自然数的成分数

给定自然数n，请考虑将顺序考虑在内时可以将n表示为自然数之和的方式的数目。术语顺序不同的两个序列定义了其和的不同组成。
例子：

Input :  4
Output : 8
Explanation  
All 8 position composition are:
4, 1+3, 3+1, 2+2, 1+1+2, 1+2+1, 2+1+1
and 1+1+1+1

Input :  8
Output : 128

一个简单的解决方案是生成所有合成并对其进行计数。
使用组合学的概念，可以证明，当考虑阶数时，任何自然数n都将具有2 ^(n-1)个不同的组成。

One way to see why the answer is 2^(n-1) directly is to write n as a sum of 1s: 
n = 1 + 1 + 1 +…+ 1 (n times).
There are (n-1) plus signs between all 1s. For every plus sign we can choose to split ( by putting a bracket) at the point or not split. Therefore answer is 2^(n-1).
For example, n = 4 
No Split 
4 = 1 + 1 + 1 + 1 [We write as single 4]
Different ways to split once 
4 = (1) + (1 + 1 + 1) [We write as 1 + 3] 
4 = (1 + 1) + (1 + 1) [We write as 2 + 2] 
4 = (1 + 1 + 1) + (1) [We write as 3 + 1]
Different ways to split twice 
4 = (1) + (1 + 1) + (1) [We write as 1 + 2 + 1] 
4 = (1 + 1) + (1) + (1) [We write as 2 + 1 + 1] 
4 = (1) + (1) + (1 + 1) [We write as 1 + 1 + 2]
Different ways to split three times 
4 = (1) + (1) + (1) + (1) [We write as 1 + 1 + 1 + 1]
Since there are (n-1) plus signs between the n 1s, there are 2^(n-1) ways of choosing where to split the sum, and hence 2^(n-1) possible sums .

C++

// C++ program to find the total number of
// compositions of a natural number
#include
using namespace std;
 
#define ull unsigned long long
 
ull countCompositions(ull n)
{
    // Return 2 raised to power (n-1)
    return (1L) << (n-1);
}
 
// Driver Code
int main()
{
    ull n = 4;
    cout << countCompositions(n) << "\n";
    return 0;
}

Java

// Java program to find
// the total number of
// compositions of a
// natural number
import java.io.*;
import java.util.*;
 
class GFG
{
public static int countCompositions(int n)
{
    // Return 2 raised
    // to power (n-1)
    return 1 << (n - 1);
}
 
// Driver Code
public static void main(String args[])
{
    int n = 4;
    System.out.print(countCompositions(n));
}
}
 
// This code is contributed by
// Akanksha Rai(Abby_akku)

Python

# Python code to find the total number of
# compositions of a natural number
def countCompositions(n):
 
    # function to return the total number
    # of composition of n
    return (2**(n-1))
 
# Driver Code
print(countCompositions(4))

C#

// C# program to find the
// total number of compositions
// of a natural number
using System;
 
class GFG
{
public static int countCompositions(int n)
{
    // Return 2 raised
    // to power (n-1)
    return 1 << (n - 1);
}
 
// Driver Code
public static void Main()
{
    int n = 4;
    Console.Write(countCompositions(n));
}
}
 
// This code is contributed by mits

Javascript

 
// javascript program to find
// the total number of
// compositions of a
// natural number
 
function countCompositions(n)
{
    // Return 2 raised
    // to power (n-1)
    return 1 << (n - 1);
}
 
// Driver Code
 
    var n = 4;
    document.write(countCompositions(n));
 
 
 
// This code is contributed by 29AjayKumar

输出：

8