Najmanjše število, deljivo s prvimi n števili
Preizkusite na GfG Practice 
Izhod
Glede na številko n poišči najmanjše število, ki je enakomerno deljivo z vsakim številom od 1 do n.
Primeri:
Input : n = 4 Output : 12 Explanation : 12 is the smallest numbers divisible by all numbers from 1 to 4 Input : n = 10 Output : 2520 Input : n = 20 Output : 232792560
Če pozorno opazujete leta mora biti LCM števil od 1 do n .
Če želite najti LCM števil od 1 do n -
- Inicializiraj ans = 1.
- Ponovite vsa števila od i = 1 do i = n.
Pri i-ti ponovitvi ans = LCM(1 2 …….. i) . To je mogoče enostavno narediti kot LCM(1 2 …. i) = LCM(ans i) .
Tako moramo pri i-ti ponovitvi samo narediti -
ans = LCM(ans i) = ans * i / gcd(ans i) [Using the below property a*b = gcd(ab) * lcm(ab)]C++
Opomba: V kodi C++ odgovor hitro preseže omejitev celih števil, celo dolgo in dolgo omejitev.
Spodaj je implementacija logike.
Java// C++ program to find smallest number evenly divisible by // all numbers 1 to n #includeusing namespace std ; // Function returns the lcm of first n numbers long long lcm ( long long n ) { long long ans = 1 ; for ( long long i = 1 ; i <= n ; i ++ ) ans = ( ans * i ) / ( __gcd ( ans i )); return ans ; } // Driver program to test the above function int main () { long long n = 20 ; cout < < lcm ( n ); return 0 ; } Python// Java program to find the smallest number evenly divisible by // all numbers 1 to n class GFG { static long gcd ( long a long b ) { if ( a % b != 0 ) return gcd ( b a % b ); else return b ; } // Function returns the lcm of first n numbers static long lcm ( long n ) { long ans = 1 ; for ( long i = 1 ; i <= n ; i ++ ) ans = ( ans * i ) / ( gcd ( ans i )); return ans ; } // Driver program to test the above function public static void main ( String [] args ) { long n = 20 ; System . out . println ( lcm ( n )); } }C## Python program to find the smallest number evenly # divisible by all number 1 to n import math # Returns the lcm of first n numbers def lcm ( n ): ans = 1 for i in range ( 1 n + 1 ): ans = int (( ans * i ) / math . gcd ( ans i )) return ans # main n = 20 print ( lcm ( n ))Javascript// C# program to find smallest number // evenly divisible by // all numbers 1 to n using System ; public class GFG { static long gcd ( long a long b ) { if ( a % b != 0 ) return gcd ( b a % b ); else return b ; } // Function returns the lcm of first n numbers static long lcm ( long n ) { long ans = 1 ; for ( long i = 1 ; i <= n ; i ++ ) ans = ( ans * i ) / ( gcd ( ans i )); return ans ; } // Driver program to test the above function static public void Main (){ long n = 20 ; Console . WriteLine ( lcm ( n )); } //This code is contributed by akt_mit }PHP// Javascript program to find the smallest number evenly divisible by // all numbers 1 to n function gcd ( a b ) { if ( a % b != 0 ) return gcd ( b a % b ); else return b ; } // Function returns the lcm of first n numbers function lcm ( n ) { let ans = 1 ; for ( let i = 1 ; i <= n ; i ++ ) ans = ( ans * i ) / ( gcd ( ans i )); return ans ; } // function call let n = 20 ; console . log ( lcm ( n ));
Izhod232792560Časovna kompleksnost: O(n log2n) ker je kompleksnost _gcd(ab) v c++ log2n in se zažene n-krat v zanki.
Pomožni prostor: O(1)
Zgornja rešitev dobro deluje za en vnos. Če pa imamo več vhodov, je dobro, da za shranjevanje vseh prafaktorjev uporabimo Eratostenovo sito. Oglejte si spodnji članek za pristop, ki temelji na situ.Pristop : [Uporaba Eratostenovo sito ]
Da bi učinkoviteje rešili problem iskanja najmanjšega števila, deljivega s prvimi 'n' števili, lahko uporabimo Eratostenovo sito za vnaprejšnje izračunavanje praštevil do 'n'. Nato lahko ta praštevila uporabimo za učinkovitejši izračun najmanjšega skupnega večkratnika (LCM), tako da upoštevamo največje potence vsakega praštevila, ki so manjše ali enake 'n'.
Pristop po korakih:
- Generirajte praštevila do n: Uporabite Eratostenovo sito, da poiščete vsa praštevila do 'n'.
