计算不包含连续数字的有序集合的数量

给定一个正整数N 。计算任何大小的有序集合的总数，使得集合中不存在连续数字并且所有数字 <= N。

Ordered Set: Arrays in which all numbers are distinct, and order of elements is taken into consideration, For e.g. {1, 2, 3} is different from {1, 3, 2}. 
 

例子：

Input: N = 3 
Output: 5 
Explanation: 
The ordered sets are: 
{1}, {2}, {3}, {1, 3}, {3, 1}

Input: N = 6 
Output: 50 
 

天真的方法：

• 我们将使用递归来解决这个问题。如果我们获得有序集合的计数 C，其中元素按升序/降序排列且集合大小为 S，则此大小的总计数将

$C * S!$

• .我们将为每个有序的不同尺寸集执行此操作。
• 迭代所有有序集的大小，并为每个大小找到有序集的计数并乘以大小的阶乘 ( S! )。在每个递归步骤中，我们有两个选择——
  1. 包含当前元素 x 并移动到下一个具有现在可以包含的最大元素的位置为 x – 2。
  2. 不包含当前元素 x 并停留在当前位置，其中最大元素现在可以包含为 x – 1。
• 所以递归关系是：

countSets(x, pos) = countSets(x-2, pos-1) + countSets(x-1, pos)

C++

// C++ program to Count the number of
// ordered sets not containing
// consecutive numbers
#include 
using namespace std;
 
// Function to calculate the count
// of ordered set for a given size
int CountSets(int x, int pos)
{
    // Base cases
    if (x <= 0) {
        if (pos == 0)
            return 1;
        else
            return 0;
    }
    if (pos == 0)
        return 1;
 
    int answer = CountSets(x - 1, pos)
              + CountSets(x - 2, pos - 1);
     
    return answer;
}
// Function returns the count
// of all ordered sets
int CountOrderedSets(int n)
{
    // Prestore the factorial value
    int factorial[10000];
    factorial[0] = 1;
 
    for (int i = 1; i < 10000; i++)
        factorial[i] = factorial[i - 1] * i;
 
    int answer = 0;
 
    // Iterate all ordered set sizes and find
    // the count for each one maximum ordered
    // set size will be smaller than N as all
    // elements are distinct and non consecutive
    for (int i = 1; i <= n; i++) {
 
        // Multiply ny size! for all the
        // arrangements because sets are ordered
        int sets = CountSets(n, i) * factorial[i];
         
        // Add to total answer
        answer = answer + sets;
    }
    return answer;
}
// Driver code
int main()
{
    int N = 3;
 
    cout << CountOrderedSets(N);
 
    return 0;
}

Java

// Java program to count the number 
// of ordered sets not containing
// consecutive numbers
class GFG{
 
// Function to calculate the count
// of ordered set for a given size
static int CountSets(int x, int pos)
{
     
    // Base cases
    if (x <= 0)
    {
        if (pos == 0)
            return 1;
        else
            return 0;
    }
    if (pos == 0)
        return 1;
 
    int answer = CountSets(x - 1, pos) +
                 CountSets(x - 2, pos - 1);
    return answer;
}
 
// Function returns the count
// of all ordered sets
static int CountOrderedSets(int n)
{
     
    // Prestore the factorial value
    int []factorial = new int[10000];
    factorial[0] = 1;
 
    for(int i = 1; i < 10000; i++)
       factorial[i] = factorial[i - 1] * i;
 
    int answer = 0;
 
    // Iterate all ordered set sizes and find
    // the count for each one maximum ordered
    // set size will be smaller than N as all
    // elements are distinct and non consecutive
    for(int i = 1; i <= n; i++)
    {
        
       // Multiply ny size! for all the
       // arrangements because sets are ordered
       int sets = CountSets(n, i) * factorial[i];
        
       // Add to total answer
       answer = answer + sets;
    }
    return answer;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 3;
 
    System.out.print(CountOrderedSets(N));
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 program to count the number of
# ordered sets not containing
# consecutive numbers
 
