Dado un número entero N , la tarea es encontrar los valores XOR y OR de todos los números de Armstrong de N dígitos .
Ejemplos
Entrada: N = 3
Salida: XOR = 271, OR = 511
153, 370, 371, 407 son los números armstrong de tres dígitosEntrada: N = 4
Salida: XOR = 880, OR = 10098
Acercarse:
- Encuentre el número inicial y final del número de Armstrong de N dígitos por:
Starting N-digit Armstrong number = pow(10, n - 1) Ending N-digit Armstrong number = pow(10, n) - 1
- Repita los números de Armstrong de N dígitos desde el número inicial hasta el número final y verifique si ese número es Armstrong o no.
- Si el número es Armstrong, entonces tome XOR y OR de ese número por separado.
- De lo contrario, continúe con la siguiente iteración e imprima el valor de XOR y OR después de todas las iteraciones.
A continuación se muestra la implementación del enfoque anterior:
C++14
// C++ program to find the XOR
// and OR of all Armstrong numbers
// of N digits
#include <bits/stdc++.h>
using namespace std;
// Function to check if a number
// is Armstrong or not
bool isArmstrong(int x, int n)
{
int sum1 = 0;
int temp = x;
while (temp > 0) {
int digit = temp % 10;
sum1 += (int)pow(digit, n);
temp /= 10;
}
return sum1 == x;
}
// Function to find XOR of all
// N-digits Armstrong number
void CalculateXORandOR(int n)
{
// To store the XOR and OR of all
// Armstrong number
int CalculateXOR = 0;
int CalculateOR = 0;
// Starting N-digit
// Armstrong number
int start = (int)pow(10, n - 1);
// Ending N-digit
// Armstrong number
int end = (int)pow(10, n) - 1;
// Iterate over starting and
// ending number
for (int i = start; i < end + 1; i++)
{
// To check if i is
// Armstrong or not
if (isArmstrong(i, n)) {
CalculateXOR = CalculateXOR ^ i;
CalculateOR = CalculateOR | i;
}
}
// Print the XOR and OR of all
// Armstrong number
cout << "XOR = " << CalculateXOR << endl;
cout << "OR = " << CalculateOR << endl;
}
// Driver Code
int main()
{
int n = 4;
CalculateXORandOR(n);
}
// This code is contributed by shivanisinghss2110
Java
// Java program to find the XOR
// and OR of all Armstrong numbers
// of N digits
class GFG
{
// Function to check if a number
// is Armstrong or not
static boolean isArmstrong(int x, int n) {
int sum1 = 0;
int temp = x;
while (temp > 0) {
int digit = temp % 10;
sum1 += Math.pow(digit, n);
temp /= 10;
}
return sum1 == x;
}
// Function to find XOR of all
// N-digits Armstrong number
static void CalculateXORandOR(int n) {
// To store the XOR and OR of all
// Armstrong number
int CalculateXOR = 0;
int CalculateOR = 0;
// Starting N-digit
// Armstrong number
int start = (int) Math.pow(10, n - 1);
// Ending N-digit
// Armstrong number
int end = (int) (Math.pow(10, n)) - 1;
// Iterate over starting and
// ending number
for (int i = start; i < end + 1; i++) {
// To check if i is
// Armstrong or not
if (isArmstrong(i, n)) {
CalculateXOR = CalculateXOR ^ i;
CalculateOR = CalculateOR | i;
}
}
// Print the XOR and OR of all
// Armstrong number
System.out.println("XOR = " + CalculateXOR);
System.out.println("OR = " + CalculateOR);
}
// Driver Code
public static void main(String[] args) {
int n = 4;
CalculateXORandOR(n);
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 program to find the XOR
# and OR of all Armstrong numbers
# of N digits
# Function to check if a number
# is Armstrong or not
def isArmstrong (x, n):
