Find ceil of a/b without using ceil() function

14 Jun, 2022
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) = 2 

Input : a = 10, b = 2
Output : 5 
Explanation: a/b = ceil(10/2) = 5 
 
The problem can be solved using ceiling function, but the ceiling function does not work when integers are passed as parameters. Hence there are 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++
Java
Python3
C#
PHP
Javascript
// C++ program to find ceil(a/b)// without using ceil() function#include <cmath>#include <iostream>using namespace std; // Driver functionint 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() functionimport java.io.*; class GFG{    // Driver Code    public static void main(String args[])    {        // taking input 1        int a = 4;        int b = 3, val = 0;        if((a % b) != 0)            val = (a / b) +                  (a % b);        else            val = (a / b);        System.out.println("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);        System.out.println("The ceiling " +                       "value of 6/3 is " +                                      val);    }} // This code is contributed by// Manish Shaw(manishshaw1)
# Python3 program to find ceil(a/b)# without using ceil() functionimport math # Driver Code # taking input 1a = 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 2a = 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() functionusing 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
 
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++
Java
Python3
C#
PHP
Javascript
// C++ program to find ceil(a/b)// without using ceil() function#include <cmath>#include <iostream>using namespace std; // Driver functionint 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 Codepublic 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() functionimport math # Driver Code# taking input 1a = 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 2a = 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() functionusing 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)
jaideeppyne1997
vt_m
manishshaw1
Mithun Kumar
GauravRajput1
Rajput-Ji
singhh3010
number-theory
Mathematical
number-theory
Mathematical
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