Consultas de LCM de rango
Dada una matriz arr[] de enteros de tamaño N y una matriz de consultas Q query[] donde cada consulta es de tipo [L R] que denota el rango del índice L al índice R, la tarea es encontrar el MCM de todos los números del rango para todas las consultas.
Ejemplos:
Aporte: arreglo[] = {5 7 5 2 10 12 11 17 14 1 44}
consulta[] = {{2 5} {5 10} {0 10}}
Producción: 6015708 78540
Explicación: En la primera consulta MCM(5 2 10 12) = 60
En la segunda consulta MCM(12 11 17 14 1 44) = 15708
En la última consulta LCM(5 7 5 2 10 12 11 17 14 1 44) = 78540Aporte: arreglo[] = {2 4 8 16} consulta[] = {{2 3} {0 1}}
Producción: 16 4
Enfoque ingenuo: El enfoque se basa en la siguiente idea matemática:
Matemáticamente MCM(l r) = MCM(arr[l] arr[l+1] . . . arr[r-1] arr[r]) y
MCM(a b) = (a*b) / MCD(ab)
Así que recorra la matriz para cada consulta y calcule la respuesta utilizando la fórmula anterior para LCM.
Complejidad del tiempo: O(N * Q)
Espacio Auxiliar: O(1)
Consultas RangeLCM usando Árbol de segmentos :
Como el número de consultas puede ser grande, la solución ingenua no sería práctica. Este tiempo se puede reducir
No hay ninguna operación de actualización en este problema. Entonces, inicialmente podemos construir un árbol de segmentos y usarlo para responder las consultas en tiempo logarítmico.
Cada nodo del árbol debe almacenar el valor LCM para ese segmento en particular y podemos usar la misma fórmula anterior para combinar los segmentos.
Siga los pasos que se mencionan a continuación para implementar la idea:
- Construya un árbol de segmentos a partir de la matriz dada.
- Recorra las consultas. Para cada consulta:
- Encuentre ese rango particular en el árbol de segmentos.
- Utilice la fórmula mencionada anteriormente para combinar los segmentos y calcular el MCM para ese rango.
- Imprime la respuesta para ese segmento.
A continuación se muestra la implementación del enfoque 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>
Producción
60 15708 78540
Complejidad del tiempo: O(Log N * Log n) donde N es el número de elementos de la matriz. El otro log n denota el tiempo necesario para encontrar el LCM. Esta vez la complejidad es para cada consulta. La complejidad del tiempo total es O(N + Q*Log N*log n). Esto se debe a que se requiere tiempo O(N) para construir el árbol y luego responder las consultas.
Espacio Auxiliar: O(N) donde N es el número de elementos de la matriz. Este espacio es necesario para almacenar el árbol de segmentos.
Tema relacionado: Árbol de segmentos
Enfoque n.° 2: usar las matemáticas
Primero definimos una función auxiliar lcm() para calcular el mínimo común múltiplo de dos números. Luego, para cada consulta, iteramos a través del subarreglo de arr definido por el rango de consulta y calculamos el LCM usando la función lcm(). El valor LCM se almacena en una lista que se devuelve como resultado final.
Árbol de segmentos
Enfoque n.° 2: usar las matemáticas
Algoritmo
Árbol de segmentos
Enfoque n.° 2: usar las matemáticas
1. Defina una función auxiliar mcm(a b) para calcular el mínimo común múltiplo de dos números.
2. Defina una función range_lcm_queries(consultas arr) que tome una matriz arr y una lista de consultas de rangos de consulta como entrada.
3. Cree una lista de resultados vacía para almacenar los valores de LCM para cada consulta.
4. Para cada consulta en consultas, extraiga los índices izquierdo y derecho ly r.
5. Establezca lcm_val en el valor de arr[l].
6. Para cada índice i en el rango l+1 a r actualice lcm_val para que sea el LCM de lcm_val y arr[i] usando la función lcm().
7. Agregue lcm_val a la lista de resultados.
8. Devuelve la lista de resultados.
Enfoque n.° 2: usar las matemáticas
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
Producción
[60 15708 78540]
Complejidad del tiempo: O(log(mín(ab))). Para cada rango de consulta, iteramos a través de un subarreglo de tamaño O(n) donde n es la longitud de arr. Por lo tanto, la complejidad temporal de la función general es O(qn log(min(a_i))) donde q es el número de consultas y a_i es el i-ésimo elemento de arr.
Complejidad espacial: O(1) ya que solo almacenamos unos pocos números enteros a la vez. No se considera el espacio utilizado por la entrada arr y las consultas.