sum1 = 0
temp = x
while temp > 0:
digit = temp % 10
sum1 += digit **n
temp //= 10
return sum1 == x
# Function to find XOR of all
# N-digits Armstrong number
def CalculateXORandOR(n) :
# To store the XOR and OR of all
# Armstrong number
CalculateXOR = 0
CalculateOR = 0
# Starting N-digit
# Armstrong number
start = 10 ** (n - 1)
# Ending N-digit
# Armstrong number
end = (10**n) - 1
# Iterate over starting and
# ending number
for i in range( start, end + 1) :
# To check if i is
# Armstrong or not
if (isArmstrong(i, n)) :
CalculateXOR = CalculateXOR ^ i
CalculateOR = CalculateOR | i
# Print the XOR and OR of all
# Armstrong number
print("XOR = ", CalculateXOR)
print("OR = ", CalculateOR)
# Driver Code
if __name__ == "__main__" :
n = 4;
CalculateXORandOR(n);
C#
// C# program to find the XOR
// and OR of all Armstrong numbers
// of N digits
using System;
class GFG
{
// Function to check if a number
// is Armstrong or not
static bool isArmstrong(int x, int n) {
int sum1 = 0;
int temp = x;
while (temp > 0) {
int digit = temp % 10;
sum1 += (int)Math.Pow(digit, n);
temp /= 10;
}
return sum1 == x;
}
// Function to find XOR of all
// N-digits Armstrong number
static void CalculateXORandOR(int n) {
// To store the XOR and OR of all
// Armstrong number
int CalculateXOR = 0;
int CalculateOR = 0;
// Starting N-digit
// Armstrong number
int start = (int) Math.Pow(10, n - 1);
// Ending N-digit
// Armstrong number
int end = (int) (Math.Pow(10, n)) - 1;
// Iterate over starting and
// ending number
for (int i = start; i < end + 1; i++) {
// To check if i is
// Armstrong or not
if (isArmstrong(i, n)) {
CalculateXOR = CalculateXOR ^ i;
CalculateOR = CalculateOR | i;
}
}
// Print the XOR and OR of all
// Armstrong number
Console.WriteLine("XOR = " + CalculateXOR);
Console.WriteLine("OR = " + CalculateOR);
}
// Driver Code
public static void Main(String[] args) {
int n = 4;
CalculateXORandOR(n);
}
}
// This code is contributed by PrinciRaj1992
Javascript
<script>
// Javascript program to find the XOR
// and OR of all Armstrong numbers
// of N digits
// Function to check if a number
// is Armstrong or not
function isArmstrong(x, n)
{
let sum1 = 0;
let temp = x;
while (temp > 0)
{
let digit = temp % 10;
sum1 += Math.pow(digit, n);
temp = parseInt(temp / 10, 10);
}
return (sum1 == x);
}
// Function to find XOR of all
// N-digits Armstrong number
function CalculateXORandOR(n)
{
// To store the XOR and OR of all
// Armstrong number
let CalculateXOR = 0;
let CalculateOR = 0;
// Starting N-digit
// Armstrong number
let start = Math.pow(10, n - 1);
// Ending N-digit
// Armstrong number
let end = (Math.pow(10, n)) - 1;
// Iterate over starting and
// ending number
for(let i = start; i < end + 1; i++)
{
// To check if i is
// Armstrong or not
if (isArmstrong(i, n))
{
CalculateXOR = CalculateXOR ^ i;
CalculateOR = CalculateOR | i;
}
}
// Print the XOR and OR of all
// Armstrong number
document.write("XOR = " + CalculateXOR + "</br>");
document.write("OR = " + CalculateOR + "</br>");
}
// Driver code
let n = 4;
CalculateXORandOR(n);
// This code is contributed by divyeshrabadiya07
</script>
Producción:
XOR = 880 OR = 10098
Complejidad del tiempo: O((10 n – 10 n-1 ) * log 10 n)
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por SHUBHAMSINGH10 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA