Interval de consultes LCM
Donada una matriu arr[] de nombres enters de mida N i una matriu de consultes Q query[] on cada consulta és de tipus [L R] que denota l'interval des de l'índex L fins a l'índex R, la tasca és trobar el LCM de tots els números de l'interval per a totes les consultes.
Exemples:
Entrada: arr[] = {5 7 5 2 10 12 11 17 14 1 44}
consulta[] = {{2 5} {5 10} {0 10}}
Sortida: 6015708 78540
Explicació: A la primera consulta MCM(5 2 10 12) = 60
A la segona consulta MCM(12 11 17 14 1 44) = 15708
A l'última consulta MCM(5 7 5 2 10 12 11 17 14 1 44) = 78540Entrada: arr[] = {2 4 8 16} consulta[] = {{2 3} {0 1}}
Sortida: 16 4
Enfocament ingenu: L'enfocament es basa en la idea matemàtica següent:
Matemàticament LCM(l r) = LCM(arr[l] arr[l+1] . . . arr[r-1] arr[r]) i
MCM(a b) = (a*b) / MCD(ab)
Per tant, travessa la matriu per a cada consulta i calcula la resposta utilitzant la fórmula anterior per a LCM.
Complexitat temporal: O(N * Q)
Espai auxiliar: O(1)
Consultes RangeLCM utilitzant Arbre de segments :
Com que el nombre de consultes pot ser gran, la solució ingènua seria poc pràctica. Aquest temps es pot reduir
No hi ha cap operació d'actualització en aquest problema. Així que inicialment podem construir un arbre de segments i utilitzar-lo per respondre les consultes en temps logarítmic.
Cada node de l'arbre hauria d'emmagatzemar el valor LCM d'aquest segment en concret i podem utilitzar la mateixa fórmula que l'anterior per combinar els segments.
Seguiu els passos esmentats a continuació per implementar la idea:
- Construeix un arbre de segments a partir de la matriu donada.
- Recorre les consultes. Per a cada consulta:
- Trobeu aquest rang concret a l'arbre de segments.
- Utilitzeu la fórmula esmentada anteriorment per combinar els segments i calcular el LCM per a aquest rang.
- Imprimeix la resposta d'aquest segment.
A continuació es mostra la implementació de l'enfocament anterior.
C++ // LCM of given range queries using Segment Tree #include using namespace std ; #define MAX 1000 // allocate space for tree int tree [ 4 * MAX ]; // declaring the array globally int arr [ MAX ]; // Function to return gcd of a and b int gcd ( int a int b ) { if ( a == 0 ) return b ; return gcd ( b % a a ); } // utility function to find lcm int lcm ( int a int b ) { return a * b / gcd ( a b ); } // Function to build the segment tree // Node starts beginning index of current subtree. // start and end are indexes in arr[] which is global void build ( int node int start int end ) { // If there is only one element in current subarray if ( start == end ) { tree [ node ] = arr [ start ]; return ; } int mid = ( start + end ) / 2 ; // build left and right segments build ( 2 * node start mid ); build ( 2 * node + 1 mid + 1 end ); // build the parent int left_lcm = tree [ 2 * node ]; int right_lcm = tree [ 2 * node + 1 ]; tree [ node ] = lcm ( left_lcm right_lcm ); } // Function to make queries for array range )l r). // Node is index of root of current segment in segment // tree (Note that indexes in segment tree begin with 1 // for simplicity). // start and end are indexes of subarray covered by root // of current segment. int query ( int node int start int end int l int r ) { // Completely outside the segment returning // 1 will not affect the lcm; if ( end < l || start > r ) return 1 ; // completely inside the segment if ( l <= start && r >= end ) return tree [ node ]; // partially inside int mid = ( start + end ) / 2 ; int left_lcm = query ( 2 * node start mid l r ); int right_lcm = query ( 2 * node + 1 mid + 1 end l r ); return lcm ( left_lcm right_lcm ); } // driver function to check the above program int main () { // initialize the array arr [ 0 ] = 5 ; arr [ 1 ] = 7 ; arr [ 2 ] = 5 ; arr [ 3 ] = 2 ; arr [ 4 ] = 10 ; arr [ 5 ] = 12 ; arr [ 6 ] = 11 ; arr [ 7 ] = 17 ; arr [ 8 ] = 14 ; arr [ 9 ] = 1 ; arr [ 10 ] = 44 ; // build the segment tree build ( 1 0 10 ); // Now we can answer each query efficiently // Print LCM of (2 5) cout < < query ( 1 0 10 2 5 ) < < endl ; // Print LCM of (5 10) cout < < query ( 1 0 10 5 10 ) < < endl ; // Print LCM of (0 10) cout < < query ( 1 0 10 0 10 ) < < endl ; return 0 ; }
Java // LCM of given range queries // using Segment Tree class GFG { static final int MAX = 1000 ; // allocate space for tree static int tree [] = new int [ 4 * MAX ] ; // declaring the array globally static int arr [] = new int [ MAX ] ; // Function to return gcd of a and b static int gcd ( int a int b ) { if ( a == 0 ) { return b ; } return gcd ( b % a a ); } // utility function to find lcm static int lcm ( int a int b ) { return a * b / gcd ( a b ); } // Function to build the segment tree // Node starts beginning index // of current subtree. start and end // are indexes in arr[] which is global static void build ( int node int start int end ) { // If there is only one element // in current subarray if ( start == end ) { tree [ node ] = arr [ start ] ; return ; } int mid = ( start + end ) / 2 ; // build left and right segments build ( 2 * node start mid ); build ( 2 * node + 1 mid + 1 end ); // build the parent int left_lcm = tree [ 2 * node ] ; int right_lcm = tree [ 2 * node + 1 ] ; tree [ node ] = lcm ( left_lcm right_lcm ); } // Function to make queries for // array range )l r). Node is index // of root of current segment in segment // tree (Note that indexes in segment // tree begin with 1 for simplicity). // start and end are indexes of subarray // covered by root of current segment. static int query ( int node int start int end int l int r ) { // Completely outside the segment returning // 1 will not affect the lcm; if ( end < l || start > r ) { return 1 ; } // completely inside the segment if ( l <= start && r >= end ) { return tree [ node ] ; } // partially inside int mid = ( start + end ) / 2 ; int left_lcm = query ( 2 * node start mid l r ); int right_lcm = query ( 2 * node + 1 mid + 1 end l r ); return lcm ( left_lcm right_lcm ); } // Driver code public static void main ( String [] args ) { // initialize the array arr [ 0 ] = 5 ; arr [ 1 ] = 7 ; arr [ 2 ] = 5 ; arr [ 3 ] = 2 ; arr [ 4 ] = 10 ; arr [ 5 ] = 12 ; arr [ 6 ] = 11 ; arr [ 7 ] = 17 ; arr [ 8 ] = 14 ; arr [ 9 ] = 1 ; arr [ 10 ] = 44 ; // build the segment tree build ( 1 0 10 ); // Now we can answer each query efficiently // Print LCM of (2 5) System . out . println ( query ( 1 0 10 2 5 )); // Print LCM of (5 10) System . out . println ( query ( 1 0 10 5 10 )); // Print LCM of (0 10) System . out . println ( query ( 1 0 10 0 10 )); } } // This code is contributed by 29AjayKumar
Python # LCM of given range queries using Segment Tree MAX = 1000 # allocate space for tree tree = [ 0 ] * ( 4 * MAX ) # declaring the array globally arr = [ 0 ] * MAX # Function to return gcd of a and b def gcd ( a : int b : int ): if a == 0 : return b return gcd ( b % a a ) # utility function to find lcm def lcm ( a : int b : int ): return ( a * b ) // gcd ( a b ) # Function to build the segment tree # Node starts beginning index of current subtree. # start and end are indexes in arr[] which is global def build ( node : int start : int end : int ): # If there is only one element # in current subarray if start == end : tree [ node ] = arr [ start ] return mid = ( start + end ) // 2 # build left and right segments build ( 2 * node start mid ) build ( 2 * node + 1 mid + 1 end ) # build the parent left_lcm = tree [ 2 * node ] right_lcm = tree [ 2 * node + 1 ] tree [ node ] = lcm ( left_lcm right_lcm ) # Function to make queries for