Dado un entero N , la tarea es contar todos los enteros menores o iguales a N que siguen la propiedad donde la suma de sus dígitos elevados a la potencia (comenzando desde 1 y aumentado en 1 cada vez) es igual al entero mismo es decir, si D 1 D 2 D 3 …D N es un número de N dígitos, entonces para que satisfaga la propiedad dada (D 1 1 + D 2 2 + D 3 3 + … + D N N ) debe ser igual a D 1 D 2 re 3 …ren _
Ejemplos:
Entrada: N = 100
Salida: 11
0 1 = 0
1 1 = 1
2 1 = 2
…
9 1 = 9
8 1 + 9 2 = 8 + 81 = 89
Entrada: N = 200
Salida: 13
Enfoque: Inicialice el conteo = 0 y para cada número de 0 a N , encuentre la suma de los dígitos elevados a la potencia creciente y, si la suma resultante es igual al número en sí, incremente el conteo . Finalmente, imprima el conteo .
A continuación se muestra la implementación del enfoque anterior:
CPP
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the
// count of digits of n
int countDigits(int n)
{
int cnt = 0;
while (n > 0) {
cnt++;
n /= 10;
}
return cnt;
}
// Function to return the sum of
// increasing powers of N
int digitPowSum(int n)
{
// To store the required answer
int sum = 0;
// Count of digits in n which will
// be the power of the last digit
int pw = countDigits(n);
// While there are digits left
while (n > 0) {
// Get the last digit
int d = n % 10;
// Add the last digit after raising
// it to the required power
sum += pow(d, pw);
// Decrement the power for
// the previous digit
pw--;
// Remove the last digit
n /= 10;
}
return sum;
}
// Function to return the count
// of integers which satisfy
// the given conditions
int countNum(int n)
{
int count = 0;
for (int i = 0; i <= n; i++) {
// If current element satisfies
// the given condition
if (i == digitPowSum(i)) {
count++;
}
}
return count;
}
// Driver code
int main()
{
int n = 200;
cout << countNum(n);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to return the
// count of digits of n
static int countDigits(int n)
{
int cnt = 0;
while (n > 0)
{
cnt++;
n /= 10;
}
return cnt;
}
// Function to return the sum of
// increasing powers of N
static int digitPowSum(int n)
{
// To store the required answer
int sum = 0;
// Count of digits in n which will
// be the power of the last digit
int pw = countDigits(n);
// While there are digits left
while (n > 0)
{
// Get the last digit
int d = n % 10;
// Add the last digit after raising
// it to the required power
sum += Math.pow(d, pw);
// Decrement the power for
// the previous digit
pw--;
// Remove the last digit
n /= 10;
}
return sum;
}
// Function to return the count
// of integers which satisfy
// the given conditions
static int countNum(int n)
{
int count = 0;
for (int i = 0; i <= n; i++)
{
// If current element satisfies
// the given condition
if (i == digitPowSum(i))
{
count++;
}
}
return count;
}
// Driver code
public static void main (String[] args)
{
int n = 200;
System.out.println(countNum(n));
}
}
// This code is contributed by AnkitRai01
Python
# Python3 implementation of the approach # Function to return the # count of digits of n def countDigits(n): cnt = 0 while (n > 0): cnt += 1 n //= 10 return cnt # Function to return the sum of # increasing powers of N def digitPowSum(n): # To store the required answer sum = 0 # Count of digits in n which will # be the power of the last digit pw = countDigits(n) # While there are digits left while (n > 0): # Get the last digit d = n % 10 # Add the last digit after raising # it to the required power sum += pow(d, pw) # Decrement the power for # the previous digit pw -= 1 # Remove the last digit n //= 10 return sum # Function to return the count # of integers which satisfy # the given conditions def countNum(n): count = 0 for i in range(n + 1): # If current element satisfies # the given condition if (i == digitPowSum(i)): count += 1 return count # Driver code n = 200 print(countNum(n)) # This code is contributed by mohit kumar 29
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the
// count of digits of n
static int countDigits(int n)
{
int cnt = 0;
while (n > 0)
{
cnt++;
n /= 10;
}
return cnt;
}
// Function to return the sum of
// increasing powers of N
static int digitPowSum(int n)
{
// To store the required answer
int sum = 0;
// Count of digits in n which will
// be the power of the last digit
int pw = countDigits(n);
// While there are digits left
while (n > 0)
{
// Get the last digit
int d = n % 10;
// Add the last digit after raising
// it to the required power
sum += (int) Math.Pow(d, pw);
// Decrement the power for
// the previous digit
pw--;
// Remove the last digit
n /= 10;
}
return sum;
}
// Function to return the count
// of integers which satisfy
// the given conditions
static int countNum(int n)
{
int count = 0;
for (int i = 0; i <= n; i++)
{
// If current element satisfies
// the given condition
if (i == digitPowSum(i))
count++;
}
return count;
}
// Driver code
public static void Main (String[] args)
{
int n = 200;
Console.WriteLine(countNum(n));
}
}
// This code is contributed by
// sanjeev2552
Javascript
<script>
// JavaScript implementation of the approach
// Function to return the
// count of digits of n
function countDigits(n)
{
let cnt = 0;
while (n > 0) {
cnt++;
n = Math.floor(n / 10);
}
return cnt;
}
// Function to return the sum of
// increasing powers of N
function digitPowSum(n)
{
// To store the required answer
let sum = 0;
// Count of digits in n which will
// be the power of the last digit
let pw = countDigits(n);
// While there are digits left
while (n > 0) {
// Get the last digit
let d = n % 10;
// Add the last digit after raising
// it to the required power
sum += Math.pow(d, pw);
// Decrement the power for
// the previous digit
pw--;
// Remove the last digit
n = Math.floor(n / 10);
}
return sum;
}
// Function to return the count
// of integers which satisfy
// the given conditions
function countNum(n)
{
let count = 0;
for (let i = 0; i <= n; i++) {
// If current element satisfies
// the given condition
if (i == digitPowSum(i)) {
count++;
}
}
return count;
}
// Driver code
let n = 200;
document.write(countNum(n));
// This code is contributed by Surbhi Tyagi.
</script>
Producción:
13
Complejidad del tiempo: O(n * 2log 10 n)
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por jrolofmeister y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA