Given a and b, find the ceiling value of a/b without using ceiling function.
Examples:
Input: a = 5, b = 4
Output: 2
Explanation: a/b = ceil(5/4) = 2Input: a = 10, b = 2
Output: 5
Explanation: a/b = ceil(10/2) = 5
The problem can be solved using the ceiling function, but the ceiling function does not work when integers are passed as parameters. Hence there are the following 2 approaches below to find the ceiling value.
Approach 1:
ceilVal = (a / b) + ((a % b) != 0)
a/b returns the integer division value, and ((a % b) != 0) is a checking condition which returns 1 if we have any remainder left after the division of a/b, else it returns 0. The integer division value is added with the checking value to get the ceiling value.
Given below is the illustration of the above approach:
// C++ program to find ceil(a/b) // without using ceil() function #include <cmath> #include <iostream> using namespace std;
// Driver function int main()
{ // taking input 1
int a = 4;
int b = 3;
int val = (a / b) + ((a % b) != 0);
cout << "The ceiling value of 4/3 is "
<< val << endl;
// example of perfect division
// taking input 2
a = 6;
b = 3;
val = (a / b) + ((a % b) != 0);
cout << "The ceiling value of 6/3 is "
<< val << endl;
return 0;
} |
// Java program to find // ceil(a/b) without // using ceil() function class GFG
{ // Driver Code
public static void main(String args[])
{
// taking input 1
System.out.println( "The ceiling " +
"value of 4/3 is " +
intCeil( 4 , 3 ));
System.out.println( "The ceiling " +
"value of 5/3 is " +
intCeil( 5 , 3 ));
// example of perfect
// division taking input 2
System.out.println( "The ceiling " +
"value of 6/3 is " +
intCeil( 6 , 3 ));
}
public static int intCeil( int a, int b) {
if (a % b != 0 ) {
return (a / b) + 1 ;
} else {
return (a / b);
}
}
} // This code is contributed by // Manish Shaw(manishshaw1) // and Derek Chen-Becker |
# Python3 program to find ceil(a/b) # without using ceil() function import math
# Driver Code # taking input 1 a = 4 ;
b = 3 ;
val = (a / b) + ((a % b) ! = 0 );
print ( "The ceiling value of 4/3 is" ,
math.floor(val));
# example of perfect division # taking input 2 a = 6 ;
b = 3 ;
val = int ((a / b) + ((a % b) ! = 0 ));
print ( "The ceiling value of 6/3 is" , val);
# This code is contributed by mits |
// C# program to find ceil(a/b) // without using ceil() function using System;
class GFG
{ // Driver Code
static void Main()
{
// taking input 1
int a = 4;
int b = 3, val = 0;
if ((a % b) != 0)
val = (a / b) + (a % b);
else
val = (a / b);
Console.WriteLine( "The ceiling " +
"value of 4/3 is " +
val);
// example of perfect
// division taking input 2
a = 6;
b = 3;
if ((a % b) != 0)
val = (a / b) + (a % b);
else
val = (a / b);
Console.WriteLine( "The ceiling " +
"value of 6/3 is " +
val);
}
} // This code is contributed by // Manish Shaw(manishshaw1) |
<?php // PHP program to find ceil(a/b) // without using ceil() function // Driver function // taking input 1
$a = 4;
$b = 3;
$val = ( $a / $b ) + (( $a % $b ) != 0);
echo "The ceiling value of 4/3 is "
, floor ( $val ) , "\n" ;
// example of perfect division
// taking input 2
$a = 6;
$b = 3;
$val = ( $a / $b ) + (( $a % $b ) != 0);
echo "The ceiling value of 6/3 is "
, $val ;
// This code is contributed by anuj_67. ?> |
<script> // javascript program to find // ceil(a/b) without // using ceil() function // Driver Code
// taking input 1
var a = 4;
var b = 3, val = 0;
if ((a % b) != 0)
val = parseInt(a / b) + (a % b);
else
val = parseInt(a / b);
document.write( "The ceiling " + "value of 4/3 is " + val+ "<br/>" );
// example of perfect
// division taking input 2
a = 6;
b = 3;
if ((a % b) != 0)
val = parseInt(a / b) + (a % b);
else
val = parseInt(a / b);
document.write( "The ceiling " + "value of 6/3 is " + val);
// This code is contributed by gauravrajput1 </script> |
The ceiling value of 4/3 is 2 The ceiling value of 6/3 is 2
Time Complexity: O(1)
Auxiliary Space: O(1)
Approach 2:
ceilVal = (a+b-1) / b
Using simple maths, we can add the denominator to the numerator and subtract 1 from it and then divide it by denominator to get the ceiling value.
Given below is the illustration of the above approach:
// C++ program to find ceil(a/b) // without using ceil() function #include <cmath> #include <iostream> using namespace std;
// Driver function int main()
{ // taking input 1
int a = 4;
int b = 3;
int val = (a + b - 1) / b;
cout << "The ceiling value of 4/3 is "
<< val << endl;
// example of perfect division
// taking input 2
a = 6;
b = 3;
val = (a + b - 1) / b;
cout << "The ceiling value of 6/3 is "
<< val << endl;
return 0;
} |
// Java program to find ceil(a/b) // without using ceil() function class GFG {
// Driver Code public static void main(String args[])
{ // taking input 1
int a = 4 ;
int b = 3 ;
int val = (a + b - 1 ) / b;
System.out.println( "The ceiling value of 4/3 is "
+ val);
// example of perfect division
// taking input 2
a = 6 ;
b = 3 ;
val = (a + b - 1 ) / b;
System.out.println( "The ceiling value of 6/3 is "
+ val );
} } // This code is contributed by Jaideep Pyne |
# Python3 program to find # math.ceil(a/b) without # using math.ceil() function import math
# Driver Code # taking input 1 a = 4 ;
b = 3 ;
val = (a + b - 1 ) / b;
print ( "The ceiling value of 4/3 is " ,
math.floor(val));
# example of perfect division # taking input 2 a = 6 ;
b = 3 ;
val = (a + b - 1 ) / b;
print ( "The ceiling value of 6/3 is " ,
math.floor(val));
# This code is contributed by mits |
// C# program to find ceil(a/b) // without using ceil() function using System;
class GFG {
// Driver Code
public static void Main()
{
// taking input 1
int a = 4;
int b = 3;
int val = (a + b - 1) / b;
Console.WriteLine( "The ceiling"
+ " value of 4/3 is " + val);
// example of perfect division
// taking input 2
a = 6;
b = 3;
val = (a + b - 1) / b;
Console.WriteLine( "The ceiling"
+ " value of 6/3 is " + val );
}
} // This code is contributed by anuj_67. |
<?php // PHP program to find ceil(a/b) // without using ceil() function // Driver function
// taking input 1
$a = 4;
$b = 3;
$val = ( $a + $b - 1) / $b ;
echo "The ceiling value of 4/3 is "
, floor ( $val ) , "\n" ;
// example of perfect division
// taking input 2
$a = 6;
$b = 3;
$val = ( $a + $b - 1) / $b ;
echo "The ceiling value of 6/3 is "
, floor ( $val ) ;
// This code is contributed by anuj_67. ?> |
<script> // javascript program to find ceil(a/b) // without using ceil() function // Driver Code
// taking input 1
var a = 4;
var b = 3;
var val = (a + b - 1) / b;
document.write( "The ceiling value of 4/3 is " + val+ "<br/>" );
// example of perfect division
// taking input 2
a = 6;
b = 3;
val = parseInt((a + b - 1) / b);
document.write( "The ceiling value of 6/3 is " + val);
// This code contributed by Rajput-Ji </script> |
The ceiling value of 4/3 is 2 The ceiling value of 6/3 is 2
Time Complexity: O(1)
Auxiliary Space: O(1)