集合上既不是自反也不是自反的关系数

给定一个正整数N ，任务是在一组前N个自然数上找到既不是自反也不是反自反的关系的数量。由于关系的数量可能非常大，因此将其打印为模 10 ⁹ + 7。

A relation R on a set A is called reflexive, if no (a, a) € R holds for every element a € A.
For Example: If set A = {a, b} then R = {(a, b), (b, a)} is irreflexive relation.

例子：

Input: N = 2
Output: 8
Explanation: Considering the set {1, 2}, the total possible relations that are neither reflexive nor irreflexive are:

1. {(1, 1)}
2. {(1, 1), (1, 2)}
3. {(1, 1), (2, 1)}
4. {(1, 1), (1, 2), (2, 1)}
5. {(2, 2)}
6. {(2, 2), (2, 1)}
7. {(2, 2), (1, 2)}
8. {(2, 2), (2, 1), (1, 2)}

Therefore, the total count is 8.

Input: N = 3
Output: 384

方法：可以根据以下观察解决给定的问题：

• 集合A上的关系R是笛卡尔积的子集 一个集合，即A * A与N ^{2 个}元素。
• 如果关系不包含至少一对(x, x) ，则关系将是非自反的，如果它包含至少一对(x, x)其中x € R ，则关系将是非自反的。
• 可以得出结论，如果该关系包含至少一对(x, x)并且最多(N – 1)对(x, x) ，则该关系将是非自反和非自反的。
• 在N对(x, x)中，选择除0和N – 1之外的任意对的可能性总数为(2 ^N – 2) 。对于剩余的(N ² – N) 个元素，每个元素都有两个选择，即在子集中包含或排除它。

从上面的观察，在一组前N个自然数上既不是自反也不是反自反的关系总数 是（谁）给的$(2^N - 2)*(2^{N^2 - N})$ .

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
const int mod = 1000000007;
 
// Function to calculate x^y
// modulo 10^9 + 7 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 res
        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 value of x^y
    return res;
}
 
// Function to count the number
// of relations that are neither
// reflexive nor irreflexive
void countRelations(int N)
{
    // Return the resultant count
    cout << (power(2, N) - 2)
                * power(2, N * N - N);
}
 
// Driver Code
int main()
{
    int N = 2;
    countRelations(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
static int mod = 1000000007;
 
// Function to calculate x^y
// modulo 10^9 + 7 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 res
        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 value of x^y
    return res;
}
 
// Function to count the number
// of relations that are neither
// reflexive nor irreflexive
static void countRelations(int N)
{
     
    // Return the resultant count
    System.out.print((power(2, N) - 2) *
                      power(2, N * N - N));
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 2;
     
    countRelations(N);
}
}
 
// This code is contributed by susmitakundugoaldanga

Python3

# Python program for the above approach
mod = 1000000007
 
# Function to calculate x^y
# modulo 10^9 + 7 in O(log y)
def power(x, y):
 
    # 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 res
        if (y % 2 == 1):
            res = (res * x) % mod
 
        # Divide y by 2
        y = y >> 1
 
        # Update the value of x
        x = (x * x) % mod
 
    # Return the value of x^y
    return res
 
# Function to count the number
# of relations that are neither
# reflexive nor irreflexive
def countRelations(N):
 
    # Return the resultant count
    print((power(2, N) - 2) * power(2, N * N - N))
 
# Driver Code
N = 2
countRelations(N)
 
# This code is contributed by abhinavjain194

C#

// C# program for the above approach
using System;
 
class GFG{
 
static int mod = 1000000007;
 
// Function to calculate x^y
// modulo 10^9 + 7 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 res
        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 value of x^y
    return res;
}
 
// Function to count the number
// of relations that are neither
// reflexive nor irreflexive
static void countRelations(int N)
{
     
    // Return the resultant count
    Console.Write((power(2, N) - 2) *
                   power(2, N * N - N));
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 2;
    countRelations(N);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript program for the above approach
 
var mod = 1000000007;
 
// Function to calculate x^y
// modulo 10^9 + 7 in O(log y)
function power(x, y)
{
    // Stores the result of (x^y)
    var 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 res
        if (y %2 != 0)
            res = (res * x) % mod;
 
        // Divide y by 2
        y = y >> 1;
 
        // Update the value of x
        x = (x * x) % mod;
    }
 
    // Return the value of x^y
    return res;
}
 
// Function to count the number
// of relations that are neither
// reflexive nor irreflexive
function countRelations(N)
{
    // Return the resultant count
    document.write((power(2, N) - 2)
                * power(2, (N * N) - N));
}
 
// Driver Code
var N = 2;
countRelations(N);
 

输出：

8

时间复杂度： O(log N)
辅助空间： O(1)