# Function to calculate the count
# of ordered set for a given size
def CountSets(x, pos):
     
    # Base cases
    if (x <= 0):
        if (pos == 0):
            return 1
        else:
            return 0
    if (pos == 0):
        return 1
 
    answer = (CountSets(x - 1, pos) +
              CountSets(x - 2, pos - 1))
               
    return answer
     
# Function returns the count
# of all ordered sets
def CountOrderedSets(n):
     
    # Prestore the factorial value
    factorial = [1 for i in range(10000)]
    factorial[0] = 1
 
    for i in range(1, 10000, 1):
        factorial[i] = factorial[i - 1] * i
 
    answer = 0
 
    # Iterate all ordered set sizes and find
    # the count for each one maximum ordered
    # set size will be smaller than N as all
    # elements are distinct and non consecutive
    for i in range(1, n + 1, 1):
         
        # Multiply ny size! for all the
        # arrangements because sets are ordered
        sets = CountSets(n, i) * factorial[i]
         
        # Add to total answer
        answer = answer + sets
         
    return answer
 
# Driver code
if __name__ == '__main__':
     
    N = 3
    print(CountOrderedSets(N))
 
# This code is contributed by Samarth

C#

// C# program to count the number
// of ordered sets not containing
// consecutive numbers
using System;
class GFG{
 
// Function to calculate the count
// of ordered set for a given size
static int CountSets(int x, int pos)
{
     
    // Base cases
    if (x <= 0)
    {
        if (pos == 0)
            return 1;
        else
            return 0;
    }
    if (pos == 0)
        return 1;
 
    int answer = CountSets(x - 1, pos) +
                 CountSets(x - 2, pos - 1);
    return answer;
}
 
// Function returns the count
// of all ordered sets
static int CountOrderedSets(int n)
{
     
    // Prestore the factorial value
    int []factorial = new int[10000];
    factorial[0] = 1;
 
    for(int i = 1; i < 10000; i++)
    factorial[i] = factorial[i - 1] * i;
 
    int answer = 0;
 
    // Iterate all ordered set sizes and find
    // the count for each one maximum ordered
    // set size will be smaller than N as all
    // elements are distinct and non consecutive
    for(int i = 1; i <= n; i++)
    {
         
        // Multiply ny size! for all the
        // arrangements because sets are ordered
        int sets = CountSets(n, i) * factorial[i];
             
        // Add to total answer
        answer = answer + sets;
    }
    return answer;
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 3;
 
    Console.Write(CountOrderedSets(N));
}
}
 
// This code is contributed by sapnasingh4991

Javascript

 
// Javascript program to count the number 
// of ordered sets not containing
// consecutive numbers   
 
// Function to calculate the count
// of ordered set for a given size
    function CountSets(x , pos)
    {
 
        // Base cases
        if (x <= 0) {
            if (pos == 0)
                return 1;
            else
                return 0;
        }
        if (pos == 0)
            return 1;
 
        var answer = CountSets(x - 1, pos) +
        CountSets(x - 2, pos - 1);
        return answer;
    }
 
    // Function returns the count
    // of all ordered sets
    function CountOrderedSets(n) {
 
        // Prestore the factorial value
        var factorial = Array(10000).fill(0);
        factorial[0] = 1;
 
        for (i = 1; i < 10000; i++)
            factorial[i] = factorial[i - 1] * i;
 
        var answer = 0;
 
        // Iterate all ordered set sizes and find
        // the count for each one maximum ordered
        // set size will be smaller than N as all
        // elements are distinct and non consecutive
        for (i = 1; i <= n; i++) {
 
            // Multiply ny size! for all the
            // arrangements because sets are ordered
            var sets = CountSets(n, i) * factorial[i];
 
            // Add to total answer
            answer = answer + sets;
        }
        return answer;
    }
 
    // Driver code
     
        var N = 3;
 
        document.write(CountOrderedSets(N));
 
// This code contributed by umadevi9616
 

输出：

5

时间复杂度： O(2 ^N )

有效的方法：

• 在递归方法中，我们多次求解子问题，即它遵循动态规划中的重叠子问题属性。所以我们可以使用记忆表或缓存来提高解决方案的效率。

C++

// C++ program to Count the number
// of ordered sets not containing
// consecutive numbers
#include 
using namespace std;
 
// DP table
int dp[500][500];
 
