how to always round up to the next integer [duplicate]

This question already has answers here: How can I ensure that a division of integers is always rounded up? (10 answers) Closed 6 years ago.

i am trying to find total pages in building a pager on a website (so i want the result to开发者_JAVA技巧 be an integer. i get a list of records and i want to split into 10 per page (the page count)

when i do this:

list.Count() / 10

or

list.Count() / (decimal)10

and the list.Count() =12, i get a result of 1.

How would I code it so i get 2 in this case (the remainder should always add 1)

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Math.Ceiling((double)list.Count() / 10);

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(list.Count() + 9) / 10

Everything else here is either overkill or simply wrong (except for bestsss' answer, which is awesome). We do not want the overhead of a function call (Math.Truncate(), Math.Ceiling(), etc.) when simple math is enough.

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OP's question generalizes (pigeonhole principle) to:

How many boxes do I need to store x objects if only y objects fit into each box?

The solution:

1. derives from the realization that the last box might be partially empty, and
2. is (x + y - 1) ÷ y using integer division.

You'll recall from 3^rd grade math that integer division is what we're doing when we say 5 ÷ 2 = 2.

Floating-point division is when we say 5 ÷ 2 = 2.5, but we don't want that here.

Many programming languages support integer division. In languages derived from C, you get it automatically when you divide int types (short, int, long, etc.). The remainder/fractional part of any division operation is simply dropped, thus:

5 / 2 == 2

Replacing our original question with x = 5 and y = 2 we have:

How many boxes do I need to store 5 objects if only 2 objects fit into each box?

The answer should now be obvious: 3 boxes -- the first two boxes hold two objects each and the last box holds one.

(x + y - 1) ÷ y =
(5 + 2 - 1) ÷ 2 =
6 ÷ 2 =
3

So for the original question, x = list.Count(), y = 10, which gives the solution using no additional function calls:

(list.Count() + 9) / 10

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A proper benchmark or how the number may lie

Following the argument about Math.ceil(value/10d) and (value+9)/10 I ended up coding a proper non-dead code, non-interpret mode benchmark. I've been telling that writing micro benchmark is not an easy task. The code below illustrates this:

00:21:40.109 starting up....
00:21:40.140 doubleCeil: 19444599
00:21:40.140 integerCeil: 19444599
00:21:40.140 warming up...
00:21:44.375 warmup doubleCeil: 194445990000
00:21:44.625 warmup integerCeil: 194445990000
00:22:27.437 exec doubleCeil: 1944459900000, elapsed: 42.806s
00:22:29.796 exec integerCeil: 1944459900000, elapsed: 2.363s

The benchmark is in Java since I know well how Hotspot optimizes and ensures it's a fair result. With such results, no statistics, noise or anything can taint it.

Integer ceil is insanely much faster.

The code

package t1;

import java.math.BigDecimal;

import java.util.Random;

public class Div {
    static int[] vals;

    static long doubleCeil(){
        int[] v= vals;
        long sum = 0;
        for (int i=0;i<v.length;i++){
            int value = v[i];
            sum+=Math.ceil(value/10d);
        }
        return sum;
    }

    static long integerCeil(){      
        int[] v= vals;
        long sum = 0;
        for (int i=0;i<v.length;i++){
            int value = v[i];
            sum+=(value+9)/10;
        }
        return sum;     
    }

    public static void main(String[] args) {
        vals = new  int[7000];
        Random r= new Random(77);
        for (int i = 0; i < vals.length; i++) {
            vals[i] = r.nextInt(55555);
        }
        log("starting up....");

        log("doubleCeil: %d", doubleCeil());
        log("integerCeil: %d", integerCeil());
        log("warming up...");       

        final int warmupCount = (int) 1e4;
        log("warmup doubleCeil: %d", execDoubleCeil(warmupCount));
        log("warmup integerCeil: %d", execIntegerCeil(warmupCount));

        final int execCount = (int) 1e5;

        {       
        long time = System.nanoTime();
        long s = execDoubleCeil(execCount);
        long elapsed = System.nanoTime() - time;
        log("exec doubleCeil: %d, elapsed: %.3fs",  s, BigDecimal.valueOf(elapsed, 9));
        }

        {
        long time = System.nanoTime();
        long s = execIntegerCeil(execCount);
        long elapsed = System.nanoTime() - time;
        log("exec integerCeil: %d, elapsed: %.3fs",  s, BigDecimal.valueOf(elapsed, 9));            
        }
    }

    static long execDoubleCeil(int count){
        long sum = 0;
        for(int i=0;i<count;i++){
            sum+=doubleCeil();
        }
        return sum;
    }


    static long execIntegerCeil(int count){
        long sum = 0;
        for(int i=0;i<count;i++){
            sum+=integerCeil();
        }
        return sum;
    }

    static void log(String msg, Object... params){
        String s = params.length>0?String.format(msg, params):msg;
        System.out.printf("%tH:%<tM:%<tS.%<tL %s%n", new Long(System.currentTimeMillis()), s);
    }   
}

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This will also work:

c = (count - 1) / 10 + 1;

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I think the easiest way is to divide two integers and increase by one :

int r = list.Count() / 10;
r += (list.Count() % 10 == 0 ? 0 : 1);

No need of libraries or functions.

edited with the right code.

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You can use Math.Ceiling

http://msdn.microsoft.com/en-us/library/system.math.ceiling%28v=VS.100%29.aspx

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Xform to double (and back) for a simple ceil?

list.Count()/10 + (list.Count()%10 >0?1:0) - this bad, div + mod

edit 1st: on a 2n thought that's probably faster (depends on the optimization): div * mul (mul is faster than div and mod)

int c=list.Count()/10;
if (c*10<list.Count()) c++;

edit2 scarpe all. forgot the most natural (adding 9 ensures rounding up for integers)

(list.Count()+9)/10

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Check by using mod - if there is a remainder, simply increment the value by one.