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Multiplica un nombre amb 10 sense utilitzar l'operador de multiplicació

Donat un nombre, la tasca és multiplicar-lo per 10 sense utilitzar l'operador de multiplicació?
Exemples: 
 

Input : n = 50 Output: 500 // multiplication of 50 with 10 is = 500 Input : n = 16 Output: 160 // multiplication of 16 with 10 is = 160


 

scan.next java


A solució senzilla perquè aquest problema és executar un bucle i afegir n amb si mateix 10 vegades. Aquí hem de fer 10 operacions. 



C++ 
// C++ program to multiply a number with 10 // without using multiplication operator #include   using namespace std; // Function to find multiplication of n with // 10 without using multiplication operator int multiplyTen(int n) {  int sum=0;  // Running a loop and add n with itself 10 times  for(int i=0;i<10;i++)  {  sum=sum+n;  }  return sum; } // Driver program to run the case int main() {  int n = 50;  cout << multiplyTen(n);  return 0; }
Java 
// Java program to multiply a number with 10 // without using multiplication operator import java.util.*; public class GFG {  // Function to find multiplication of n with  // 10 without using multiplication operator  public static int multiplyTen(int n) {  int sum = 0;  // Running a loop and add n with itself 10 times  for (int i = 0; i < 10; i++) {  sum = sum + n;  }  return sum;  }    // Driver program to run the case  public static void main(String[] args) {  int n = 50;  System.out.println(multiplyTen(n));  } } // This code is contributed by Prasad Kandekar(prasad264)
Python3 
# python program to multiply a number with 10 # without using multiplication operator # Function to find multiplication of n with # 10 without using multiplication operator def multiplyTen(n): sum = 0 # Running a loop and add n with itself 10 times for i in range(10): sum += n return sum # Driver code n = 50 print(multiplyTen(n)) # This code is contributed by Prasad Kandekar(prasad264)
C# 
// C# program to multiply a number with 10 // without using multiplication operator using System; public class GFG {    // Function to find multiplication of n with  // 10 without using multiplication operator  static int MultiplyTen(int n) {  int sum = 0;    // Running a loop and add n with itself 10 times  for (int i = 0; i < 10; i++) {  sum += n;  }  return sum;  }    // Driver program to run the case  static void Main(string[] args) {  int n = 50;  Console.WriteLine(MultiplyTen(n));  } } // This code is contributed by Prasad Kandekar(prasad264)
JavaScript 
// Javascript program to multiply a number with 10 // without using multiplication operator // Function to find multiplication of n with // 10 without using multiplication operator function multiplyTen(n) {  let sum = 0;  // Running a loop and add n with itself 10 times  for (let i = 0; i < 10; i++) {  sum += n;  }  return sum; } // Driver code let n = 50; console.log(multiplyTen(n)); // This code is contributed by Prasad Kandekar(prasad264)

Sortida 
500

Complexitat temporal: O(1)

Espai auxiliar: O(1)


A millor solució és utilitzar la manipulació de bits. Hem de multiplicar n per 10 és a dir; n*10 podem escriure això com n*(2+8) = n*2 + n*8 i com que no se'ns permet utilitzar l'operador de multiplicació, podem fer-ho mitjançant l'operador de bits de desplaçament a l'esquerra. Així n*10 = n<<1 + n<<3.
 

C++ 
// C++ program to multiply a number with 10 using // bitwise operators #include   using namespace std; // Function to find multiplication of n with // 10 without using multiplication operator int multiplyTen(int n) {  return (n<<1) + (n<<3); } // Driver program to run the case int main() {  int n = 50;  cout << multiplyTen(n);  return 0; }
Java 
// Java Code to Multiply a number with 10 // without using multiplication operator import java.util.*; class GFG {    // Function to find multiplication of n   // with 10 without using multiplication  // operator  public static int multiplyTen(int n)  {  return (n << 1) + (n << 3);  }    /* Driver program to test above function */  public static void main(String[] args)   {  int n = 50;  System.out.println(multiplyTen(n));    } }   // This code is contributed by Arnav Kr. Mandal.
Python 3 
# Python 3 program to multiply a  # number with 10 using bitwise # operators # Function to find multiplication # of n with 10 without using # multiplication operator def multiplyTen(n): return (n << 1) + (n << 3) # Driver program to run the case n = 50 print (multiplyTen(n)) # This code is contributed by  # Smitha
C# 
// C# Code to Multiply a number with 10 // without using multiplication operator using System; class GFG {    // Function to find multiplication of n   // with 10 without using multiplication  // operator  public static int multiplyTen(int n)  {  return (n << 1) + (n << 3);  }    // Driver Code  public static void Main()   {  int n = 50;  Console.Write(multiplyTen(n));    } }   // This code is contributed by Nitin Mittal.
PHP 
 // PHP program to multiply a  // number with 10 using // bitwise operators // Function to find multiplication  // of n with 10 without using  // multiplication operator function multiplyTen($n) { return ($n << 1) + ($n << 3); } // Driver Code $n = 50; echo multiplyTen($n); // This code is contributed by nitin mittal.  ?>
JavaScript 
<script> // JavaScript program to multiply a number with 10 using  // bitwise operators  // Function to find multiplication of n with  // 10 without using multiplication operator  function multiplyTen(n)  {   return (n<<1) + (n<<3);  }  // Driver program to run the case   let n = 50;   document.write(multiplyTen(n));    // This code is contributed by Surbhi Tyagi. </script>

Sortida:  

500

Complexitat temporal: O(1)

concatenar una cadena java

Espai auxiliar: O(1)