一组 N 个元素上的不对称关系数

原文:https://www . geesforgeks . org/n 元素集合上的非对称关系数/

给定一个正整数 N ，任务是在一组 N 元素中找到关系T7】的个数。由于关系数可以很大，打印出来模 10 9 +7 。

集合 A 上的关系 R 称为非对称当且仅当 x R y 存在时，则yT10RT13x为每一个 (x，y) € A 。 比如:如果设置 A = {a，b}，那么 R = {(a，b)} 就是不对称关系。

*示例:*

*输入:* N = 2 输出: 3 说明:考虑集合{1，2}，可能的不对称关系总数为{{}、{(1，2)}、{(2，1)}}。

*输入:*N = 5 T3】输出: 59049

*方法:*给定的问题可以基于以下观察来解决:

  • 集合 A 上的 A 关系 R 是集合的 笛卡尔乘积的子集，即 A * A 与 N 2 元素。
  • 笛卡尔乘积中存在总共 N 对类型 (x，x) ，其中任何 (x，x) 都不应包含在子集内。
  • 现在，剩下一个是笛卡儿积的(N2–N)元素。****
  • 为了满足非对称关系的性质，有三种可能性，要么只包括类型为 (x，y) 的，要么只包括类型为 (y，x) 的，要么不从单个组进入子集。
  • 因此，可能的不对称关系的总数等于3(N2–N)/2。****

因此，想法是打印3(N2–N)/2模 10 9 + 7 的值作为结果。

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 (10^9 + 7)
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 final
    // value of x ^ y
    return res;
}

// Function to count the number of
// asymmetric relations in a set
// consisting of N elements
int asymmetricRelation(int N)
{
    // Return the resultant count
    return power(3, (N * N - N) / 2);
}

// Driver Code
int main()
{
    int N = 2;
    cout << asymmetricRelation(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.*;

class GFG{

final static int mod = 1000000007;

// Function to calculate
// x^y modulo (10^9 + 7)
public 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 % 2 == 1)
            res = (res * x) % mod;

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

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

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

// Function to count the number of
// asymmetric relations in a set
// consisting of N elements
public static int asymmetricRelation(int N)
{

    // Return the resultant count
    return power(3, (N * N - N) / 2);
}

// Driver code
public static void main (String[] args)
{
    int N = 2;

    System.out.print(asymmetricRelation(N));
}
}

// This code is contributed by user_qa7r

Python 3

# Python3 program for the above approach
mod = 1000000007

# Function to calculate
# x^y modulo (10^9 + 7)
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 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 final
    # value of x ^ y
    return res

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

    # Return the resultant count
    return power(3, (N * N - N) // 2)

# Driver Code
if __name__ == '__main__':

    N = 2

    print(asymmetricRelation(N))

# This code is contributed by SURENDRA_GANGWAR

C#

// C# program for the above approach
using System;

class GFG{

const int mod = 1000000007;

// Function to calculate
// x^y modulo (10^9 + 7)
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 final
    // value of x ^ y
    return res;
}

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

    // Return the resultant count
    return power(3, (N * N - N) / 2);
}

// Driver Code
public static void Main(string[] args)
{
    int N = 2;
    Console.WriteLine(asymmetricRelation(N));
}
}

// This code is contributed by ukasp

java 描述语言

<script>

// Javascript program for the above approach

var mod = 1000000007;

// Function to calculate
// x^y modulo (10^9 + 7)
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 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 final
    // value of x ^ y
    return res;
}

// Function to count the number of
// asymmetric relations in a set
// consisting of N elements
function asymmetricRelation( N)
{
    // Return the resultant count
    return power(3, (N * N - N) / 2);
}

// Driver Code
var N = 2;
document.write( asymmetricRelation(N));

</script>

**Output: 

3**

*时间复杂度: O(Nlog N) 辅助空间:* O(1)*

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