Dado N como el número de dígitos, la tarea es encontrar todos los Números Autobiográficos cuya longitud sea igual a N.
Un número autobiográfico es un número tal que el primer dígito cuenta cuántos ceros hay en él, el segundo dígito cuenta cuántos hay y así sucesivamente.
Por ejemplo, 1210 tiene 1 cero, 2 unos, 1 dos y 0 tres.
Ejemplos:
Entrada: N = 4
Salida: 1210, 2020
Entrada: N = 5
Salida: 21200
Enfoque: cualquier número con N dígitos se encuentra en el rango [10 (n-1) , 10 n -1] . Entonces, cada número en este rango se itera y verifica si es un número autobiográfico o no.
- Convertir el número en una string
- iterar a través de cada dígito y almacenarlo en una variable.
- Luego ejecute un ciclo interno, compare el iterador del ciclo externo con cada dígito del ciclo interno y, si son iguales, incremente el recuento de ocurrencias del dígito.
- Luego verifique la igualdad entre el recuento de ocurrencias y la variable en la que se almacena cada dígito para que podamos saber si el número actual es autobiográfico o no.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation to find
// Autobiographical numbers with length N
#include <bits/stdc++.h>
using namespace std;
// Function to return if the
// number is autobiographical or not
bool isAutoBio(int num)
{
string autoStr;
int index, number, i, j, cnt;
// Converting the integer
// number to string
autoStr = to_string(num);
for (int i = 0;
i < autoStr.size();
i++) {
// Extracting each character
// from each index one by one
// and converting into an integer
index = autoStr.at(i) - '0';
// Initialise count as 0
cnt = 0;
for (j = 0; j < autoStr.size(); j++) {
number = autoStr.at(j) - '0';
// Check if it is equal to the
// index i if true then
// increment the count
if (number == i)
// It is an
// Autobiographical
// number
cnt++;
}
// Return false if the count and
// the index number are not equal
if (index != cnt)
return false;
}
return true;
}
// Function to print autobiographical number
// with given number of digits
void findAutoBios(int n)
{
int high, low, i, flag = 0;
// Left boundary of interval
low = pow(10, n - 1);
// Right boundary of interval
high = pow(10, n) - 1;
for (i = low; i <= high; i++) {
if (isAutoBio(i)) {
flag = 1;
cout << i << ", ";
}
}
// Flag = 0 implies that the number
// is not an autobiographical no.
if (!flag)
cout << "There is no "
<< "Autobiographical number"
<< " with " << n
<< " digits\n";
}
// Driver Code
int main()
{
int N = 0;
findAutoBios(N);
N = 4;
findAutoBios(N);
return 0;
}
Java
// Java implementation to find
// Autobiographical numbers with length N
import java.util.*;
import java.lang.Math;
public class autobio {
public static boolean isAutoBio(int num)
{
String autoStr;
int index, number, i, j, cnt;
// Converting the integer
// number to string
autoStr = Integer.toString(num);
for (i = 0; i < autoStr.length(); i++) {
// Extracting each character
// from each index one by one
// and converting into an integer
index = Integer.parseInt(autoStr.charAt(i) + "");
// initialize count as 0
cnt = 0;
for (j = 0; j < autoStr.length(); j++) {
number = Integer.parseInt(autoStr.charAt(j) + "");
// Check if it is equal to the
// index i if true then
// increment the count
if (number == i)
// It is an
// Autobiographical
// number
cnt++;
}
// Return false if the count and
// the index number are not equal
if (cnt != index)
return false;
}
return true;
}
// Function to print autobiographical number
// with given number of digits
public static void findAutoBios(double n)
{
// both the boundaries are taken double, so as
// to satisfy Math.pow() function's signature
double high, low;
int i, flag = 0;
// Left boundary of interval
low = Math.pow(10.0, n - 1);
// Right boundary of interval
high = Math.pow(10.0, n) - 1.0;
for (i = (int)low; i <= (int)high; i++)
if (isAutoBio(i)) {
flag = 1;
System.out.print(i + ", ");
}
// Flag = 0 implies that the number
// is not an autobiographical no.