- Izračunajte LCM z uporabo teh praštevil: Za vsako praštevilo določite največjo potenco tega praštevila, ki je manjše ali enako 'n'. Pomnožite te najvišje moči skupaj, da dobite LCM
Spodaj je izvedba zgornjega pristopa:
C++ #include #include #include using namespace std ; // Function to generate all prime numbers up to n using the // Sieve of Eratosthenes vector < int > sieve_of_eratosthenes ( int n ) { vector < bool > is_prime ( n + 1 true ); int p = 2 ; while ( p * p <= n ) { if ( is_prime [ p ]) { for ( int i = p * p ; i <= n ; i += p ) { is_prime [ i ] = false ; } } ++ p ; } vector < int > prime_numbers ; for ( int p = 2 ; p <= n ; ++ p ) { if ( is_prime [ p ]) { prime_numbers . push_back ( p ); } } return prime_numbers ; } // Function to find the smallest number divisible by all // numbers from 1 to n long long smallest_multiple ( int n ) { vector < int > primes = sieve_of_eratosthenes ( n ); long long lcm = 1 ; for ( int prime : primes ) { // Calculate the highest power of the prime that is // <= n int power = 1 ; while ( pow ( prime power + 1 ) <= n ) { ++ power ; } lcm *= pow ( prime power ); } return lcm ; } int main () { int n = 20 ; cout < < smallest_multiple ( n ) < < endl ; return 0 ; }
Java import java.util.ArrayList ; import java.util.List ; public class SmallestMultiple { // Function to generate all prime numbers up to n using // the Sieve of Eratosthenes public static List < Integer > sieveOfEratosthenes ( int n ) { boolean [] isPrime = new boolean [ n + 1 ] ; for ( int i = 0 ; i <= n ; i ++ ) { isPrime [ i ] = true ; } int p = 2 ; while ( p * p <= n ) { if ( isPrime [ p ] ) { for ( int i = p * p ; i <= n ; i += p ) { isPrime [ i ] = false ; } } p ++ ; } List < Integer > primeNumbers = new ArrayList <> (); for ( int i = 2 ; i <= n ; i ++ ) { if ( isPrime [ i ] ) { primeNumbers . add ( i ); } } return primeNumbers ; } // Function to find the smallest number divisible by all // numbers from 1 to n public static long smallestMultiple ( int n ) { List < Integer > primes = sieveOfEratosthenes ( n ); long lcm = 1 ; for ( int prime : primes ) { // Calculate the highest power of the prime that // is <= n int power = 1 ; while ( Math . pow ( prime power + 1 ) <= n ) { power ++ ; } lcm *= Math . pow ( prime power ); } return lcm ; } public static void main ( String [] args ) { int n = 20 ; System . out . println ( smallestMultiple ( n )); } }
Python import math def sieve_of_eratosthenes ( n ): '''Generate all prime numbers up to n.''' is_prime = [ True ] * ( n + 1 ) p = 2 while ( p * p <= n ): if ( is_prime [ p ] == True ): for i in range ( p * p n + 1 p ): is_prime [ i ] = False p += 1 prime_numbers = [ p for p in range ( 2 n + 1 ) if is_prime [ p ]] return prime_numbers def smallest_multiple ( n ): '''Find the smallest number divisible by all numbers from 1 to n.''' primes = sieve_of_eratosthenes ( n ) lcm = 1 for prime in primes : # Calculate the highest power of the prime that is <= n power = 1 while prime ** ( power + 1 ) <= n : power += 1 lcm *= prime ** power return lcm # Example usage: n = 20 print ( smallest_multiple ( n ))
JavaScript // Function to generate all prime numbers up to n using the // Sieve of Eratosthenes function sieveOfEratosthenes ( n ) { let isPrime = new Array ( n + 1 ). fill ( true ); let p = 2 ; while ( p * p <= n ) { if ( isPrime [ p ]) { for ( let i = p * p ; i <= n ; i += p ) { isPrime [ i ] = false ; } } p ++ ; } let primeNumbers = []; for ( let p = 2 ; p <= n ; p ++ ) { if ( isPrime [ p ]) { primeNumbers . push ( p ); } } return primeNumbers ; } // Function to find the smallest number divisible by all // numbers from 1 to n function smallestMultiple ( n ) { let primes = sieveOfEratosthenes ( n ); let lcm = 1 ; for ( let prime of primes ) { // Calculate the highest power of the prime that is // <= n let power = 1 ; while ( Math . pow ( prime power + 1 ) <= n ) { power ++ ; } lcm *= Math . pow ( prime power ); } return lcm ; } // Example usage: let n = 20 ; console . log ( smallestMultiple ( n ));
Izhod
The smallest number divisible by all numbers from 1 to 20 is 232792560
Časovna zapletenost: O(nloglogn)
Pomožni prostor: O(n)