array range )l r). # Node is index of root of current segment in segment # tree (Note that indexes in segment tree begin with 1 # for simplicity). # start and end are indexes of subarray covered by root # of current segment. def query ( node : int start : int end : int l : int r : int ): # Completely outside the segment # returning 1 will not affect the lcm; if end < l or start > r : return 1 # completely inside the segment if l <= start and r >= end : return tree [ node ] # partially inside mid = ( start + end ) // 2 left_lcm = query ( 2 * node start mid l r ) right_lcm = query ( 2 * node + 1 mid + 1 end l r ) return lcm ( left_lcm right_lcm ) # Driver Code if __name__ == '__main__' : # initialize the array arr [ 0 ] = 5 arr [ 1 ] = 7 arr [ 2 ] = 5 arr [ 3 ] = 2 arr [ 4 ] = 10 arr [ 5 ] = 12 arr [ 6 ] = 11 arr [ 7 ] = 17 arr [ 8 ] = 14 arr [ 9 ] = 1 arr [ 10 ] = 44 # build the segment tree build ( 1 0 10 ) # Now we can answer each query efficiently # Print LCM of (2 5) print ( query ( 1 0 10 2 5 )) # Print LCM of (5 10) print ( query ( 1 0 10 5 10 )) # Print LCM of (0 10) print ( query ( 1 0 10 0 10 )) # This code is contributed by # sanjeev2552
C# // LCM of given range queries // using Segment Tree using System ; using System.Collections.Generic ; class GFG { static readonly int MAX = 1000 ; // allocate space for tree static int [] tree = new int [ 4 * MAX ]; // declaring the array globally static int [] arr = new int [ MAX ]; // Function to return gcd of a and b static int gcd ( int a int b ) { if ( a == 0 ) { return b ; } return gcd ( b % a a ); } // utility function to find lcm static int lcm ( int a int b ) { return a * b / gcd ( a b ); } // Function to build the segment tree // Node starts beginning index // of current subtree. start and end // are indexes in []arr which is global static void build ( int node int start int end ) { // If there is only one element // in current subarray if ( start == end ) { tree [ node ] = arr [ start ]; return ; } int mid = ( start + end ) / 2 ; // build left and right segments build ( 2 * node start mid ); build ( 2 * node + 1 mid + 1 end ); // build the parent int left_lcm = tree [ 2 * node ]; int right_lcm = tree [ 2 * node + 1 ]; tree [ node ] = lcm ( left_lcm right_lcm ); } // Function to make queries for // array range )l r). Node is index // of root of current segment in segment // tree (Note that indexes in segment // tree begin with 1 for simplicity). // start and end are indexes of subarray // covered by root of current segment. static int query ( int node int start int end int l int r ) { // Completely outside the segment // returning 1 will not affect the lcm; if ( end < l || start > r ) { return 1 ; } // completely inside the segment if ( l <= start && r >= end ) { return tree [ node ]; } // partially inside int mid = ( start + end ) / 2 ; int left_lcm = query ( 2 * node start mid l r ); int right_lcm = query ( 2 * node + 1 mid + 1 end l r ); return lcm ( left_lcm right_lcm ); } // Driver code public static void Main ( String [] args ) { // initialize the array arr [ 0 ] = 5 ; arr [ 1 ] = 7 ; arr [ 2 ] = 5 ; arr [ 3 ] = 2 ; arr [ 4 ] = 10 ; arr [ 5 ] = 12 ; arr [ 6 ] = 11 ; arr [ 7 ] = 17 ; arr [ 8 ] = 14 ; arr [ 9 ] = 1 ; arr [ 10 ] = 44 ; // build the segment tree build ( 1 0 10 ); // Now we can answer each query efficiently // Print LCM of (2 5) Console . WriteLine ( query ( 1 0 10 2 5 )); // Print LCM of (5 10) Console . WriteLine ( query ( 1 0 10 5 10 )); // Print LCM of (0 10) Console . WriteLine ( query ( 1 0 10 0 10 )); } } // This code is contributed by Rajput-Ji
