Ceil conterpart for Math.floorDiv in Java?

Question

Is there any ceil for Math.floorDiv()

How to calculate it in the fastest way with what we have?

UPDATE

The code for floorDiv() follows:

  public static long floorDiv(long x, long y) { long r = x / y; // if the signs are different and modulo not zero, round down if ((x ^ y) < 0 && (r * y != x)) { r--; } return r; } 

Is it possible to encode ceil same way?

UPDATE 2

I have seen this answer https://stackoverflow.com/a/166268/2126 , but there seem to be too many unnecessary operations.

Answer 1

There are none in the Math class, but you can easily compute it

 long ceilDiv(long x, long y){ return -Math.floorDiv(-x,y); } 

For example, ceilDiv(1,2) = -floorDiv(-1,2) = -(-1) = 1 (the correct answer).

Answer 2

I would also use floorMod negation, but if you are going to define your own function, you can simply adapt the code above:

 public static int ceilDiv(int x, int y) { int r = x / y; // if the signs are the same and modulo not zero, round up if ((x ^ y) >= 0 && (r * y != x)) r++; return r; } 

Answer 3

You can use the floorDiv function and play with it:

 int ceilDiv(int x, int y) { return Math.floorDiv(x, y) + (x % y == 0 ? 0 : 1) }