Étant donné un nombre n, trouver le nième numéro intelligent (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. Exemples :
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 37Recommandé : veuillez le résoudre sur PRATIQUE d'abord avant de passer à la solution.
L'idée est basée sur Tamis d'Ératosthène . Nous utilisons un tableau pour utiliser un tableau prime[] afin de garder une trace des nombres premiers. Nous utilisons également le même tableau pour suivre le nombre de facteurs premiers observés jusqu'à présent. Chaque fois que le décompte atteint 3, nous ajoutons le nombre au résultat.
java entier en chaîne
- Prenez un tableau primes[] et initialisez-le avec 0.
- Maintenant, nous savons que le premier nombre premier est i = 2, alors marquez primes[2] = 1, c'est-à-dire ; primes[i] = 1 indique que « i » est un nombre premier.
- Parcourez maintenant le tableau primes[] et marquez tous les multiples de « i » par la condition primes[j] -= 1 où « j » est un multiple de « i » et vérifiez la condition primes[j]+3 = 0 car chaque fois que les nombres premiers[j] deviennent -3, cela indique qu'auparavant, il s'agissait d'un multiple de trois facteurs premiers distincts. Si condition primes[j]+3=0 devient vrai, cela signifie que « j » est un numéro intelligent, alors stockez-le dans un tableau result[].
- Maintenant, triez le résultat du tableau[] et renvoyez le résultat[n-1].
Vous trouverez ci-dessous la mise en œuvre de l’idée ci-dessus.
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>
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Complexité temporelle : O(MAX)
Espace auxiliaire : O(MAX)