Expresar 2^N en exactamente N+1 potencias de 2

Dado un entero positivo N , la tarea es expresar 2^N en la suma de potencias de 2 y exactamente en N+1 términos. Imprime esos términos N+1 .

Ejemplo:

Entrada : N = 4
Salida : 1 1 2 4 8
Explicación : 2^4 = 2^0 + 2^0 + 2^1 + 2^2 + 2^3 = 1 + 1 + 2 + 4 + 8

Entrada : N = 5
Salida : 1 1 2 4 8 16  

Enfoque: Como sabemos que:

2^0 + 2^1 +…. 2^(N-1) = 2^N-1

Por lo tanto, sumar 1 al principio de la serie de potencias de 2 dará como resultado 2^N exactamente en N+1 términos.

1 + 2^0 + 2^1 +…. 2^(N-1) = 2^N

 

A continuación se muestra la implementación del enfoque anterior.

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find exactly N + 1 terms
// of powers of 2 series,  whose sum is 2^N
void powerOf2(long long N)
{
    cout << 1 << ' ';
    for (int i = 0; i < N; ++i) {
        cout << (long long)pow(2, i) << ' ';
    }
}
 
// Driver Code
int main()
{
    long long N = 5;
 
    powerOf2(N);
}

// Java program for the above approach
import java.util.*;
public class GFG
{
// Function to find exactly N + 1 terms
// of powers of 2 series,  whose sum is 2^N
static void powerOf2(long N)
{
    System.out.print(1);
    for (int i = 0; i < N; ++i) {
        System.out.print(" " + (long)Math.pow(2, i));
    }
}
 
// Driver Code
public static void main(String args[])
{
    long N = 5;
    powerOf2(N);
}
}
// This code is contributed by Samim Hossain Mondal

// C# program for the above approach
using System;
class GFG
{
// Function to find exactly N + 1 terms
// of powers of 2 series,  whose sum is 2^N
static void powerOf2(long N)
{
    Console.Write(1);
    for (int i = 0; i < N; ++i) {
        Console.Write(" " + (long)Math.Pow(2, i));
    }
}
 
// Driver Code
public static void Main()
{
    long N = 5;
    powerOf2(N);
}
}
// This code is contributed by Samim Hossain Mondal

# Python3 program for the above approach
import math
 
# Function to find exactly N + 1 terms
# of powers of 2 series,  whose sum is 2^N
def powerOf2(N) :
 
    print(1,end= ' ');
    for i in range(N) :
        print(int(math.pow(2, i)),end = ' ');
 
# Driver Code
if __name__ == "__main__" :
 
    N = 5;
 
    powerOf2(N);
     
    # This code is contributed by AnkThon

<script>
// JavaScript program for the above approach
 
// Function to find exactly N + 1 terms
// of powers of 2 series,  whose sum is 2^N
function powerOf2(N)
{
    document.write(1 + ' ');
    for (var i = 0; i < N; ++i) {
        document.write(Math.pow(2, i) + ' ');
    }
}
 
// Driver Code
N = 5;
powerOf2(N);
 
// This code is contributed by AnkThon
</script>

1 1 2 4 8 16 

Complejidad temporal: O(N)
Espacio auxiliar: O(1)