📜  计数存在于N个分区中的数字

📅  最后修改于: 2021-06-25 12:23:16             🧑  作者: Mango

给定整数N ,任务是对N的有序整数分区中的数字进行计数。
例子:

方法:可以根据以下观察结果解决问题:

N个分区精确地划分为k个分区的方法计数=

\binom{n-1}{k-1}

因此, N的有序整数分区中的数字计数为

下面是上述方法的实现:

C++
// C++ program to implement
// the above approach 
#include 
using namespace std;
 
// Function to count of numbers in
// ordered partitions of N
int CtOfNums(int N)
{
  
    // Stores count the numbers in
    // ordered integer partitions
    int res = (N + 1) * (1 << (N - 2));
  
    return round(res);
}
  
// Driver Code
int main()
{
    int N = 3;
     
    cout << CtOfNums(N);
}
 
// This code is contributed by code_hunt


Java
// Java program to implement
// the above approach 
import java.io.*;
 
class GFG{
  
// Function to count of numbers in
// ordered partitions of N
static int CtOfNums(int N)
{
  
    // Stores count the numbers in
    // ordered integer partitions
    int res = (N + 1) * (1 << (N - 2));
  
    return Math.round(res);
}
  
// Driver Code
public static void main (String[] args)
{
    int N = 3;
  
    System.out.print(CtOfNums(N));
}
}
 
// This code is contributed by code_hunt


Python3
# Python3 program to implement
# the above approach
 
# Function to count of numbers in
# ordered partitions of N
def CtOfNums(N):
 
    # Stores count the numbers in
    # ordered integer partitions
    res = (N + 1) * (1<<(N - 2))
 
    return round(res)
 
# Driver code
if __name__ == '__main__':
    N = 3
    print(CtOfNums(N))


C#
// C# program to implement
// the above approach 
using System;
 
class GFG{
  
// Function to count of numbers in
// ordered partitions of N
static int CtOfNums(int N)
{
  
    // Stores count the numbers in
    // ordered integer partitions
    double res = (N + 1) * (1 << (N - 2));
  
    return (int)Math.Round(res);
}
  
// Driver Code
public static void Main ()
{
    int N = 3;
  
    Console.Write(CtOfNums(N));
}
}
 
// This code is contributed by code_hunt


Javascript


输出:
8

时间复杂度: O(log 2 N)
辅助空间: O(1)