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Keith Nummer

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Eine n-stellige Zahl x wird Keith-Zahl genannt, wenn sie in einer speziellen Reihenfolge (unten definiert) erscheint, die aus ihren Ziffern generiert wird. Die spezielle Sequenz hat die ersten n Terme als Ziffern von x und andere Terme werden rekursiv als Summe der vorherigen n Terme ausgewertet.
Die Aufgabe besteht darin, herauszufinden, ob eine bestimmte Zahl eine Keith-Zahl ist oder nicht.
Beispiele:  
 

Input : x = 197 Output : Yes 197 has 3 digits so n = 3 The number is Keith because it appears in the special sequence that has first three terms as 1 9 7 and remaining terms evaluated using sum of previous 3 terms. 1 9 7 17 33 57 107   197   ..... Input : x = 12 Output : No The number is not Keith because it doesn't appear in the special sequence generated using its digits. 1 2 3 5 8 13 21 ..... Input : x = 14 Output : Yes 14 is a Keith number since it appears in the sequence 1 4 5 9   14   ...


 

Java vs. C++
Empfohlene Praxis Keith Nummer Probieren Sie es aus!


Algorithmus:  
 



  1. Speichern Sie die 'n' Ziffern der gegebenen Zahl 'x' in einem Array 'terms'.
  2. Schleife zum Erzeugen der nächsten Terme der Sequenz und zum Hinzufügen der vorherigen „n“ Terme.
  3. Speichern Sie weiterhin die next_terms aus Schritt 2 im Array „terms“.
  4. Wenn der nächste Term gleich x wird, ist x eine Keith-Zahl. Wenn der nächste Term größer als x wird, ist x keine Keith-Zahl.


 

