计数存在于N个分区中的数字

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：

Input: N = 3 
Output: 8 
Integer partitions of N(=3) are {{1 + 1 + 1}, {1 + 2}, {2 + 1}, {3}}. 
Numbers in integer partition of N are:{1, 1, 1, 1, 2, 2, 1, 3} 
Therefore, the count of numbers in integer partitions of N(=3) is 8.

Input: N = 4 
Output: 20 

方法：可以根据以下观察结果解决问题：

将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

    $\begin{align*} = \sum_{k=1}^nk\binom{n-1}{k-1}&=\sum_{k=0}^{n-1}(k+1)\binom{n-1}k\\ &=(n-1)\sum_{k=0}^{n-1}\binom{n-2}{k-1}+\sum_{k=0}^{n-1}\binom{n-1}k\\ &=(n-1)2^{n-2}+2^{n-1}\\ &=(n+1)2^{n-2}\, . \end{align*}$

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

 
// Javascript program to implement
// the above approach 
 
// Function to count of numbers in
// ordered partitions of N
function CtOfNums(N)
{
  
    // Stores count the numbers in
    // ordered integer partitions
    var res = (N + 1) * (1 << (N - 2));
  
    return Math.round(res);
}
  
// Driver Code
var N = 3;
document.write(CtOfNums(N));
 

输出：

8

时间复杂度： O(log ₂ N)
辅助空间： O(1)