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📜  在给定素数P下具有等于其自身的模逆的计数数组元素

📅  最后修改于: 2021-04-22 01:24:31             🧑  作者: Mango

给定大小为N且质数为P的数组arr [] ,任务是对数组的元素进行计数,以使元素在模P下的模乘逆数等于元素本身。

例子:

天真的方法:最简单的方法是解决此问题,即遍历数组并打印数组元素的数量,以使元素在模P下的模乘逆数等于元素本身。

时间复杂度: O(N * log P)
辅助空间: O(1)

高效方法:为了优化上述方法,该思想基于以下观察结果:

请按照以下步骤解决问题:

  • 初始化一个变量,例如cntElem,以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 。如果发现是真的,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


输出:
2

时间复杂度: O(N)
辅助空间: O(1)