JavaScript < script > // LCM of given range queries using Segment Tree const MAX = 1000 // allocate space for tree var tree = new Array ( 4 * MAX ); // declaring the array globally var arr = new Array ( MAX ); // Function to return gcd of a and b function gcd ( a b ) { if ( a == 0 ) return b ; return gcd ( b % a a ); } //utility function to find lcm function lcm ( a b ) { return Math . floor ( a * b / gcd ( a b )); } // Function to build the segment tree // Node starts beginning index of current subtree. // start and end are indexes in arr[] which is global function build ( node start end ) { // If there is only one element in current subarray if ( start == end ) { tree [ node ] = arr [ start ]; return ; } let mid = Math . floor (( start + end ) / 2 ); // build left and right segments build ( 2 * node start mid ); build ( 2 * node + 1 mid + 1 end ); // build the parent let left_lcm = tree [ 2 * node ]; let right_lcm = tree [ 2 * node + 1 ]; tree [ node ] = lcm ( left_lcm right_lcm ); } // Function to make queries for array range )l r). // Node is index of root of current segment in segment // tree (Note that indexes in segment tree begin with 1 // for simplicity). // start and end are indexes of subarray covered by root // of current segment. function query ( node start end l r ) { // Completely outside the segment returning // 1 will not affect the lcm; if ( end < l || start > r ) return 1 ; // completely inside the segment if ( l <= start && r >= end ) return tree [ node ]; // partially inside let mid = Math . floor (( start + end ) / 2 ); let left_lcm = query ( 2 * node start mid l r ); let right_lcm = query ( 2 * node + 1 mid + 1 end l r ); return lcm ( left_lcm right_lcm ); } //driver function to check the above program //initialize the array arr [ 0 ] = 5 ; arr [ 1 ] = 7 ; arr [ 2 ] = 5 ; arr [ 3 ] = 2 ; arr [ 4 ] = 10 ; arr [ 5 ] = 12 ; arr [ 6 ] = 11 ; arr [ 7 ] = 17 ; arr [ 8 ] = 14 ; arr [ 9 ] = 1 ; arr [ 10 ] = 44 ; // build the segment tree build ( 1 0 10 ); // Now we can answer each query efficiently // Print LCM of (2 5) document . write ( query ( 1 0 10 2 5 ) + '
' ); // Print LCM of (5 10) document . write ( query ( 1 0 10 5 10 ) + '
' ); // Print LCM of (0 10) document . write ( query ( 1 0 10 0 10 ) + '
' ); // This code is contributed by Manoj. < /script>
Sortida
60 15708 78540
Complexitat temporal: O(Log N * Log n) on N és el nombre d'elements de la matriu. L'altre log n denota el temps necessari per trobar el LCM. Aquesta complexitat temporal és per a cada consulta. La complexitat del temps total és O(N + Q*Log N*log n) això és perquè es requereix temps O(N) per construir l'arbre i després per respondre les consultes.
Espai auxiliar: O(N) on N és el nombre d'elements de la matriu. Aquest espai és necessari per emmagatzemar l'arbre de segments.
Tema relacionat: Arbre de segments
Enfocament núm. 2: Ús de les matemàtiques
Primer definim una funció auxiliar lcm() per calcular el mínim comú múltiple de dos nombres. A continuació, per a cada consulta, iterem a través del subbarray d'arr definit per l'interval de consulta i calculem el LCM mitjançant la funció lcm(). El valor LCM s'emmagatzema en una llista que es retorna com a resultat final.
Arbre de segments
Enfocament núm. 2: Ús de les matemàtiques
Algorisme
Arbre de segments
Enfocament núm. 2: Ús de les matemàtiques
1. Defineix una funció auxiliar mcm(a b) per calcular el mínim comú múltiple de dos nombres.
2. Definiu una funció range_lcm_queries(arr queries) que pren com a entrada una matriu arr i una llista de consultes d'intervals de consulta.
3. Creeu una llista de resultats buida per emmagatzemar els valors de LCM per a cada consulta.
4. Per a cada consulta en consultes extreu els índexs esquerre i dret l i r.
5. Estableix lcm_val al valor de arr[l].
6. Per a cada índex i del rang l+1 a r actualitzeu lcm_val perquè sigui el MCM de lcm_val i arr[i] mitjançant la funció lcm().