if (flag == 0)
System.out.println("There is no Autobiographical Number"
+ "with " + (int)n + " digits");
}
// Driver Code
public static void main(String[] args)
{
double N = 0;
findAutoBios(N);
N = 4;
findAutoBios(N);
}
}
Python3
# Python implementation to find
# Autobiographical numbers with length N
from math import pow
# Function to return if the
# number is autobiographical or not
def isAutoBio(num):
# Converting the integer
# number to string
autoStr = str(num)
for i in range(0, len(autoStr)):
# Extracting each character
# from each index one by one
# and converting into an integer
index = int(autoStr[i])
# Initialize count as 0
cnt = 0
for j in range(0, len(autoStr)):
number = int(autoStr[j])
# Check if it is equal to the
# index i if true then
# increment the count
if number == i:
# It is an
# Autobiographical
# number
cnt += 1
# Return false if the count and
# the index number are not equal
if cnt != index:
return False
return True
# Function to print autobiographical number
# with given number of digits
def findAutoBios(n):
# Left boundary of interval
low = int(pow(10, n-1))
# Right boundary of interval
high = int(pow(10, n) - 1)
flag = 0
for i in range(low, high + 1):
if isAutoBio(i):
flag = 1
print(i, end =', ')
# Flag = 0 implies that the number
# is not an autobiographical no.
if flag == 0:
print("There is no Autobiographical Number with "+ str(n) + " digits")
# Driver Code
if __name__ == "__main__":
N = 0
findAutoBios(N)
N = 4
findAutoBios(N)
C#
// C# implementation to find
// Autobiographical numbers with length N
using System;
class autobio {
public static bool isAutoBio(int num)
{
String autoStr;
int index, number, i, j, cnt;
// Converting the integer
// number to string
autoStr = num.ToString();
for (i = 0; i < autoStr.Length; i++) {
// Extracting each character
// from each index one by one
// and converting into an integer
index = Int32.Parse(autoStr[i] + "");
// initialize count as 0
cnt = 0;
for (j = 0; j < autoStr.Length; j++) {
number = Int32.Parse(autoStr[j] + "");
// Check if it is equal to the
// index i if true then
// increment the count
if (number == i)
// It is an
// Autobiographical
// number
cnt++;
}
// Return false if the count and
// the index number are not equal
if (cnt != index)
return false;
}
return true;
}
// Function to print autobiographical number
// with given number of digits
public static void findAutoBios(double n)
{
// both the boundaries are taken double, so as
// to satisfy Math.Pow() function's signature
double high, low;
int i, flag = 0;
// Left boundary of interval
low = Math.Pow(10.0, n - 1);
// Right boundary of interval
high = Math.Pow(10.0, n) - 1.0;
for (i = (int)low; i <= (int)high; i++)
if (isAutoBio(i)) {
flag = 1;
Console.Write(i + ", ");
}
// Flag = 0 implies that the number
// is not an autobiographical no.
if (flag == 0)
Console.WriteLine("There is no Autobiographical Number"
+ "with " + (int)n + " digits");
}
// Driver Code
public static void Main(String[] args)
{
double N = 0;
findAutoBios(N);
N = 4;
findAutoBios(N);
}
}
// This code is contributed by sapnasingh4991
Javascript
<script>
// JavaScript implementation to find
// Autobiographical numbers with length N
// Function to return if the
// number is autobiographical or not
function isAutoBio(num){
// Converting the integer
// number to string
let autoStr = num.toString()
for(let i=0;i<autoStr.length;i++){
// Extracting each character
// from each index one by one
// and converting into an integer
let index = parseInt(autoStr[i])
// Initialize count as 0
let cnt = 0
for(let j=0;j<autoStr.length;j++){
let number = parseInt(autoStr[j])
// Check if it is equal to the
// index i if true then
// increment the count
if(number == i){
// It is an
// Autobiographical
// number
cnt += 1
}
}
// Return false if the count and
// the index number are not equal
if(cnt != index){
return false
}
}
return true
}
// Function to print autobiographical number
// with given number of digits
function findAutoBios(n){
// Left boundary of interval
let low = parseInt(Math.pow(10, n-1))
// Right boundary of interval
let high = parseInt(Math.pow(10, n) - 1)
let flag = 0
for(let i=low;i<high+1;i++){
if(isAutoBio(i)){
flag = 1
document.write(i,', ')
}
}
// Flag = 0 implies that the number
// is not an autobiographical no.
if(flag == 0)
document.write("There is no Autobiographical Number with "+ n + " digits","</br>")
}
// Driver Code
let N = 0
findAutoBios(N)
N = 4
findAutoBios(N)
// This code is contributed by shinjanpatra
</script>
Producción:
There is no Autobiographical number with 0 digits 1210, 2020
Complejidad del tiempo: O(10 n – 10 n-1 )
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por saikiranjagini y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA