Givet et tal er opgaven at kontrollere, om et tal er deleligt med 16 eller ej. Indtastningsnummeret kan være stort, og det er muligvis ikke muligt at gemme, selvom vi bruger long long int.
Eksempler:
Input : n = 1128 Output : No Input : n = 11216 Output : Yes Input : n = 1124273542764284287 Output : No
Da inputtal kan være meget stort, kan vi ikke bruge n % 16 til at kontrollere, om et tal er deleligt med 16 eller ej, især i sprog som C/C++. Ideen er baseret på følgende fakta.
for hver java
A number is divisible by 16 if number formed by last four digits of it is divisible by 16.
Illustration:
For example let us consider 769616 Number formed by last four digits = 9616 Since 9522 is divisible by 16 answer is YES.
Hvordan virker dette?
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 using namespace std; // Function to find that // number divisible by 16 or not 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 int main() { string str = '769528'; check(str)? cout << 'Yes' : cout << 'No '; return 0; }
Java // 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.
Python3 # 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# // 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 // 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. ?> JavaScript <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>
Produktion:
No
Tidskompleksitet: O(1)
Hjælpeplads: O(1)
En anden tilgang (ved at bruge AND bitwise operator):
For at kontrollere, om et stort tal er deleligt med 16 eller ej uden at bruge modulo-operatoren, kan vi tjekke de sidste 4 bit af tallet. Hvis disse bits alle er 0'er, er tallet deleligt med 16 ellers er det ikke.
Dette skyldes, at 16 er repræsenteret i binært som 0b10000, hvilket betyder, at den har en 1 i den 5. bit position og alle 0'er i de nederste 4 bit. Derfor, hvis et tal er deleligt med 16, skal det have alle 0'er i de nederste 4 bit.
Nedenfor er implementeringen af ovenstående tilgang:
stdin i cC++
#include using namespace std; // Function to check if a number is divisible by 16 bool is_divisible_by_16(int num) { int 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 } int main() { int num = 769528; if (is_divisible_by_16(num)) { cout << 'Yes' << endl; } else { cout << 'No' << endl; } return 0; }
Java 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'); } } }
Python3 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')
C# 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'); } } }
JavaScript 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'); }
Produktion
No
Tidskompleksitet: O(1)
Hjælpeplads: O(1)
I denne kode bruger vi den bitvise AND-operator & med det binære tal 0b1111 (som har alle 1'er i de nederste 4 bits og 0'er i de øverste bits) til at udtrække de sidste 4 bit af inputnummeret num. Så kontrollerer vi, om disse 4 bit alle er 0'er eller ej. Hvis de alle er 0'er, returnerer funktionen Sand (hvilket betyder, at tallet er deleligt med 16), ellers returnerer den Falsk.