Al N-lea număr inteligent
Având în vedere un număr n, găsiți al-al-lea număr inteligent (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. Exemple:
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 37Recomandat: Vă rugăm să rezolvați problema PRACTICA mai întâi înainte de a trece la soluție.
Ideea se bazează pe Sita lui Eratosthenes . Folosim o matrice pentru a folosi o matrice prime[] pentru a ține evidența numerelor prime. De asemenea, folosim aceeași matrice pentru a ține evidența numărului de factori primi observați până acum. Ori de câte ori numărul ajunge la 3, adăugăm numărul rezultat.
- Luați o matrice prime[] și inițializați-o cu 0.
- Acum știm că primul număr prim este i = 2, deci marcați primii[2] = 1, adică; prime[i] = 1 indică faptul că „i” este un număr prim.
- Acum parcurgeți matricea primes[] și marcați toți multiplii lui „i” prin condiția primes[j] -= 1 unde „j” este multiplu al lui „i” și verificați condiția primes[j]+3 = 0 deoarece ori de câte ori primii[j] devin -3 indică faptul că anterior a fost multiplu a trei factori primi diferiți. Dacă starea numere prime[j]+3=0 devine adevărat, ceea ce înseamnă că „j” este un număr inteligent, așa că stocați-l într-un rezultat de matrice[].
- Acum sortați rezultatul matricei[] și returnați rezultatul[n-1].
Mai jos este implementarea ideii de mai sus.
C++ // C++ implementation to find n'th smart number #include using namespace std ; // Limit on result const int MAX = 3000 ; // Function to calculate n'th smart number int smartNumber ( int n ) { // Initialize all numbers as not prime int primes [ MAX ] = { 0 }; // iterate to mark all primes and smart number vector < int > result ; // 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 . push_back ( j ); } } } // Sort all smart numbers sort ( result . begin () result . end ()); // return n'th smart number return result [ n -1 ]; } // Driver program to run the case int main () { int n = 50 ; cout < < smartNumber ( n ); return 0 ; }
Java // 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 # 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# // 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 // 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 $MAX 0 ); // 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 ?>
JavaScript < 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 (( a b )=> 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>
Ieșire:
273
Complexitatea timpului: O(MAX)
Spațiu auxiliar: O(MAX)
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