Duotas skaičius, užduotis yra patikrinti, ar skaičius dalijasi iš 16, ar ne. Įvesties skaičius gali būti didelis ir jo gali būti neįmanoma išsaugoti, net jei naudosime long long int.
Pavyzdžiai:
Input : n = 1128 Output : No Input : n = 11216 Output : Yes Input : n = 1124273542764284287 Output : No
Kadangi įvestas skaičius gali būti labai didelis, negalime naudoti n % 16, kad patikrintume, ar skaičius dalijasi iš 16, ypač tokiomis kalbomis kaip C/C++. Idėja paremta sekančiu faktu.
nuskaityti.nextstring java
A number is divisible by 16 if number formed by last four digits of it is divisible by 16.
Iliustracija:
For example let us consider 769616 Number formed by last four digits = 9616 Since 9522 is divisible by 16 answer is YES.
Kaip tai veikia?
Let us consider 76952 we can write it as 76942 = 7*10000 + 6*1000 + 9*100 + 5*10 + 2 The proof is based on below observation: Remainder of 10i divided by 16 is 0 if i greater than or equal to four. Note that 10000 100000... etc lead to remainder 0 when divided by 16. So remainder of '7*10000 + 6*1000 + 9*100 + 5*10 + 2' divided by 16 is equivalent to remainder of following : 0 + 6*1000 + 9*100 + 5*10 + 2 = 6952 Therefore we can say that the whole number is divisible by 16 if 6952 is divisible by 16.C++
// C++ program to find if a number // is divisible by 16 or not #include
// Java program to find if a number // is divisible by 16 or not import java.io.*; class GFG { // Function to find that // number divisible by 16 or not static boolean check(String str) { int n = str.length(); // Empty string if (n == 0 && n == 1) return false; // If there is double digit if (n == 2) return (((str.charAt(n-2)-'0')*10 + (str.charAt(n-1)-'0'))%16 == 0); // If there is triple digit if(n == 3) return ( ((str.charAt(n-3)-'0')*100 + (str.charAt(n-2)-'0')*10 + (str.charAt(n-1)-'0'))%16 == 0); // If number formed by last // four digits is divisible by 16. int last = str.charAt(n-1) - '0'; int second_last = str.charAt(n-2) - '0'; int third_last = str.charAt(n-3) - '0'; int fourth_last = str.charAt(n-4) - '0'; return ((fourth_last*1000 + third_last*100 + second_last*10 + last) % 16 == 0); } // Driver code public static void main(String args[]) { String str = '769528'; if(check(str)) System.out.println('Yes'); else System.out.println('No '); } } // This code is contributed by Nikita Tiwari.
# Python 3 program to find # if a number is divisible # by 16 or not # Function to find that # number divisible by # 16 or not def check(st) : n = len(st) # Empty string if (n == 0 and n == 1) : return False # If there is double digit if (n == 2) : return ((int)(st[n-2])*10 + ((int)(st[n-1])%16 == 0)) # If there is triple digit if(n == 3) : return ( ((int)(st[n-3])*100 + (int)(st[n-2])*10 + (int)(st[n-1]))%16 == 0) # If number formed by last # four digits is divisible # by 16. last = (int)(st[n-1]) second_last = (int)(st[n-2]) third_last = (int)(st[n-3]) fourth_last = (int)(st[n-4]) return ((fourth_last*1000 + third_last*100 + second_last*10 + last) % 16 == 0) # Driver code st = '769528' if(check(st)) : print('Yes') else : print('No') # This code is contributed by Nikita Tiwari.
// C# program to find if a number // is divisible by 16 or not using System; class GFG { // Function to find that number // divisible by 16 or not static bool check(String str) { int n = str.Length; // Empty string if (n == 0 && n == 1) return false; // If there is double digit if (n == 2) return (((str[n - 2] - '0') * 10 + (str[n - 1] - '0')) % 16 == 0); // If there is triple digit if(n == 3) return (((str[n - 3] - '0') * 100 + (str[n - 2] - '0') * 10 + (str[n - 1] - '0')) % 16 == 0); // If number formed by last // four digits is divisible by 16. int last = str[n - 1] - '0'; int second_last = str[n - 2] - '0'; int third_last = str[n - 3] - '0'; int fourth_last = str[n - 4] - '0'; return ((fourth_last * 1000 + third_last * 100 + second_last * 10 + last) % 16 == 0); } // Driver code public static void Main() { String str = '769528'; if(check(str)) Console.Write('Yes'); else Console.Write('No '); } } // This code is contributed by Nitin Mittal.
