr[j] = arr[i] + arr[j] 
  
# Function to return the count of pairs(i, j) 
# such that arr[i] * arr[j] = arr[i] + arr[j] 
def countPairs(arr, n) : 
  
    count = 0; 
  
    for i in range(n - 1) :
        for j in range(i + 1, n) :
  
            # Increment count if condition satisfy 
            if (arr[i] * arr[j] == arr[i] + arr[j]) :
                count += 1; 
  
    # Return count of pairs 
    return count; 
  
# Driver code 
if __name__ == "__main__" : 
  
    arr = [ 2, 0, 3, 2, 0 ]; 
    n = len(arr); 
  
    # Get and print count of pairs 
    print(countPairs(arr, n)); 
      
# This code is contributed by AnkitRai01

C#

// C# program to count pairs (i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
  
using System;
class GFG {
    // Function to return the count of pairs(i, j)
    // such that arr[i] * arr[j] = arr[i] + arr[j]
    static long countPairs(int[] arr, int n)
    {
        long count = 0;
  
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
  
                // Increment count if condition satisfy
                if (arr[i] * arr[j] == arr[i] + arr[j])
                    count++;
            }
        }
  
        // Return count of pairs
        return count;
    }
  
    // Driver code
    public static void Main(string[] args)
    {
  
        int[] arr = { 2, 0, 3, 2, 0 };
        int n = arr.Length;
  
        // Get and print count of pairs
        Console.WriteLine(countPairs(arr, n));
    }
}

输出：

2

时间复杂度： O(n ² )

高效的解决方案：
以arr [i]为x且arr [j]为y，我们可以将给定条件重写为以下方程式。

xy = x + y
xy - x - y = 0
xy - x - y + 1 = 1
x(y - 1) -(y - 1) = 1 
(x - 1)(y - 1) = 1

Case 1:
x - 1 = 1 i.e x = 2
y - 1 = 1 i.e y = 2

Case 2:
x - 1 = -1 i.e x = 0
y - 1 = -1 i.e y = 0

因此，现在我们知道，只有当arr [i] = arr [j] = 0或arr [i] = arr时，条件arr [i] * arr [j] = arr [i] + arr [j]才能满足[j] = 2 。
我们需要做的只是计算2和0的出现次数。然后我们可以使用公式获得对数

(count * (count - 1)) / 2

下面是上述方法的实现：

CPP

// C++ program to count pairs (i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
  
#include 
using namespace std;
  
// Function to return the count of pairs(i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
long countPairs(int arr[], int n)
{
  
    int countZero = 0;
    int countTwo = 0;
  
    // Count number of 0's and 2's in the array
    for (int i = 0; i < n; i++) {
        if (arr[i] == 0)
            countZero++;
  
        else if (arr[i] == 2)
            countTwo++;
    }
  
    // Total pairs due to occurrence of 0's
    long pair0 = (countZero * (countZero - 1)) / 2;
  
    // Total pairs due to occurrence of 2's
    long pair2 = (countTwo * (countTwo - 1)) / 2;
  
    // Return count of all pairs
    return pair0 + pair2;
}
  
// Driver code
int main()
{
  
    int arr[] = { 2, 0, 3, 2, 0 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Get and print count of pairs
    cout << countPairs(arr, n);
  
    return 0;
}

Java

// Java program to count pairs (i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
  
class GFG {
    // Function to return the count of pairs(i, j)
    // such that arr[i] * arr[j] = arr[i] + arr[j]
    static long countPairs(int arr[], int n)
    {
  
        int countZero = 0;
        int countTwo = 0;
  
        // Count number of 0's and 2's in the array
        for (int i = 0; i < n; i++) {
            if (arr[i] == 0)
                countZero++;
  
            else if (arr[i] == 2)
                countTwo++;
        }
  
        // Total pairs due to occurrence of 0's
        long pair0 = (countZero * (countZero - 1)) / 2;
  
        // Total pairs due to occurrence of 2's
        long pair2 = (countTwo * (countTwo - 1)) / 2;
  
        // Return count of all pairs
        return pair0 + pair2;
    }
  
    // Driver code
    public static void main(String[] args)
    {
  
        int arr[] = { 2, 0, 3, 2, 0 };
        int n = arr.length;
  
        // Get and print count of pairs
        System.out.println(countPairs(arr, n));
    }
}

Python3

# Python3 program to count pairs (i, j) 
# such that arr[i] * arr[j] = arr[i] + arr[j] 
  
# Function to return the count of pairs(i, j) 
# such that arr[i] * arr[j] = arr[i] + arr[j] 
def countPairs(arr, n): 
  
    countZero = 0; 
    countTwo = 0; 
  
    # Count number of 0's and 2's in the array 
    for i in range(n) : 
        if (arr[i] == 0) :
            countZero += 1; 
  
        elif (arr[i] == 2) :
            countTwo += 1;
  
    # Total pairs due to occurrence of 0's 
    pair0 = (countZero * (countZero - 1)) // 2; 
  
    # Total pairs due to occurrence of 2's 
    pair2 = (countTwo * (countTwo - 1)) // 2; 
  
    # Return count of all pairs 
    return pair0 + pair2; 
  
# Driver code 
if __name__ == "__main__" : 
  
    arr = [ 2, 0, 3, 2, 0 ]; 
    n = len(arr); 
  
    # Get and print count of pairs 
    print(countPairs(arr, n)); 
  
# This code is contributed by AnkitRai01

C＃

// C# program to count pairs (i, j)
// such that arr[i] * arr[j] = arr[i] + arr[j]
  
using System;
class GFG {
    // Function to return the count of pairs(i, j)
    // such that arr[i] * arr[j] = arr[i] + arr[j]
    static long countPairs(int[] arr, int n)
    {
  
        int countZero = 0;
        int countTwo = 0;
  
        // Count number of 0's and 2's in the array
        for (int i = 0; i < n; i++) {
            if (arr[i] == 0)
                countZero++;
  
            else if (arr[i] == 2)
                countTwo++;
        }
  
        // Total pairs due to occurrence of 0's
        long pair0 = (countZero * (countZero - 1)) / 2;
  
        // Total pairs due to occurrence of 2's
        long pair2 = (countTwo * (countTwo - 1)) / 2;
  
        // Return count of all pairs
        return pair0 + pair2;
    }
  
    // Driver code
    public static void Main(string[] args)
    {
  
        int[] arr = { 2, 0, 3, 2, 0 };
        int n = arr.Length;
  
        // Get and print count of pairs
        Console.WriteLine(countPairs(arr, n));
    }
}

输出：

2

时间复杂度： O(n)