Dado un número entero N, la tarea es encontrar los números que, elevados a la potencia de 2 y finalmente sumados, dan el número entero N.
Ejemplo :
C++
// CPP program to find the
// blocks for given number.
#include <bits/stdc++.h>
using namespace std;
void block(long int x)
{
vector<long int> v;
// Converting the decimal number
// into its binary equivalent.
cout << "Blocks for " << x << " : ";
while (x > 0)
{
v.push_back(x % 2);
x = x / 2;
}
// Displaying the output when
// the bit is '1' in binary
// equivalent of number.
for (int i = 0; i < v.size(); i++)
{
if (v[i] == 1)
{
cout << i;
if (i != v.size() - 1)
cout << ", ";
}
}
cout << endl;
}
// Driver Function
int main()
{
block(71307);
block(1213);
block(29);
block(100);
return 0;
}
Java
// Java program to find the
// blocks for given number.
import java.util.*;
class GFG {
static void block(long x)
{
ArrayList<Integer> v = new ArrayList<Integer>();
// Convert decimal number to
// its binary equivalent
System.out.print("Blocks for "+x+" : ");
while (x > 0)
{
v.add((int)x % 2);
x = x / 2;
}
// Displaying the output when
// the bit is '1' in binary
// equivalent of number.
for (int i = 0; i < v.size(); i++)
{
if (v.get(i) == 1)
{
System.out.print(i);
if (i != v.size() - 1)
System.out.print( ", ");
}
}
System.out.println();
}
// Driver Code
public static void main(String args[])
{
block(71307);
block(1213);
block(29);
block(100);
}
}
// This code is contributed by Arnab Kundu.
Python3
# Python3 program to find the
# blocks for given number.
def block(x):
v = []
# Converting the decimal number
# into its binary equivalent.
print ("Blocks for %d : " %x, end="")
while (x > 0):
v.append(int(x % 2))
x = int(x / 2)
# Displaying the output when
# the bit is '1' in binary
# equivalent of number.
for i in range(0, len(v)):
if (v[i] == 1):
print (i, end = "")
if (i != len(v) - 1):
print (", ", end = "")
print ("\n")
block(71307)
block(1213)
block(29)
block(100)
# This code is contributed by Manish
# Shaw (manishshaw1)
C#
// C# program to find the
// blocks for given number.
using System;
using System.Collections.Generic;
class GFG {
static void block(long x)
{
List<int> v = new List<int>();
// Convert decimal number to
// its binary equivalent
Console.Write("Blocks for " + x + " : ");
while (x > 0)
{
v.Add((int)x % 2);
x = x / 2;
}
// Displaying the output when
// the bit is '1' in binary
// equivalent of number.
for (int i = 0; i < v.Count; i++)
{
if (v[i] == 1)
{
Console.Write(i);
if (i != v.Count - 1)
Console.Write(", ");
}
}
Console.WriteLine();
}
// Driver Code here
public static void Main()
{
block(71307);
block(1213);
block(29);
block(100);
}
}
// This code is contributed by Ajit.
PHP
<?php
// PHP program to find the
// blocks for given number.
function block($x)
{
$v = array();
// Convert decimal number to
// its binary equivalent
echo 'Blocks for ' .$x.' : ';
while ($x > 0)
{
array_push($v,intval($x % 2));
$x = intval($x / 2);
}
// Displaying the output when
// the bit is '1' in binary
// equivalent of number.
for ($i = 0; $i < sizeof($v); $i++)
{
if ($v[$i] == 1)
{
print $i;
if ($i != sizeof($v) - 1)
echo ', ';
}
}
echo "\n";
}
// Driver Code
block(71307);
block(1213);
block(29);
block(100);
// This code is contributed
// by Manish Shaw (manishshaw1)
?>
Javascript
<script>
// Javascript program to find the
// blocks for given number.
function block(x)
{
let v = [];
// Convert decimal number to
// its binary equivalent
document.write("Blocks for " + x + " : ");
while (x > 0)
{
v.push(x % 2);
x = parseInt(x / 2, 10);
}
// Displaying the output when
// the bit is '1' in binary
// equivalent of number.
for(let i = 0; i < v.length; i++)
{
if (v[i] == 1)
{
document.write(i);
if (i != v.length - 1)
document.write(", ");
}
}
document.write("</br>");
}
// Driver code
block(71307);
block(1213);
block(29);
block(100);
// This code is contributed by mukesh07
</script>
Publicación traducida automáticamente
Artículo escrito por sumit.chauhan y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA