Given an integer, write a function that calculates ?7n/8? (ceiling of 7n/8) without using division and multiplication operators.
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Method 1:
The idea is to first calculate floor of n/8, i.e., ?n/8? using right shift bitwise operator. The expression n>>3 produces the same.
If we subtract ?n/8? from n, we get ?7n/8?
Below is the implementation of above idea :
// C++ program to evaluate ceil(7n/8) // without using * and / #include <iostream> using namespace std;
int multiplyBySevenByEight( int n)
{ // Note the inner bracket here. This is needed
// because precedence of '-' operator is higher
// than '<<'
return (n - (n >> 3));
} // Driver code int main()
{ int n = 9;
cout << multiplyBySevenByEight(n);
return 0;
} // This code is contributed by khushboogoyal499 |
// C program to evaluate ceil(7n/8) without using * and / #include <stdio.h> int multiplyBySevenByEight(unsigned int n)
{ /* Note the inner bracket here. This is needed
because precedence of '-' operator is higher
than '<<' */
return (n - (n >> 3));
} /* Driver program to test above function */ int main()
{ unsigned int n = 9;
printf ( "%d" , multiplyBySevenByEight(n));
return 0;
} |
// Java program to evaluate ceil(7n/8) // without using * and import java.io.*;
class GFG {
static int multiplyBySevenByEight( int n)
{
/* Note the inner bracket here. This is needed
because precedence of '-' operator is higher
than '<<' */
return (n - (n >> 3 ));
}
// Driver code
public static void main(String args[])
{
int n = 9 ;
System.out.println(multiplyBySevenByEight(n));
}
} // This code is contributed by Anshika Goyal. |
# Python program to evaluate ceil(7n/8) without using * and / def multiplyBySevenByEight(n):
# Note the inner bracket here. This is needed
# because precedence of '-' operator is higher
# than '<<'
return (n - (n >> 3 ))
# Driver program to test above function */ n = 9
print (multiplyBySevenByEight(n))
# This code is contributed by # Smitha Dinesh Semwal |
// C# program to evaluate ceil(7n/8) // without using * and using System;
public class GFG {
static int multiplyBySevenByEight( int n)
{
/* Note the inner bracket here.
This is needed because precedence
of '-' operator is higher than
'<<' */
return (n - (n >> 3));
}
// Driver code
public static void Main()
{
int n = 9;
Console.WriteLine(multiplyBySevenByEight(n));
}
} // This code is contributed by Sam007. |
<script> // JavaScript program to evaluate ceil(7n/8) without using * and / function multiplyBySevenByEight(n)
{ /* Note the inner bracket here. This is needed
because precedence of '-' operator is higher
than '<<' */
return (n - (n>>3));
} /* Driver program to test above function */ let n = 9;
document.write(multiplyBySevenByEight(n));
// This code is contributed by Surbhi Tyagi. </script> |
<?php // PHP program to evaluate ceil // (7n/8) without using * and function multiplyBySevenByEight( $n )
{ // Note the inner bracket here.
// This is needed because
// precedence of '-' operator
// is higher than '<<'
return ( $n - ( $n >> 3));
} // Driver Code $n = 9;
echo multiplyBySevenByEight( $n );
// This code is contributed by Ajit ?> |
Output :
8
Time Complexity: O(1)
Auxiliary Space: O(1)
Method 2 (Always matches with 7*n/8):
The above method doesn’t always produce result same as “printf(“%u”, 7*n/8)”. For example, the value of expression 7*n/8 is 13 for n = 15, but above program produces 14. Below is modified version that always matches 7*n/8. The idea is to first multiply the number with 7, then divide by 8 as it happens in expression 7*n/8.
