Dado un número n, encuentre el enésimo número inteligente (1<=n<=1000). Smart number is a number which has at least three distinct prime factors. We are given an upper limit on value of result as MAX For example 30 is 1st smart number because it has 2 3 5 as it's distinct prime factors. 42 is 2nd smart number because it has 2 3 7 as it's distinct prime factors. Ejemplos:
Input : n = 1 Output: 30 // three distinct prime factors 2 3 5 Input : n = 50 Output: 273 // three distinct prime factors 3 7 13 Input : n = 1000 Output: 2664 // three distinct prime factors 2 3 37Recomendado: Resuélvelo en PRÁCTICA primero antes de pasar a la solución.
La idea se basa en Tamiz de Eratóstenes . Usamos una matriz para usar una matriz prima [] para realizar un seguimiento de los números primos. También utilizamos la misma matriz para realizar un seguimiento del recuento de factores primos visto hasta ahora. Cada vez que la cuenta llega a 3 sumamos el número al resultado.
- Tome una matriz de números primos [] e inicialícela con 0.
- Ahora sabemos que el primer número primo es i = 2, así que marque los primos [2] = 1, es decir; primos[i] = 1 indica que 'i' es un número primo.
- Ahora recorra la matriz de primos[] y marque todos los múltiplos de 'i' según la condición primos[j] -= 1 donde 'j' es múltiplo de 'i' y verifique la condición primos[j]+3 = 0 porque siempre que los primos[j] se convierten en -3 indica que anteriormente había sido múltiplo de tres factores primos distintos. si condición primos[j]+3=0 se vuelve verdadero, lo que significa que 'j' es un número inteligente, así que guárdelo en un resultado de matriz [].
- Ahora ordene el resultado de la matriz [] y devuelva el resultado [n-1].
A continuación se muestra la implementación de la idea anterior.
C++
// C++ implementation to find n'th smart number #include
// Java implementation to find n'th smart number import java.util.*; import java.lang.*; class GFG { // Limit on result static int MAX = 3000; // Function to calculate n'th smart number public static int smartNumber(int n) { // Initialize all numbers as not prime Integer[] primes = new Integer[MAX]; Arrays.fill(primes new Integer(0)); // iterate to mark all primes and smart // number Vector<Integer> result = new Vector<>(); // Traverse all numbers till maximum // limit for (int i = 2; i < MAX; i++) { // 'i' is maked as prime number // because it is not multiple of // any other prime if (primes[i] == 0) { primes[i] = 1; // mark all multiples of 'i' // as non prime for (int j = i*2; j < MAX; j = j+i) { primes[j] -= 1; // If i is the third prime // factor of j then add it // to result as it has at // least three prime factors. if ( (primes[j] + 3) == 0) result.add(j); } } } // Sort all smart numbers Collections.sort(result); // return n'th smart number return result.get(n-1); } // Driver program to run the case public static void main(String[] args) { int n = 50; System.out.println(smartNumber(n)); } } // This code is contributed by Prasad Kshirsagar
# Python3 implementation to find # n'th smart number # Limit on result MAX = 3000; # Function to calculate n'th # smart number def smartNumber(n): # Initialize all numbers as not prime primes = [0] * MAX; # iterate to mark all primes # and smart number result = []; # Traverse all numbers till maximum limit for i in range(2 MAX): # 'i' is maked as prime number because # it is not multiple of any other prime if (primes[i] == 0): primes[i] = 1; # mark all multiples of 'i' as non prime j = i * 2; while (j < MAX): primes[j] -= 1; # If i is the third prime factor of j # then add it to result as it has at # least three prime factors. if ( (primes[j] + 3) == 0): result.append(j); j = j + i; # Sort all smart numbers result.sort(); # return n'th smart number return result[n - 1]; # Driver Code n = 50; print(smartNumber(n)); # This code is contributed by mits
// C# implementation to find n'th smart number using System.Collections.Generic; class GFG { // Limit on result static int MAX = 3000; // Function to calculate n'th smart number public static int smartNumber(int n) { // Initialize all numbers as not prime int[] primes = new int[MAX]; // iterate to mark all primes and smart // number List<int> result = new List<int>(); // Traverse all numbers till maximum // limit for (int i = 2; i < MAX; i++) { // 'i' is maked as prime number // because it is not multiple of // any other prime if (primes[i] == 0) { primes[i] = 1; // mark all multiples of 'i' // as non prime for (int j = i*2; j < MAX; j = j+i) { primes[j] -= 1; // If i is the third prime // factor of j then add it // to result as it has at // least three prime factors. if ( (primes[j] + 3) == 0) result.Add(j); } } } // Sort all smart numbers result.Sort(); // return n'th smart number return result[n-1]; } // Driver program to run the case public static void Main() { int n = 50; System.Console.WriteLine(smartNumber(n)); } } // This code is contributed by mits
// PHP implementation to find n'th smart number // Limit on result $MAX = 3000; // Function to calculate n'th smart number function smartNumber($n) { global $MAX; // Initialize all numbers as not prime $primes=array_fill(0$MAX0); // iterate to mark all primes and smart number $result=array(); // Traverse all numbers till maximum limit for ($i=2; $i<$MAX; $i++) { // 'i' is maked as prime number because // it is not multiple of any other prime if ($primes[$i] == 0) { $primes[$i] = 1; // mark all multiples of 'i' as non prime for ($j=$i*2; $j<$MAX; $j=$j+$i) { $primes[$j] -= 1; // If i is the third prime factor of j // then add it to result as it has at // least three prime factors. if ( ($primes[$j] + 3) == 0) array_push($result$j); } } } // Sort all smart numbers sort($result); // return n'th smart number return $result[$n-1]; } // Driver program to run the case $n = 50; echo smartNumber($n); // This code is contributed by mits ?> <script> // JavaScript implementation to find n'th smart number // Limit on result const MAX = 3000; // Function to calculate n'th smart number function smartNumber(n) { // Initialize all numbers as not prime let primes = new Array(MAX).fill(0); // iterate to mark all primes and smart number let result = []; // Traverse all numbers till maximum limit for (let i=2; i<MAX; i++) { // 'i' is maked as prime number because // it is not multiple of any other prime if (primes[i] == 0) { primes[i] = 1; // mark all multiples of 'i' as non prime for (let j=i*2; j<MAX; j=j+i) { primes[j] -= 1; // If i is the third prime factor of j // then add it to result as it has at // least three prime factors. if ( (primes[j] + 3) == 0) result.push(j); } } } // Sort all smart numbers result.sort((ab)=>a-b); // return n'th smart number return result[n-1]; } // Driver program to run the case let n = 50; document.write(smartNumber(n)); // This code is contributed by shinjanpatra </script>
Producción:
273
Complejidad del tiempo: O(MAX)
Espacio auxiliar: O(MAX)