7. Afegiu lcm_val a la llista de resultats.
8. Retorna la llista de resultats.
Enfocament núm. 2: Ús de les matemàtiques
C++ Java #include
Python /*package whatever //do not write package name here */ import java.util.ArrayList ; import java.util.List ; public class GFG { public static int gcd ( int a int b ) { if ( b == 0 ) return a ; return gcd ( b a % b ); } public static int lcm ( int a int b ) { return a * b / gcd ( a b ); } public static List < Integer > rangeLcmQueries ( List < Integer > arr List < int []> queries ) { List < Integer > results = new ArrayList <> (); for ( int [] query : queries ) { int l = query [ 0 ] ; int r = query [ 1 ] ; int lcmVal = arr . get ( l ); for ( int i = l + 1 ; i <= r ; i ++ ) { lcmVal = lcm ( lcmVal arr . get ( i )); } results . add ( lcmVal ); } return results ; } public static void main ( String [] args ) { List < Integer > arr = List . of ( 5 7 5 2 10 12 11 17 14 1 44 ); List < int []> queries = List . of ( new int [] { 2 5 } new int [] { 5 10 } new int [] { 0 10 }); List < Integer > results = rangeLcmQueries ( arr queries ); for ( int result : results ) { System . out . print ( result + ' ' ); } System . out . println (); } }
C# from math import gcd def lcm ( a b ): return a * b // gcd ( a b ) def range_lcm_queries ( arr queries ): results = [] for query in queries : l r = query lcm_val = arr [ l ] for i in range ( l + 1 r + 1 ): lcm_val = lcm ( lcm_val arr [ i ]) results . append ( lcm_val ) return results # example usage arr = [ 5 7 5 2 10 12 11 17 14 1 44 ] queries = [( 2 5 ) ( 5 10 ) ( 0 10 )] print ( range_lcm_queries ( arr queries )) # output: [60 15708 78540]
JavaScript using System ; using System.Collections.Generic ; class GFG { // Function to calculate the greatest common divisor (GCD) // using Euclidean algorithm static int GCD ( int a int b ) { if ( b == 0 ) return a ; return GCD ( b a % b ); } // Function to calculate the least common multiple (LCM) // using GCD static int LCM ( int a int b ) { return a * b / GCD ( a b ); } static List < int > RangeLcmQueries ( List < int > arr List < Tuple < int int >> queries ) { List < int > results = new List < int > (); foreach ( var query in queries ) { int l = query . Item1 ; int r = query . Item2 ; int lcmVal = arr [ l ]; for ( int i = l + 1 ; i <= r ; i ++ ) { lcmVal = LCM ( lcmVal arr [ i ]); } results . Add ( lcmVal ); } return results ; } static void Main () { List < int > arr = new List < int > { 5 7 5 2 10 12 11 17 14 1 44 }; List < Tuple < int int >> queries = new List < Tuple < int int >> { Tuple . Create ( 2 5 ) Tuple . Create ( 5 10 ) Tuple . Create ( 0 10 ) }; List < int > results = RangeLcmQueries ( arr queries ); foreach ( var result in results ) { Console . Write ( result + ' ' ); } Console . WriteLine (); } }
// JavaScript Program for the above approach // function to find out gcd function gcd ( a b ) { if ( b === 0 ) { return a ; } return gcd ( b a % b ); } // function to find out lcm function lcm ( a b ) { return ( a * b ) / gcd ( a b ); } function rangeLcmQueries ( arr queries ) { const results = []; for ( const query of queries ) { const l = query [ 0 ]; const r = query [ 1 ]; let lcmVal = arr [ l ]; for ( let i = l + 1 ; i <= r ; i ++ ) { lcmVal = lcm ( lcmVal arr [ i ]); } results . push ( lcmVal ); } return results ; } // Driver code to test above function const arr = [ 5 7 5 2 10 12 11 17 14 1 44 ]; const queries = [[ 2 5 ] [ 5 10 ] [ 0 10 ]]; const results = rangeLcmQueries ( arr queries ); for ( const result of results ) { console . log ( result + ' ' ); } console . log (); // THIS CODE IS CONTRIBUTED BY PIYUSH AGARWAL
Sortida
[60 15708 78540]
Complexitat temporal: O(log(min(ab))). Per a cada rang de consulta, iterem a través d'un subbarray de mida O(n) on n és la longitud d'arr. Per tant, la complexitat temporal de la funció global és O(qn log(min(a_i))) on q és el nombre de consultes i a_i és l'i-è element de arr.
Complexitat espacial: O(1) ja que només emmagatzemem uns quants nombres enters alhora. L'espai utilitzat per l'entrada arr i les consultes no es considera.