// C++ program to evaluate 7n/8 without using * and / #include<iostream> using namespace std;
int multiplyBySevenByEight(unsigned int n)
{ /* Step 1) First multiply number by 7 i.e. 7n = (n << 3) -n
* Step 2) Divide result by 8 */
return ((n << 3) -n) >> 3;
} /* Driver program to test above function */ int main()
{ unsigned int n = 15;
cout<< " " << multiplyBySevenByEight(n);
return 0;
} // This code is contributed by shivanisinghss2110 |
// C program to evaluate 7n/8 without using * and / #include<stdio.h> int multiplyBySevenByEight(unsigned int n)
{ /* Step 1) First multiply number by 7 i.e. 7n = (n << 3) -n
* Step 2) Divide result by 8 */
return ((n << 3) -n) >> 3;
} /* Driver program to test above function */ int main()
{ unsigned int n = 15;
printf ( "%u" , multiplyBySevenByEight(n));
return 0;
} |
// Java program to evaluate 7n/8 // without using * and / import java.io.*;
class GFG
{ static int multiplyBySevenByEight( int n)
{
// Step 1) First multiply number
// by 7 i.e. 7n = (n << 3) -n
// * Step 2) Divide result by 8
return ((n << 3 ) -n) >> 3 ;
}
// Driver program
public static void main(String args[])
{
int n = 15 ;
System.out.println(multiplyBySevenByEight(n));
}
} // This code is contributed by Anshika Goyal. |
# Python3 program to evaluate 7n/8 # without using * and / def multiplyBySevenByEight(n):
#Step 1) First multiply number
# by 7 i.e. 7n = (n << 3) -n
# Step 2) Divide result by 8
return ((n << 3 ) - n) >> 3 ;
# Driver code n = 15 ;
print (multiplyBySevenByEight(n));
#this code is contributed by sam007. |
// C# program to evaluate 7n/8 // without using * and / using System;
public class GFG {
static int multiplyBySevenByEight( int n)
{
// Step 1) First multiply number
// by 7 i.e. 7n = (n << 3) -n
// * Step 2) Divide result by 8
return ((n << 3) -n) >> 3;
}
// Driver program
public static void Main()
{
int n = 15;
Console.WriteLine(
multiplyBySevenByEight(n));
}
} // This code is contributed by Sam007. |
<script> // Javascript program to evaluate 7n/8 // without using * and / function multiplyBySevenByEight(n)
{ // Step 1) First multiply number
// by 7 i.e. 7n = (n << 3) -n
// * Step 2) Divide result by 8
return ((n << 3) -n) >> 3;
} // Driver code var n = 15;
document.write(multiplyBySevenByEight(n)); // This code is contributed by shikhasingrajput </script> |
<?php // PHP program to evaluate 7n/8 // without using * and / function multiplyBySevenByEight( $n )
{ /* Step 1) First multiply number by
7 i.e. 7n = (n << 3) - n
Step 2) Divide result by 8 */
return (( $n << 3) - $n ) >> 3;
} // Driver Code
$n = 15;
echo multiplyBySevenByEight( $n );
// This code is contributed by anuj_67. ?> |
Output :
13
Time Complexity: O(1)
Auxiliary Space: O(1)
Note : There is difference between outcomes of two methods. The method 1 produces ceil(7n/8), but method two produces integer value of 7n/8. For example, for n = 15, outcome of first method is 14, but for second method is 13.
Thanks to Narendra Kangralkar for suggesting this method.
Approach:
- Multiply the number n by 7 using the following steps.
- Left shift n by 3 positions (multiply n by 8).
- Subtract n from the result obtained in the previous step.
- To calculate the ceiling of 7n/8, add 7 to the result obtained in step 1.
- Right shift the result obtained in step 2 by 3 positions to divide it by 8.
Below is the implementation of this approach:
#include <iostream> using namespace std;
// Function to calculate ceil(7n/8) without using * and / int calculateCeiling( int n) {
int result = ((n << 3) - n + 7) >> 3;
return result;
} // Driver code int main() {
int n = 9;
cout << calculateCeiling(n) << endl;
return 0;
} |
// Java code of the above approach import java.util.*;
public class GFG {
// Function to calculate ceil(7n/8) without using * and /
static int calculateCeiling( int n)
{
int result = ((n << 3 ) - n + 7 ) >> 3 ;
return result;
}
public static void main(String[] args)
{
int n = 9 ;
System.out.println(calculateCeiling(n));
}
} // This code is contributed by Susobhan Akhuli |
# // Python code of the above approach # Function to calculate ceil(7n/8) without using * and / def calculateCeiling(n):
result = ((n << 3 ) - n + 7 ) >> 3
return result
# Driver code if __name__ = = "__main__" :
n = 9
print (calculateCeiling(n))
# This code is contributed by Susobhan Akhuli |
using System;
class Program
{ // Function to calculate ceil(7n/8) without using * and /
static int CalculateCeiling( int n)
{
// Shift n left by 3 bits (equivalent to multiplying by 8)
// Subtract n from the result
// Add 7 to the result
// Shift the result right by 3 bits (equivalent to dividing by 8)
int result = ((n << 3) - n + 7) >> 3;
return result;
}
// Driver code
static void Main( string [] args)
{
int n = 9;
Console.WriteLine(CalculateCeiling(n));
}
} |
// JavaScript code for the above approach // Function to calculate ceil(7n/8) without using * and / function calculateCeiling(n) {
// Calculate the result using bitwise operations
let result = ((n << 3) - n + 7) >> 3;
return result;
} let n = 9; document.write(calculateCeiling(n)); // This code is contributed by Susobhan Akhuli |
Output:
8
Time Complexity: O(1), as it only involves a few bitwise operations regardless of the input value n.
Space Complexity: O(1) since it does not require any additional space that grows with the input size.
This article is contributed by Rajeev.