ceil conterpart for Math.floorDiv in Java?

问题

Is there any ceil counterpart for Math.floorDiv()

How to calculate it fastest way with what we have?

UPDATE

The code for floorDiv() is 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;
    }

Can we code ceil the similar way?

UPDATE 2

I saw this answer https://stackoverflow.com/a/7446742/258483 but it seems to have too many unnecessary operations.

回答1:

There is none in the Math class, but you can easily calculate it

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

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

回答2:

I'd also just use the negation of floorMod, but if you are going to define your own function, you could simply adapt the above code:

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;
}

回答3:

You can make use of the floorDiv function and fiddle with that:

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

来源：https://stackoverflow.com/questions/27643616/ceil-conterpart-for-math-floordiv-in-java