Drukājiet visus n-ciparu skaitļus ar nepāra un pāra ciparu summas starpību kā 1
Dots vesels skaitlis n kas attēlo ciparu skaitu. Uzdevums ir izdrukāt visu n-ciparu skaitļi tā, ka absolūtā atšķirība starp ciparu summu pāra pozīcijās un nepāra pozīcijās ir precīza 1 .
Piezīme : numurs nedrīkst sākties ar (nulles sākumā nav atļautas).
Piemēri:
Ievade : n = 2
Izvade : 10 12 21 23 32 34 43 45 54 56 65 67 76 78 87 89 98
Ievade : n = 3
Izvade : 100 111 120 122 131 133 142 144 153 155 164 166 175 177 186
188 197 199 210 221 230 232 241 243 252 254 263 265 274 276 285
287 296 298 320 331 340 342 351 353 362 364 373 375 384 386 395
397 430 441 450 452 461 463 472 474 483 485 494 496 540 551 560
562 571 573 582 584 593 595 650 661 670 672 681 683 692 694 760
771 780 782 791 793 870 881 890 892 980 991
[Paredzamā pieeja] Rekursijas izmantošana
C++Ideja ir, lai rekursīvi ģenerēt visus n-ciparu skaitļus izsekot summai no cipariem plkst pat un nepāra pozīcijas, izmantojot divus mainīgos. Dotajai pozīcijai mēs to aizpildām ar visiem cipariem no 0 līdz 9 un, pamatojoties uz to, vai pašreizējā pozīcija ir pāra vai nepāra, mēs palielinām pāra vai nepāra summu. Mēs atsevišķi apstrādājam 0 reģistrus, jo tie netiek skaitīti kā cipari.
Mēs esam ievērojuši nulles numerāciju, piemēram, masīvu indeksus. t.i., tiek uzskatīts, ka priekšējais (kreisais) cipars atrodas pāra pozīcijā, un cipars blakus tam tiek uzskatīts par nepāra pozīciju utt.
// C++ program to print all n-digit numbers such that // the absolute difference between the sum of digits at // even and odd positions is 1 #include using namespace std ; // Recursive function to generate numbers void findNDigitNumsUtil ( int pos int n int num int evenSum int oddSum vector < int > & res ) { // If number is formed if ( pos == n ) { // Check absolute difference condition if ( abs ( evenSum - oddSum ) == 1 ) { res . push_back ( num ); } return ; } // Digits to consider at current position for ( int d = 0 ; d <= 9 ; d ++ ) { // Skip leading 0 if ( pos == 0 && d == 0 ) { continue ; } // If position is even (0-based) add to evenSum if ( pos % 2 == 0 ) { findNDigitNumsUtil ( pos + 1 n num * 10 + d evenSum + d oddSum res ); } // If position is odd add to oddSum else { findNDigitNumsUtil ( pos + 1 n num * 10 + d evenSum oddSum + d res ); } } } // Function to prepare and collect valid numbers vector < int > findNDigitNums ( int n ) { vector < int > res ; findNDigitNumsUtil ( 0 n 0 0 0 res ); return res ; } // Driver code int main () { int n = 2 ; vector < int > res = findNDigitNums ( n ); for ( int i = 0 ; i < res . size (); i ++ ) { cout < < res [ i ] < < ' ' ; } return 0 ; }
Java // Java program to print all n-digit numbers such that // the absolute difference between the sum of digits at // even and odd positions is 1 import java.util.* ; class GfG { // Recursive function to generate numbers static void findNDigitNumsUtil ( int pos int n int num int evenSum int oddSum ArrayList < Integer > res ) { // If number is formed if ( pos == n ) { // Check absolute difference condition if ( Math . abs ( evenSum - oddSum ) == 1 ) { res . add ( num ); } return ; } // Digits to consider at current position for ( int d = 0 ; d <= 9 ; d ++ ) { // Skip leading 0 if ( pos == 0 && d == 0 ) { continue ; } // If position is even (0-based) add to evenSum if ( pos % 2 == 0 ) { findNDigitNumsUtil ( pos + 1 n num * 10 + d evenSum + d oddSum res ); } // If position is odd add to oddSum else { findNDigitNumsUtil ( pos + 1 n num * 10 + d evenSum oddSum + d res ); } } } // Function to prepare and collect valid numbers static ArrayList < Integer > findNDigitNums ( int n ) { ArrayList < Integer > res = new ArrayList <> (); findNDigitNumsUtil ( 0 n 0 0 0 res ); return res ; } // Driver code public static void main ( String [] args ) { int n = 2 ; ArrayList < Integer > res = findNDigitNums ( n ); // Print all collected valid numbers for ( int i = 0 ; i < res . size (); i ++ ) { System . out . print ( res . get ( i ) + ' ' ); } } }
