Dados dos números enteros positivos L y R . La tarea es convertir todos los números de L a R en números binarios.
Ejemplos:
Entrada: L = 1, R = 4
Salida:
1
10
11
100
Explicación: La representación binaria de los números 1, 2, 3 y 4 son:
1 = (1) 2
2 = (10) 2
3 = (11) 2
4 = (100) 2Entrada: L = 2, R = 8
Salida:
10
11
100
101
110
111
1000
Enfoque: El problema se puede resolver utilizando el siguiente enfoque.
- Atraviesa de L a R y convierte cada número a número binario.
- Guarde cada número e imprímalo al final.
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ code to implement the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to convert a number
// to its binary equivalent
vector<int> convertToBinary(int num)
{
vector<int> bits;
if (num == 0) {
bits.push_back(0);
return bits;
}
while (num != 0) {
bits.push_back(num % 2);
// Integer division
// gives quotient
num = num / 2;
}
reverse(bits.begin(), bits.end());
return bits;
}
// Function to convert all numbers
// in range [L, R] to binary
vector<vector<int> > getAllBinary(int l,
int r)
{
// Vector to store the binary
// representations of the numbers
vector<vector<int> > binary_nos;
for (int i = l; i <= r; i++) {
vector<int> bits =
convertToBinary(i);
binary_nos.push_back(bits);
}
return binary_nos;
}
// Driver code
int main()
{
int L = 2, R = 8;
vector<vector<int> > binary_nos =
getAllBinary(L, R);
for (int i = 0; i < binary_nos.size();
i++) {
for (int j = 0; j < binary_nos[i].size();
j++)
cout << binary_nos[i][j];
cout << endl;
}
return 0;
}
Java
// Java code to implement the above approach
import java.util.ArrayList;
import java.util.Collections;
class GFG {
// Function to convert a number
// to its binary equivalent
public static ArrayList<Integer> convertToBinary(int num) {
ArrayList<Integer> bits = new ArrayList<Integer>();
if (num == 0) {
bits.add(0);
return bits;
}
while (num != 0) {
bits.add(num % 2);
// Integer division
// gives quotient
num = num / 2;
}
Collections.reverse(bits);
return bits;
}
// Function to convert all numbers
// in range [L, R] to binary
public static ArrayList<ArrayList<Integer>> getAllBinary(int l, int r)
{
// Vector to store the binary
// representations of the numbers
ArrayList<ArrayList<Integer>> binary_nos = new ArrayList<ArrayList<Integer>>();
for (int i = l; i <= r; i++)
{
ArrayList<Integer> bits = convertToBinary(i);
binary_nos.add(bits);
}
return binary_nos;
}
// Driver code
public static void main(String args[])
{
int L = 2, R = 8;
ArrayList<ArrayList<Integer>> binary_nos = getAllBinary(L, R);
for (int i = 0; i < binary_nos.size(); i++) {
for (int j = 0; j < binary_nos.get(i).size(); j++)
System.out.print(binary_nos.get(i).get(j));
System.out.println("");
}
}
}
// This code is contributed by saurabh_jaiswal.
Python3
# Python 3 code to implement the above approach # Function to convert a number # to its binary equivalent def convertToBinary(num): bits = [] if (num == 0): bits.append(0) return bits while (num != 0): bits.append(num % 2) # Integer division # gives quotient num = num // 2 bits.reverse() return bits # Function to convert all numbers # in range [L, R] to binary def getAllBinary(l, r): # Vector to store the binary # representations of the numbers binary_nos = [] for i in range(l, r+1): bits = convertToBinary(i) binary_nos.append(bits) return binary_nos # Driver code if __name__ == "__main__": L = 2 R = 8 binary_nos = getAllBinary(L, R) for i in range(len(binary_nos)): for j in range(len(binary_nos[i])): print(binary_nos[i][j], end="") print() # This code is contributed by ukasp.
C#
// C# code to implement the above approach
using System;
using System.Collections.Generic;
public class GFG {
// Function to convert a number
// to its binary equivalent
public static List<int> convertToBinary(int num) {
List<int> bits = new List<int>();
if (num == 0) {
bits.Add(0);
return bits;
}
while (num != 0) {
bits.Add(num % 2);
// int division
// gives quotient
num = num / 2;
}
bits.Reverse();
return bits;
}
// Function to convert all numbers
// in range [L, R] to binary
public static List<List<int>> getAllBinary(int l, int r)
{
// List to store the binary
// representations of the numbers
List<List<int>> binary_nos = new List<List<int>>();
for (int i = l; i <= r; i++)
{
List<int> bits = convertToBinary(i);
binary_nos.Add(bits);
}
return binary_nos;
}
// Driver code
public static void Main(String []args)
{
int L = 2, R = 8;
List<List<int>> binary_nos = getAllBinary(L, R);
for (int i = 0; i < binary_nos.Count; i++) {
for (int j = 0; j < binary_nos[i].Count; j++)
Console.Write(binary_nos[i][j]);
Console.WriteLine("");
}
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// JavaScript code to implement the above approach
// Function to convert a number
// to its binary equivalent
const convertToBinary = (num) => {
let bits = [];
if (num == 0) {
bits.push(0);
return bits;
}
while (num != 0) {
bits.push(num % 2);
// Integer division
// gives quotient
num = parseInt(num / 2);
}
bits.reverse()
return bits;
}
// Function to convert all numbers
// in range [L, R] to binary
const getAllBinary = (l, r) => {
// Vector to store the binary
// representations of the numbers
let binary_nos = [];
for (let i = l; i <= r; i++) {
let bits = convertToBinary(i);
binary_nos.push(bits);
}
return binary_nos;
}
// Driver code
let L = 2, R = 8;
let binary_nos = getAllBinary(L, R);
for (let i = 0; i < binary_nos.length;
i++) {
for (let j = 0; j < binary_nos[i].length;
j++)
document.write(`${binary_nos[i][j]}`);
document.write("<br/>");
}
// This code is contributed by rakeshsahni
</script>
Producción
10 11 100 101 110 111 1000
Complejidad de tiempo: O(N * LogR) Donde N es el conteo de números en el rango [L, R]
Espacio auxiliar: O(N * logR)