// Function to calculate the count
// of ordered set for a given size
int CountSets(int x, int pos)
{
    // Base cases
    if (x <= 0) {
        if (pos == 0)
            return 1;
        else
            return 0;
    }
    if (pos == 0)
        return 1;
 
    // If subproblem has been
    // soved before
    if (dp[x][pos] != -1)
        return dp[x][pos];
 
    int answer = CountSets(x - 1, pos)
        + CountSets(x - 2, pos - 1);
 
    // Store and return answer to
    // this subproblem
    return dp[x][pos] = answer;
}
 
// Function returns the count
// of all ordered sets
int CountOrderedSets(int n)
{
    // Prestore the factorial value
    int factorial[10000];
    factorial[0] = 1;
 
    for (int i = 1; i < 10000; i++)
        factorial[i] = factorial[i - 1] * i;
 
    int answer = 0;
 
    // Initialise the dp table
    memset(dp, -1, sizeof(dp));
 
    // Iterate all ordered set sizes and find
    // the count for each one maximum ordered
    // set size will be smaller than N as all
    // elements are distinct and non consecutive.
    for (int i = 1; i <= n; i++) {
 
        // Multiply ny size! for all the
        // arrangements because sets
        // are ordered.
        int sets = CountSets(n, i) * factorial[i];
         
        // Add to total answer
        answer = answer + sets;
    }
    return answer;
}
// Driver code
int main()
{
    int N = 3;
 
    cout << CountOrderedSets(N);
 
    return 0;
}

Java

// Java program to count the number
// of ordered sets not containing
// consecutive numbers
import java.util.*;
 
class GFG{
 
// DP table
static int [][]dp = new int[500][500];
 
// Function to calculate the count
// of ordered set for a given size
static int CountSets(int x, int pos)
{
     
    // Base cases
    if (x <= 0)
    {
        if (pos == 0)
            return 1;
        else
            return 0;
    }
    if (pos == 0)
        return 1;
 
    // If subproblem has been
    // soved before
    if (dp[x][pos] != -1)
        return dp[x][pos];
 
    int answer = CountSets(x - 1, pos) +
                 CountSets(x - 2, pos - 1);
 
    // Store and return answer to
    // this subproblem
    return dp[x][pos] = answer;
}
 
// Function returns the count
// of all ordered sets
static int CountOrderedSets(int n)
{
     
    // Prestore the factorial value
    int []factorial = new int[10000];
    factorial[0] = 1;
 
    for(int i = 1; i < 10000; i++)
        factorial[i] = factorial[i - 1] * i;
 
    int answer = 0;
 
    // Initialise the dp table
    for(int i = 0; i < 500; i++)
    {
        for(int j = 0; j < 500; j++)
        {
            dp[i][j] = -1;
        }
    }
 
    // Iterate all ordered set sizes and find
    // the count for each one maximum ordered
    // set size will be smaller than N as all
    // elements are distinct and non consecutive.
    for(int i = 1; i <= n; i++)
    {
 
        // Multiply ny size! for all the
        // arrangements because sets
        // are ordered.
        int sets = CountSets(n, i) * factorial[i];
         
        // Add to total answer
        answer = answer + sets;
    }
    return answer;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 3;
 
    System.out.print(CountOrderedSets(N));
}
}
 
// This code is contributed by Rajput-Ji

蟒蛇3

# Python3 program to count the number
# of ordered sets not containing
# consecutive numbers
 
# DP table
dp = [[-1 for j in range(500)]
          for i in range(500)]
  
# Function to calculate the count
# of ordered set for a given size
def CountSets(x, pos):
 
    # Base cases
    if (x <= 0):
        if (pos == 0):
            return 1
        else:
            return 0
     
    if (pos == 0):
        return 1
  
    # If subproblem has been
    # soved before
    if (dp[x][pos] != -1):
        return dp[x][pos]
  
    answer = (CountSets(x - 1, pos) +
              CountSets(x - 2, pos - 1))
  
