numbers

I am trying to write a C code to generate all possible partitions (into 2 or more parts) with distinct elements of a given number. The sum of all the numbers of a given partition should be equal to the given number. For example, for input n = 6 , all possible partitions having 2 or more elements with distinct elements are: 1, 5 1, 2, 3 2, 4 I think a recursive approach should work, but I am unable to take care of the added constraint of distinct elements. A pseudo code or a sample code in C/C++/Java would be greatly appreciated. Thanks! Edit: If it makes things easier, I can ignore the

I've been messing around with a JavaScript console, when I suddenly decided to try this: 0x100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 == 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF Surprisingly, they're equal: Why does it happen? They're clearly different numbers (even the 0xFFFF...FFFF is one digit shorter) If I add a F to the 0xFFFF...FF , they're not equal anymore:

Is there such a thing? for eg $var = -5; echo thefunction($var); // should be 0 $var = 5; echo thefunction($var); // should be 5 Try max($var,0) , which will have the desired effect. See the manual page for more information. Not built-in but, here you have: function thefunction($var){ return ($var < 0 ? 0 : $var); } Hope this helps In PHP, checking if a integer is negative and if it is then setting it to zero is easy, but I was looking for something shorter (and potentially faster) than: if ($x < 0) $x = 0; Well, this is a very quick check and reset, but there is a function max that does this

This question already has an answer here: Way to get number of digits in an int? 28 answers I have do detect the amount of digits on a number. For example, 329586 has 6 digits. What I done, is simply parsing the number to string, and getting the string length, like: number.toString().length() But, is there a fastest way to count digits on a number? I have to use this method several times, so I think using toString() can impact performance. Thanks. Math.floor(Math.log10(number) + 1) // or just (int) Math.log10(number) + 1 For example: int number = 123456; int length = (int) Math.log10(number) +

I was solving a programming exercise and came across a problem over which I am not able to satisfactorily find a solution. The problem goes as follows: Print all unique integer partitions given an integer as input. Integer partition is a way of writing n as a sum of positive integers. for ex: Input=4 then output should be Output= 1 1 1 1 1 1 2 2 2 1 3 4 How should I think about solving this problem? I was wondering about using recursion. Can anyone provide me an algorithm for this question? Or a hint towards solution. any explanation for such kind of problems is welcome. (I am a beginner in

I am looking for the best way to convert a Number to a BigDecimal. Is this good enough? Number number; BigDecimal big = new BigDecimal(number.toString()); Can we lose precision with the toString() method ? This is fine, remember that using the constructor of BigDecimal to declare a value can be dangerous when it's not of type String. Consider the below... BigDecimal valDouble = new BigDecimal(0.35); System.out.println(valDouble); This will not print 0.35, it will infact be... 0.34999999999999997779553950749686919152736663818359375 I'd say your solution is probably the safest because of that.