在给定的质数 P 等于自身的情况下，计算具有模逆的数组元素

给定一个大小为N的数组arr[]和一个素数P ，任务是计算数组的元素，使得模 P 下元素的模乘法逆等于元素本身。

例子：

Input: arr[] = {1, 6, 4, 5}, P = 7
Output: 2
Explanation:
Modular multiplicative inverse of arr[0](=1) under modulo P(= 7) is arr[0](= 1) itself.
Modular multiplicative inverse of arr1](= 6) under modulo P(= 7) is arr[1](= 6) itself.
Therefore, the required output is 2.

Input: arr[] = {1, 3, 8, 12, 12}, P = 13
Output: 3

朴素的方法：解决这个问题的最简单的方法是遍历数组并打印数组元素的计数，使得模 P 下元素的模乘法逆等于元素本身。

时间复杂度： O(N * log P)
辅助空间： O(1)

高效的方法：为了优化上述方法，该想法基于以下观察：

If X and Y are two numbers such that (X × Y) % P = 1, then Y is modulo inverse of X.
Therefore, If Y is X itself, then (X × X) % P must be 1.

请按照以下步骤解决问题：

• 初始化一个变量，比如cntElem来存储满足给定条件的元素的数量。
• 遍历给定的数组并检查(arr[i] * arr[i]) % P 是否等于1 。如果发现为真，则将 cntElem的计数增加1 。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to get the count
// of elements that satisfy
// the given condition.
int equvInverse(int arr[],
                int N, int P)
{
    // Stores count of elements
    // that satisfy the condition
    int cntElem = 0;
 
    // Traverse the given array.
    for (int i = 0; i < N; i++) {
 
        // If square of current
        // element is equal to 1
        if ((arr[i] * arr[i]) % P
            == 1) {
            cntElem++;
        }
    }
    return cntElem;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 6, 4, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int P = 7;
    cout << equvInverse(arr, N, P);
}

Java

// Java program to implement
// the above approach
import java.io.*;
 
class GFG{
 
// Function to get the count
// of elements that satisfy
// the given condition.
static int equvInverse(int[] arr,
                       int N, int P)
{
     
    // Stores count of elements
    // that satisfy the condition
    int cntElem = 0;
 
    // Traverse the given array.
    for(int i = 0; i < N; i++)
    {
         
        // If square of current
        // element is equal to 1
        if ((arr[i] * arr[i]) % P == 1)
        {
            cntElem++;
        }
    }
    return cntElem;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] arr = { 1, 6, 4, 5 };
    int N = arr.length;
    int P = 7;
     
    System.out.println(equvInverse(arr, N, P));
}
}
 
// This code is contributed by akhilsaini

Python3

# Python3 program to implement
# the above approach
 
# Function to get the count
# of elements that satisfy
# the given condition.
def equvInverse(arr, N, P):
     
    # Stores count of elements
    # that satisfy the condition
    cntElem = 0
 
    # Traverse the given array.
    for i in range(0, N):
         
        # If square of current
        # element is equal to 1
        if ((arr[i] * arr[i]) % P == 1):
            cntElem = cntElem + 1
 
    return cntElem
 
# Driver Code
if __name__ == "__main__":
     
    arr = [ 1, 6, 4, 5 ]
    N = len(arr)
    P = 7
     
    print(equvInverse(arr, N, P))
 
# This code is contributed by akhilsaini

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to get the count
// of elements that satisfy
// the given condition.
static int equvInverse(int[] arr, int N, int P)
{
     
    // Stores count of elements
    // that satisfy the condition
    int cntElem = 0;
 
    // Traverse the given array.
    for(int i = 0; i < N; i++)
    {
         
        // If square of current
        // element is equal to 1
        if ((arr[i] * arr[i]) % P == 1)
        {
            cntElem++;
        }
    }
    return cntElem;
}
 
// Driver Code
public static void Main()
{
    int[] arr = { 1, 6, 4, 5 };
    int N = arr.Length;
    int P = 7;
     
    Console.WriteLine(equvInverse(arr, N, P));
}
}
 
// This code is contributed by akhilsaini

Javascript

// Javascript program to implement
// the above approach
 
// Function to get the count
// of elements that satisfy
// the given condition.
function equvInverse(arr, N, P)
{
    // Stores count of elements
    // that satisfy the condition
    let cntElem = 0;
 
    // Traverse the given array.
    for (let i = 0; i < N; i++) {
 
        // If square of current
        // element is equal to 1
        if ((arr[i] * arr[i]) % P
            == 1) {
            cntElem++;
        }
    }
    return cntElem;
}
 
// Driver Code
let arr = [ 1, 6, 4, 5 ];
let N = arr.length;
let P = 7;
document.write(equvInverse(arr, N, P));
 
// This code is contributed by subham348.

输出：

2

时间复杂度： O(N)
辅助空间： O(1)

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