How to round up integer division and have int resu

问题:

This question already has an answer here:

• How to round up the result of integer division? 15 answers

I just wrote a tiny method to count the number of pages for cell phone SMS. I didn't have the option to round up using Math.ceil, and honestly it seems to be very ugly.

Here is my code:

public class Main {

/**
 * @param args the command line arguments
 */
public static void main(String[] args) {
   String message = "today we stumbled upon a huge performance leak while optimizing a raycasting algorithm. Much to our surprise, the Math.floor() method took almost half of the calculation time: 3 floor operations took the same amount of time as one trilinear interpolation. Since we could not belive that the floor-method could produce such a enourmous overhead, we wrote a small test program that reproduce";

   System.out.printf("COunt is %d ",(int)messagePageCount(message));



}

public static double messagePageCount(String message){
    if(message.trim().isEmpty() || message.trim().length() == 0){
        return 0;
    } else{
        if(message.length() <= 160){
            return 1;
        } else {
            return Math.ceil((double)message.length()/153);
        }
    }
}

I don't really like this piece of code and I'm looking for a more elegant way of doing this. With this, I'm expecting 3 and not 3.0000000. Any ideas?

回答1:

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;
}

回答2:

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);

回答3:

Another one-liner that is not too complicated:

private int countNumberOfPages(int numberOfObjects, int pageSize) {
    return numberOfObjects / pageSize + (numberOfObjects % pageSize == 0 ? 0 : 1);
}

Could use long instead of int; just change the parameter types and return type.

回答4:

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.

回答5:

(message.length() + 152) / 153

This will give a "rounded up" integer.

回答6:

long numberOfPages = new BigDecimal(resultsSize).divide(new BigDecimal(pageSize), RoundingMode.UP).longValue();

回答7:

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);
    }
}

回答8:

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

回答9:

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