Duotas skaičius n, neraskite n-ojo protingo skaičiaus (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. Pavyzdžiai:
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 37Rekomenduojama: išspręskite PRAKTIKA
. Mes naudojame masyvą, norėdami naudoti masyvo pirminį skaičių[], kad galėtume sekti pirminius skaičius. Taip pat naudojame tą patį masyvą, kad stebėtume iki šiol matytų pirminių veiksnių skaičių. Kai skaičius pasiekia 3, prie rezultato pridedame skaičių.
- Paimkite masyvo pirminius skaičius [] ir inicijuokite jį 0.
- Dabar žinome, kad pirmasis pirminis skaičius yra i = 2, todėl pažymėkite pirminius skaičius[2] = 1, t.y.; pirminiai [i] = 1 rodo, kad 'i' yra pirminis skaičius.
- Dabar pereikite pirminių [] masyvą ir pažymėkite visus 'i' kartotinius sąlygų pirminiais [j] -= 1, kur 'j' yra 'i' kartotinis, ir patikrinkite sąlygos pirminius [j]+3 = 0, nes kai pirminiai [j] tampa -3, tai rodo, kad anksčiau jis buvo trijų skirtingų pirminių faktorių kartotinis. Jei sąlyga tampa tiesa, tai reiškia, kad „j“ yra išmanusis skaičius, todėl išsaugokite jį masyvo rezultate[].
- Dabar rūšiuokite masyvo rezultatą [] ir grąžinkite rezultatą [n-1].
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>
Išvestis:
273
Laiko sudėtingumas: O(MAX)
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