// PHP program to find if a number // is divisible by 16 or not // Function to find that // number divisible by 16 or not function check($str) { $n = strlen($str); // Empty string if ($n == 0 && $n == 1) return false; // If there is double digit if ($n == 2) return ((($str[$n - 2] - '0') * 10 + ($str[$n - 1] - '0')) % 16 == 0); // If there is triple digit if($n == 3) return ((($str[$n -3] - '0') * 100 + ($str[$n - 2] - '0') * 10 + ($str[$n - 1] - '0')) % 16 == 0); // If number formed by last four // digits is divisible by 16. $last = $str[$n - 1] - '0'; $second_last = $str[$n - 2] - '0'; $third_last = $str[$n - 3] - '0'; $fourth_last = $str[$n - 4] - '0'; return (($fourth_last * 1000 + $third_last * 100 + $second_last * 10 + $last) % 16 == 0); } // Driver code $str = '769528'; $x = check($str) ? 'Yes' : 'No '; echo($x); // This code is contributed by Ajit. ?> <script> // Javascript program to find if a number // is divisible by 16 or not // Function to find that number // divisible by 16 or not function check(str) { let n = str.length; // Empty string if (n == 0 && n == 1) return false; // If there is double digit if (n == 2) return (((str[n - 2] - '0') * 10 + (str[n - 1] - '0')) % 16 == 0); // If there is triple digit if(n == 3) return (((str[n - 3] - '0') * 100 + (str[n - 2] - '0') * 10 + (str[n - 1] - '0')) % 16 == 0); // If number formed by last // four digits is divisible by 16. let last = str[n - 1] - '0'; let second_last = str[n - 2] - '0'; let third_last = str[n - 3] - '0'; let fourth_last = str[n - 4] - '0'; return ((fourth_last * 1000 + third_last * 100 + second_last * 10 + last) % 16 == 0); } // Driver code let str = '769528'; if (check(str)) document.write('Yes'); else document.write('No '); // This code is contributed by decode2207 </script>
Išvestis:
No
Laiko sudėtingumas: O(1)
Pagalbinė erdvė: O(1)
Kitas metodas (naudojant AND bitų operatorių):
Norėdami patikrinti, ar didelis skaičius dalijasi iš 16, ar ne, nenaudodami modulo operatoriaus, galime patikrinti paskutinius 4 skaičiaus bitus. Jei visi šie bitai yra 0, skaičius dalijasi iš 16, kitaip jis nėra.
java prioriteto eilė
Taip yra todėl, kad 16 dvejetainiu būdu vaizduojamas kaip 0b10000, o tai reiškia, kad 5 bitų pozicijoje yra 1, o apatiniuose 4 bituose - visi 0. Todėl, jei skaičius dalijasi iš 16, jis turi turėti visus 0 apatiniuose 4 bituose.
Žemiau pateikiamas aukščiau aprašyto metodo įgyvendinimas:
C++#include
import java.io.*; public class Gfg { // Function to check if a number is divisible by 16 static boolean is_divisible_by_16(int num) { int lastFourBits = num & 0b1111; // bitwise AND with 0b1111 to get the last 4 bits return lastFourBits == 0; // check if all 4 bits are 0's } public static void main(String[] args) { int num = 769528; if (is_divisible_by_16(num)) { System.out.println('Yes'); } else { System.out.println('No'); } } }
def is_divisible_by_16(num): last_four_bits = num & 0b1111 # bitwise AND with 0b1111 to get the last 4 bits return last_four_bits == 0 # check if all 4 bits are 0's num = 769528 if(is_divisible_by_16(num)): print('Yes') else: print('No')
using System; class MainClass { static bool IsDivisibleBy16(int num) { int lastFourBits = num & 0b1111; // bitwise AND with 0b1111 to get the last 4 bits return lastFourBits == 0; // check if all 4 bits are 0's } public static void Main (string[] args) { int num = 769528; if (IsDivisibleBy16(num)) { Console.WriteLine('Yes'); } else { Console.WriteLine('No'); } } }
function is_divisible_by_16(num) { let last_four_bits = num & 0b1111; // bitwise AND with 0b1111 to get the last 4 bits return last_four_bits === 0; // check if all 4 bits are 0's } let num = 769528; if (is_divisible_by_16(num)) { console.log('Yes'); } else { console.log('No'); }
Išvestis
No
Laiko sudėtingumas: O(1)
Pagalbinė erdvė: O(1)
Šiame kode mes naudojame bitų IR operatorių & su dvejetainiu skaičiumi 0b1111 (kurio visi 1 yra apatiniuose 4 bituose ir 0 viršutiniuose bituose), kad išskirtume paskutinius 4 įvesties numerio bitus. Tada patikriname, ar visi šie 4 bitai yra 0, ar ne. Jei visi jie yra 0, funkcija grąžina True (tai reiškia, kad skaičius dalijasi iš 16), kitu atveju ji grąžina False.