Python # Python program to print all n-digit numbers such that # the absolute difference between the sum of digits at # even and odd positions is 1 # Recursive function to generate numbers def findNDigitNumsUtil ( pos n num evenSum oddSum res ): # If number is formed if pos == n : # Check absolute difference condition if abs ( evenSum - oddSum ) == 1 : res . append ( num ) return # Digits to consider at current position for d in range ( 10 ): # Skip leading 0 if pos == 0 and d == 0 : continue # If position is even (0-based) add to evenSum if pos % 2 == 0 : findNDigitNumsUtil ( pos + 1 n num * 10 + d evenSum + d oddSum res ) # If position is odd add to oddSum else : findNDigitNumsUtil ( pos + 1 n num * 10 + d evenSum oddSum + d res ) # Function to prepare and collect valid numbers def findNDigitNums ( n ): res = [] findNDigitNumsUtil ( 0 n 0 0 0 res ) return res # Driver code if __name__ == '__main__' : n = 2 res = findNDigitNums ( n ) # Print all collected valid numbers for i in range ( len ( res )): print ( res [ i ] end = ' ' )
C# // C# program to print all n-digit numbers such that // the absolute difference between the sum of digits at // even and odd positions is 1 using System ; using System.Collections.Generic ; class GfG { // Recursive function to generate numbers static void findNDigitNumsUtil ( int pos int n int num int evenSum int oddSum List < int > res ) { // If number is formed if ( pos == n ) { // Check absolute difference condition if ( Math . Abs ( evenSum - oddSum ) == 1 ) { res . Add ( num ); } return ; } // Digits to consider at current position for ( int d = 0 ; d <= 9 ; d ++ ) { // Skip leading 0 if ( pos == 0 && d == 0 ) { continue ; } // If position is even (0-based) add to evenSum if ( pos % 2 == 0 ) { findNDigitNumsUtil ( pos + 1 n num * 10 + d evenSum + d oddSum res ); } // If position is odd add to oddSum else { findNDigitNumsUtil ( pos + 1 n num * 10 + d evenSum oddSum + d res ); } } } // Function to prepare and collect valid numbers static List < int > findNDigitNums ( int n ) { List < int > res = new List < int > (); findNDigitNumsUtil ( 0 n 0 0 0 res ); return res ; } // Driver code public static void Main ( string [] args ) { int n = 2 ; List < int > res = findNDigitNums ( n ); // Print all collected valid numbers for ( int i = 0 ; i < res . Count ; i ++ ) { Console . Write ( res [ i ] + ' ' ); } } }
JavaScript // JavaScript program to print all n-digit numbers such that // the absolute difference between the sum of digits at // even and odd positions is 1 // Recursive function to generate numbers function findNDigitNumsUtil ( pos n num evenSum oddSum res ) { // If number is formed if ( pos === n ) { // Check absolute difference condition if ( Math . abs ( evenSum - oddSum ) === 1 ) { res . push ( num ); } return ; } // Digits to consider at current position for ( let d = 0 ; d <= 9 ; d ++ ) { // Skip leading 0 if ( pos === 0 && d === 0 ) { continue ; } // If position is even (0-based) add to evenSum if ( pos % 2 === 0 ) { findNDigitNumsUtil ( pos + 1 n num * 10 + d evenSum + d oddSum res ); } // If position is odd add to oddSum else { findNDigitNumsUtil ( pos + 1 n num * 10 + d evenSum oddSum + d res ); } } } // Function to prepare and collect valid numbers function findNDigitNums ( n ) { let res = []; findNDigitNumsUtil ( 0 n 0 0 0 res ); return res ; } // Driver code let n = 2 ; let res = findNDigitNums ( n ); // Print all collected valid numbers for ( let i = 0 ; i < res . length ; i ++ ) { process . stdout . write ( res [ i ] + ' ' ); }
Izvade
10 12 21 23 32 34 43 45 54 56 65 67 76 78 87 89 98
Laika sarežģītība: O(9 × 10^(n-1)), jo katram ciparam ir līdz 10 izvēlēm (izņemot pirmo, kuram ir 9).
Kosmosa sarežģītība: O(n + k) kur n ir rekursijas dziļums un k ir derīgo rezultātu skaits.
