Podano celo število n ki predstavlja število števk. Naloga je natisniti vse n-mestna števila tako, da je absolutna razlika med vsoto števk na sodih in lihih mestih natančna 1 .
Opomba : Številka se ne sme začeti z (začetne ničle niso dovoljene).
seznam v Javi
Primeri:
Vnos : n = 2
Izhod : 10 12 21 23 32 34 43 45 54 56 65 67 76 78 87 89 98
Vnos : n = 3
Izhod : 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
[Pričakovan pristop] Uporaba rekurzije
C++Ideja je, da rekurzivno ustvari vsa n-mestna števila, medtem ko sledenje vsoti števk pri celo in liho pozicije z uporabo dveh spremenljivk. Za dano pozicijo jo zapolnimo z vsemi števkami od 0 do 9 in glede na to, ali je trenutna pozicija soda ali liha, povečamo sodo ali liho vsoto. Začetne ničle obravnavamo ločeno, saj se ne štejejo kot števke.
Sledili smo številčenju, ki temelji na ničli, kot so matrični indeksi. t.j. šteje se, da je prva (skrajno leva) številka prisotna na sodem mestu, številka poleg nje pa na lihem mestu in tako naprej.
// 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
// 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 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# 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 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] + ' '); }
Izhod
10 12 21 23 32 34 43 45 54 56 65 67 76 78 87 89 98
Časovna zapletenost: O(9 × 10^(n-1)), ker ima vsaka števka do 10 izbir (razen prve, ki ima 9).
Kompleksnost prostora: O(n + k), kjer je n globina rekurzije in k število veljavnih rezultatov.