集合上不可逆关系的个数

原文:https://www . geeksforgeeks . org/非灵活关系集数/

给定一个正整数 N ，任务是找出在给定的元素集合上可以形成的不灵活的 关系 的数量。由于计数可以很大，打印到 模 10 9 + 7 。

集合 A 上的关系 R 如果没有 (a，a) € R 适用于每个元素 a € A 。 例如:如果集合 A = {a，b}，那么 R = {(a，b)，(b，a)}就是不可逆关系。

示例:

输入: N = 2 输出: 4 解释: 考虑集合{1，2}，总的可能不灵活关系为:

  • {}
  • {(1, 2)}
  • {(2, 1)}
  • {(1, 2), (2, 1)}

输入:N = 5 T3】输出: 1048576

方法:按照以下步骤解决问题:

  • 集合上的 A 关系RA是集合的 笛卡尔乘积的子集，即 A * A 与 N 2 元素。
  • 非弹性关系:集合 A 上的关系 R 称为非弹性当且仅当xT8RT12】x[(x，x)不属于 R] 对于 A 中的每个元素 x 。
  • 笛卡尔乘积中存在总共 N 对 (x，x) 的，它们不应该包含在非弹性关系中。因此，对于剩余的(N2–N)元素，每个元素都有两个选择，即在子集中包含或排除它。
  • 因此，可能的非弹性关系的总数由2(N2–N)给出。

下面是上述方法的实现:

C++

// C++ program for the above approach
#include <iostream>
using namespace std;

const int mod = 1000000007;

// Function to calculate x^y
// modulo 1000000007 in O(log y)
int power(long long x, unsigned int y)
{
    // Stores the result of x^y
    int res = 1;

    // Update x if it exceeds mod
    x = x % mod;

    // If x is divisible by mod
    if (x == 0)
        return 0;

    while (y > 0) {

        // If y is odd, then
        // multiply x with result
        if (y & 1)
            res = (res * x) % mod;

        // Divide y by 2
        y = y >> 1;

        // Update the value of x
        x = (x * x) % mod;
    }

    // Return the resultant value of x^y
    return res;
}

// Function to count the number of
// irreflixive relations in a set
// consisting of N elements
int irreflexiveRelation(int N)
{

    // Return the resultant count
    return power(2, N * N - N);
}

// Driver Code
int main()
{
    int N = 2;
    cout << irreflexiveRelation(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.io.*;
import java.util.*;
class GFG
{

static int mod = 1000000007;

// Function to calculate x^y
// modulo 1000000007 in O(log y)
static int power(int x, int y)
{

    // Stores the result of x^y
    int res = 1;

    // Update x if it exceeds mod
    x = x % mod;

    // If x is divisible by mod
    if (x == 0)
        return 0;

    while (y > 0)
    {

        // If y is odd, then
        // multiply x with result
        if ((y & 1) != 0)
            res = (res * x) % mod;

        // Divide y by 2
        y = y >> 1;

        // Update the value of x
        x = (x * x) % mod;
    }

    // Return the resultant value of x^y
    return res;
}

// Function to count the number of
// irreflixive relations in a set
// consisting of N elements
static int irreflexiveRelation(int N)
{

    // Return the resultant count
    return power(2, N * N - N);
}

// Driver Code
public static void main(String[] args)
{
    int N = 2;
    System.out.println(irreflexiveRelation(N));
}
}

// This code is contributed by code_hunt.

Python 3

# Python3 program for the above approach
mod = 1000000007

# Function to calculate x^y
# modulo 1000000007 in O(log y)
def power(x, y):
    global mod

    # Stores the result of x^y
    res = 1

    # Update x if it exceeds mod
    x = x % mod

    # If x is divisible by mod
    if (x == 0):
        return 0

    while (y > 0):

        # If y is odd, then
        # multiply x with result
        if (y & 1):
            res = (res * x) % mod

        # Divide y by 2
        y = y >> 1

        # Update the value of x
        x = (x * x) % mod

    # Return the resultant value of x^y
    return res

# Function to count the number of
# irreflixive relations in a set
# consisting of N elements
def irreflexiveRelation(N):

    # Return the resultant count
    return power(2, N * N - N)

# Driver Code
if __name__ == '__main__':
    N = 2
    print(irreflexiveRelation(N))

    # This code is contributed by mohit kumar 29.

C

// C# program for above approach
using System;

public class GFG
{

  static int mod = 1000000007;

  // Function to calculate x^y
  // modulo 1000000007 in O(log y)
  static int power(int x, int y)
  {

    // Stores the result of x^y
    int res = 1;

    // Update x if it exceeds mod
    x = x % mod;

    // If x is divisible by mod
    if (x == 0)
      return 0;

    while (y > 0)
    {

      // If y is odd, then
      // multiply x with result
      if ((y & 1) != 0)
        res = (res * x) % mod;

      // Divide y by 2
      y = y >> 1;

      // Update the value of x
      x = (x * x) % mod;
    }

    // Return the resultant value of x^y
    return res;
  }

  // Function to count the number of
  // irreflixive relations in a set
  // consisting of N elements
  static int irreflexiveRelation(int N)
  {

    // Return the resultant count
    return power(2, N * N - N);
  }

  // Driver code
  public static void Main(String[] args)
  {
    int N = 2;
    Console.WriteLine(irreflexiveRelation(N));
  }
}

// This code is contributed by sanjoy_62.

java 描述语言

<script>

// Javascript program for the above approach

let mod = 1000000007;

// Function to calculate x^y
// modulo 1000000007 in O(log y)
function power(x, y)
{

    // Stores the result of x^y
    let res = 1;

    // Update x if it exceeds mod
    x = x % mod;

    // If x is divisible by mod
    if (x == 0)
        return 0;

    while (y > 0)
    {

        // If y is odd, then
        // multiply x with result
        if ((y & 1) != 0)
            res = (res * x) % mod;

        // Divide y by 2
        y = y >> 1;

        // Update the value of x
        x = (x * x) % mod;
    }

    // Return the resultant value of x^y
    return res;
}

// Function to count the number of
// irreflixive relations in a set
// consisting of N elements
function irreflexiveRelation(N)
{

    // Return the resultant count
    return power(2, N * N - N);
}

// Driver code

    let N = 2;
    document.write(irreflexiveRelation(N));

</script>

Output: 

4

时间复杂度:O(log N) T3】辅助空间: O(1)

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