N'te intelligente Nummer
Gegeben eine Zahl n finde die n-te intelligente Zahl (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. Beispiele:
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 37Empfohlen: Bitte lösen Sie es auf ÜBEN bevor Sie mit der Lösung fortfahren.
Die Idee basiert auf Sieb des Eratosthenes . Wir verwenden ein Array, um ein Array prime[] zu verwenden, um den Überblick über Primzahlen zu behalten. Wir verwenden dasselbe Array auch, um die Anzahl der bisher gesehenen Primfaktoren zu verfolgen. Immer wenn die Zahl 3 erreicht, addieren wir die Zahl zum Ergebnis.
- Nehmen Sie ein Array primes[] und initialisieren Sie es mit 0.
- Jetzt wissen wir, dass die erste Primzahl i = 2 ist, also markieren Sie Primzahlen[2] = 1, d. h. primes[i] = 1 gibt an, dass „i“ eine Primzahl ist.
- Durchlaufen Sie nun das Array primes[] und markieren Sie alle Vielfachen von „i“ durch die Bedingung primes[j] -= 1, wobei „j“ ein Vielfaches von „i“ ist, und überprüfen Sie die Bedingung primes[j]+3 = 0, denn wann immer primes[j] zu -3 wird, bedeutet dies, dass es zuvor ein Vielfaches von drei verschiedenen Primfaktoren war. Wenn Bedingung Primzahlen[j]+3=0 wird wahr, was bedeutet, dass „j“ eine intelligente Zahl ist. Speichern Sie sie daher in einem Array-Ergebnis[].
- Sortieren Sie nun das Array-Ergebnis[] und geben Sie Ergebnis[n-1] zurück.
Nachfolgend finden Sie die Umsetzung der obigen Idee.
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>
Ausgabe:
273
Zeitkomplexität: O(MAX)
Hilfsraum: O(MAX)
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