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

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program to find ceil(a/b) ` `// without using ceil() function ` `#include ` `#include ` `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

 `// Java program to find  ` `// ceil(a/b) without  ` `// using ceil() function ` `import` `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

 `# 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#

 `// 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

 ` `

Output:

```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++

 `// C++ program to find ceil(a/b) ` `// without using ceil() function ` `#include ` `#include ` `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

 `// 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

 `# 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#

 `// 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

 ` `

Output:

```The ceiling value of 4/3 is 2
The ceiling value of 6/3 is 2
```

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