How to round up integer division and have int result in Java? [duplicate]

Answer 1

this might be helpfull,, Subtract the remainder to the legnth and make it a divisible number and then divide it with 153

int r=message.length()%153;       //Calculate the remainder by %153
return (message.length()-r)/153;  // find the pages by adding the remainder and 
                                  //then divide by 153 

Answer 2

Google's Guava library handles this in the IntMath class:

IntMath.divide(numerator, divisor, RoundingMode.CEILING);

Unlike many answers here, it handles negative numbers. It also throws an appropriate exception when attempting to divide by zero.

Answer 3

Expanding on Peter's solution, this is what I've found works for me to always round 'towards positive infinity':

public static long divideAndRoundUp(long num, long divisor) {
    if (num == 0 || divisor == 0) { return 0; }

    int sign = (num > 0 ? 1 : -1) * (divisor > 0 ? 1 : -1);

    if (sign > 0) {
        return (num + divisor - 1) / divisor;
    }
    else {
        return (num / divisor);
    }
}

Answer 4

To round up an integer division you can use

import static java.lang.Math.abs;

public static long roundUp(long num, long divisor) {
    int sign = (num > 0 ? 1 : -1) * (divisor > 0 ? 1 : -1);
    return sign * (abs(num) + abs(divisor) - 1) / abs(divisor);
}

or if both numbers are positive

public static long roundUp(long num, long divisor) {
    return (num + divisor - 1) / divisor;
}

Answer 5

Use Math.ceil() and cast the result to int:

• This is still faster than to avoid doubles by using abs().
• The result is correct when working with negatives, because -0.999 will be rounded UP to 0

Example:

(int) Math.ceil((double)divident / divisor);

Answer 6

If you want to calculate a divided by b rounded up you can use (a+(-a%b))/b