    # Store and return answer to
    # this subproblem
    dp[x][pos] = answer
     
    return answer
 
# Function returns the count
# of all ordered sets
def CountOrderedSets(n):
 
    # Prestore the factorial value
    factorial = [0 for i in range(10000)]
    factorial[0] = 1
     
    for i in range(1, 10000):
        factorial[i] = factorial[i - 1] * i
  
    answer = 0
  
    # Iterate all ordered set sizes and find
    # the count for each one maximum ordered
    # set size will be smaller than N as all
    # elements are distinct and non consecutive.
    for i in range(1, n + 1):
  
        # Multiply ny size! for all the
        # arrangements because sets
        # are ordered.
        sets = CountSets(n, i) * factorial[i]
          
        # Add to total answer
        answer = answer + sets
     
    return answer
 
# Driver code
if __name__=="__main__":
     
    N = 3
  
    print(CountOrderedSets(N))
 
# This code is contributed by rutvik_56

C＃

// C# program to count the number
// of ordered sets not containing
// consecutive numbers
using System;
class GFG{
 
// DP table
static int [,]dp = new int[500, 500];
 
// Function to calculate the count
// of ordered set for a given size
static int CountSets(int x, int pos)
{
     
    // Base cases
    if (x <= 0)
    {
        if (pos == 0)
            return 1;
        else
            return 0;
    }
    if (pos == 0)
        return 1;
 
    // If subproblem has been
    // soved before
    if (dp[x,pos] != -1)
        return dp[x, pos];
 
    int answer = CountSets(x - 1, pos) +
                 CountSets(x - 2, pos - 1);
 
    // Store and return answer to
    // this subproblem
    return dp[x, pos] = answer;
}
 
// Function returns the count
// of all ordered sets
static int CountOrderedSets(int n)
{
     
    // Prestore the factorial value
    int []factorial = new int[10000];
    factorial[0] = 1;
 
    for(int i = 1; i < 10000; i++)
        factorial[i] = factorial[i - 1] * i;
 
    int answer = 0;
 
    // Initialise the dp table
    for(int i = 0; i < 500; i++)
    {
        for(int j = 0; j < 500; j++)
        {
            dp[i, j] = -1;
        }
    }
 
    // Iterate all ordered set sizes and find
    // the count for each one maximum ordered
    // set size will be smaller than N as all
    // elements are distinct and non consecutive.
    for(int i = 1; i <= n; i++)
    {
 
        // Multiply ny size! for all the
        // arrangements because sets
        // are ordered.
        int sets = CountSets(n, i) * factorial[i];
         
        // Add to total answer
        answer = answer + sets;
    }
    return answer;
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 3;
 
    Console.Write(CountOrderedSets(N));
}
}
 
// This code is contributed by sapnasingh4991

Javascript

 
// JavaScript program to Count the number
// of ordered sets not containing
// consecutive numbers
 
// DP table
var dp = Array.from(Array(500), ()=>Array(500));
 
// Function to calculate the count
// of ordered set for a given size
function CountSets(x, pos)
{
    // Base cases
    if (x <= 0) {
        if (pos == 0)
            return 1;
        else
            return 0;
    }
    if (pos == 0)
        return 1;
 
    // If subproblem has been
    // soved before
    if (dp[x][pos] != -1)
        return dp[x][pos];
 
    var answer = CountSets(x - 1, pos)
        + CountSets(x - 2, pos - 1);
 
    // Store and return answer to
    // this subproblem
    return dp[x][pos] = answer;
}
 
// Function returns the count
// of all ordered sets
function CountOrderedSets(n)
{
    // Prestore the factorial value
    var factorial = Array(10000).fill(0);
    factorial[0] = 1;
 
    for (var i = 1; i < 10000; i++)
        factorial[i] = factorial[i - 1] * i;
 
    var answer = 0;
 
    // Initialise the dp table
    dp = Array.from(Array(500), ()=>Array(500).fill(-1));
 
    // Iterate all ordered set sizes and find
    // the count for each one maximum ordered
    // set size will be smaller than N as all
    // elements are distinct and non consecutive.
    for (var i = 1; i <= n; i++) {
 
        // Multiply ny size! for all the
        // arrangements because sets
        // are ordered.
        var sets = CountSets(n, i) * factorial[i];
         
        // Add to total answer
        answer = answer + sets;
    }
    return answer;
}
// Driver code
var N = 3;
document.write( CountOrderedSets(N));
 
 

输出：

5

时间复杂度： O(N ² )