C++
// C++ program to check if a number is Keith or not #include   using namespace std; // Returns true if x is Keith else false. bool isKeith(int x) {  // Store all digits of x in a vector 'terms'  // Also find number of digits and store in 'n'.  vector <int> terms;  int temp = x n = 0; // n is number of digits in x  while (temp > 0)  {  terms.push_back(temp%10);  temp = temp/10;  n++;  }  // To get digits in right order (from MSB to  // LSB)  reverse(terms.begin() terms.end());  // Keep finding next terms of a sequence generated  // using digits of x until we either reach x or a  // number greater than x  int next_term = 0 i = n;  while (next_term < x)  {  next_term = 0;  // Next term is sum of previous n terms  for (int j=1; j<=n; j++)  next_term += terms[i-j];  terms.push_back(next_term);  i++;  }  /* When the control comes out of the while loop  either the next_term is equal to the number  or greater than it.  If next_term is equal to x then x is a  Keith number else not */  return (next_term == x); } // Driver program int main() {  isKeith(14)? cout << 'Yesn' : cout << 'Non';  isKeith(12)? cout << 'Yesn' : cout << 'Non';  isKeith(197)? cout << 'Yesn' : cout << 'Non';  return 0; } 
Java
// Java program to check if a number is Keith or not import java.io.*;  import java.util.*; class GFG{ // Returns true if x is Keith else false. static boolean isKeith(int x) {  // Store all digits of x in a vector 'terms'  // Also find number of digits and store in 'n'.  ArrayList<Integer> terms=new ArrayList<Integer>();  int temp = x n = 0; // n is number of digits in x  while (temp > 0)  {  terms.add(temp%10);  temp = temp/10;  n++;  }  // To get digits in right order (from MSB to  // LSB)  Collections.reverse(terms);  // Keep finding next terms of a sequence generated  // using digits of x until we either reach x or a  // number greater than x  int next_term = 0 i = n;  while (next_term < x)  {  next_term = 0;  // Next term is sum of previous n terms  for (int j=1; j<=n; j++)  next_term += terms.get(i-j);  terms.add(next_term);  i++;  }  /* When the control comes out of the while loop  either the next_term is equal to the number  or greater than it.  If next_term is equal to x then x is a  Keith number else not */  return (next_term == x); } // Driver program public static void main(String[] args) { if (isKeith(14)) System.out.println('Yes'); else  System.out.println('No'); if(isKeith(12))  System.out.println('Yes'); else System.out.println('No'); if(isKeith(197))  System.out.println('Yes'); else System.out.println('No'); } } // this code is contributed by mits 
Python3
# Python3 program to check if a number  # is Keith or not  # Returns true if x is Keith  # else false.  def isKeith(x): # Store all digits of x in a vector 'terms'  # Also find number of digits and store in 'n'.  terms = []; temp = x; n = 0; # n is number of digits in x  while (temp > 0): terms.append(temp % 10); temp = int(temp / 10); n+=1; # To get digits in right order  # (from MSB to LSB)  terms.reverse(); # Keep finding next terms of a sequence  # generated using digits of x until we  # either reach x or a number greater than x  next_term = 0; i = n; while (next_term < x): next_term = 0; # Next term is sum of previous n terms  for j in range(1n+1): next_term += terms[i - j]; terms.append(next_term); i+=1; # When the control comes out of the  # while loop either the next_term is  # equal to the number or greater than it.  # If next_term is equal to x then x is a  # Keith number else not  return (next_term == x); # Driver Code  print('Yes') if (isKeith(14)) else print('No'); print('Yes') if (isKeith(12)) else print('No'); print('Yes') if (isKeith(197)) else print('No'); # This code is contributed by mits 
C#
// C# program to check if a number is Keith or not using System;  using System.Collections; class GFG{ // Returns true if x is Keith else false. static bool isKeith(int x) {  // Store all digits of x in a vector 'terms'  // Also find number of digits and store in 'n'.  ArrayList terms = new ArrayList();  int temp = x n = 0; // n is number of digits in x  while (temp > 0)  {  terms.Add(temp%10);  temp = temp/10;  n++;  }  // To get digits in right order (from MSB to  // LSB)  terms.Reverse();  // Keep finding next terms of a sequence generated  // using digits of x until we either reach x or a  // number greater than x  int next_term = 0 i = n;  while (next_term < x)  {  next_term = 0;  // Next term is sum of previous n terms  for (int j=1; j<=n; j++)  next_term += (int)terms[i-j];  terms.Add(next_term);  i++;  }  /* When the control comes out of the while loop  either the next_term is equal to the number  or greater than it.  If next_term is equal to x then x is a  Keith number else not */  return (next_term == x); } // Driver program public static void Main() { if (isKeith(14)) Console.WriteLine('Yes'); else Console.WriteLine('No'); if(isKeith(12))  Console.WriteLine('Yes'); else Console.WriteLine('No'); if(isKeith(197))  Console.WriteLine('Yes'); else Console.WriteLine('No'); } } // this code is contributed by mits 
PHP
 // PHP program to check if a number  // is Keith or not // Returns true if x is Keith // else false. function isKeith($x) { // Store all digits of x in a vector 'terms' // Also find number of digits and store in 'n'. $terms = array(); $temp = $x; $n = 0; // n is number of digits in x while ($temp > 0) { array_push($terms $temp % 10); $temp = (int)($temp / 10); $n++; } // To get digits in right order  // (from MSB to LSB) $terms=array_reverse($terms); // Keep finding next terms of a sequence  // generated using digits of x until we  // either reach x or a number greater than x $next_term = 0; $i = $n; while ($next_term < $x) { $next_term = 0; // Next term is sum of previous n terms for ($j = 1; $j <= $n; $j++) $next_term += $terms[$i - $j]; array_push($terms $next_term); $i++; } /* When the control comes out of the   while loop either the next_term is   equal to the number or greater than it.  If next_term is equal to x then x is a  Keith number else not */ return ($next_term == $x); } // Driver Code isKeith(14) ? print('Yesn') : print('Non'); isKeith(12) ? print('Yesn') : print('Non'); isKeith(197) ? print('Yesn') : print('Non'); // This code is contributed by mits ?> 
JavaScript
<script> // Javascript program to check if a number // is Keith or not // Returns true if x is Keith // else false. function isKeith(x) {  // Store all digits of x in a vector 'terms'  // Also find number of digits and store in 'n'.  let terms = [];  let temp = x;  let n = 0; // n is number of digits in x  while (temp > 0)  {  terms.push(temp % 10);  temp = parseInt(temp / 10);  n++;  }  // To get digits in right order  // (from MSB to LSB)  terms= terms.reverse();  // Keep finding next terms of a sequence  // generated using digits of x until we  // either reach x or a number greater than x  let next_term = 0;  let i = n;  while (next_term < x)  {  next_term = 0;  // Next term is sum of previous n terms  for (let j = 1; j <= n; j++)  next_term += terms[i - j];  terms.push(next_term);  i++;  }  /* When the control comes out of the  while loop either the next_term is  equal to the number or greater than it.  If next_term is equal to x then x is a  Keith number else not */  return (next_term == x); } // Driver Code isKeith(14) ? document.write('Yes  
'
) : document.write('No
'
); isKeith(12) ? document.write('Yes
'
) : document.write('No
'
); isKeith(197) ? document.write('Yes
'
) : document.write('No
'
); // This code is contributed by _saurabh_jaiswal </script>

Ausgabe:  

Yes No Yes

Zeitkomplexität: O(n^2) wobei n die Anzahl der Ziffern ist

SQL-Auswahl aus mehreren Tabellen

Hilfsraum: An)