There is no clear convention for those cases. Remember the formula for modulo: n = am + b. Usually, it is required that the remainder b is within the interval [0..(m-1)]. This makes it very easy for all natural numbers. For negativ numbers, some conventions want the remainder to be...

I solved my problem as follows: I installed phantomjs locally and run the test script available at http://www.meteorpedia.com/read/spiderable/ phantomjs phantomtest.js This gave me more details about the error: Parse Error. Then, it was a javascript file that once compiled/minified, rendered an error caused by select2. The js library that was...

% is the integer remainder operator. For example: 21 % 7 == 0 22 % 7 == 1 25 % 7 == 4 27 % 7 == 6 28 % 7 == 0 x % y != 0 is true if the integer division yields a non-zero remainder, false if...

javascript,if-statement,do-while,modulus

Looks like a simple misunderstanding of && vs || on this line: while (firstNumber < secondNumber && ((firstNumber / secondNumber) % 0.5) != 0); The loop will only continue if both statements evaluate to true. Let's take your 8/7 example: while (8 < 7 && ((8 / 7) % 0.5)...

c++,long-integer,modulus,integer-overflow

Many compilers offer a 128-bit integral type. For example, with g++ you can make a function static inline int64_t mulmod(int64_t x, int64_t y, int64_t m) { return ( (__int128_t)x * y) % m; } Aside: if you can, try to stick to unsigned integer types when you're doing modular arithmetic....

It's giving you the reminder of the division what you need is fmod, fmod — Returns the floating point remainder (modulo) of the division of the arguments echo fmod(877.5, 1); // 0.5 ...

excel,integer,excel-formula,worksheet-function,modulus

In Row3 please try: =INT(ROW()/3)&"."&MOD(ROW(),3)+1 and copy down to suit. For other rows you would need to add an offset (+n)....

The formula you provided will work. Generally speaking, for Integers >= 0 this will always be true A % B = A - [A/B] * B, where [x] denotes greatest integer <= x ...

I'm not quite sure about your algorithm, but it seems like you try to get every 3rd and 4th post not (starting at 0). The fitting code would be: if(($i % 4 == 0 || $i % 4 == 1) && $i != 0) { /* do stuff */ }...

According to the VB6/VBA documentation The modulus, or remainder, operator divides number1 by number2 (rounding floating-point numbers to integers) and returns only the remainder as result. For example, in the following expression, A (result) equals 5. A = 19 Mod 6.7 Usually, the data type of result is a Byte,...

c,signed,modulus,ida,and-operator

The optimization of x % N to x & (N-1) only works if N is a power of two. You also need to know that x is positive, otherwise there is a little difference between the bitmask operation and the remainder operation. The bitmask operation produces the remainder for the...

javascript,jquery,for-loop,wrap,modulus

Here is the answer I think you are looking for. You need to determine when the last grouping is being made. then use the modo value to only wrap that many sections. Updated fiddle var divs = $(".someclass"); var limit = 10;//divs.length; var grouper = 3 var modo = limit...

php,arrays,math,foreach,modulus

Using the modulus operator (which determines the remainder of the division) and then calculating how many times $s fits in your $c is the quickest approach I could muster. Should look something like this: foreach ($val as $c) { $mod = $c % $s; $times = ($c-$mod)/$s; } $times should...

java,integer,printf,java.util.scanner,modulus

First, in order to avoid printing on separate lines, you should avoid using the %n formatting character in your printf(). Now, how do you print the digits in the correct order? Well, since you are limited to five-digit numbers, you can do something like this: for ( int divisor =...

I believe your problem is that the return is in the wrong place. Your code currently only loops through the 2, and any odd number is obviously not divisible by 2. So, it goes into the if n%j==0, returns True, and since a return breaks out of the loop, stops...

algorithm,for-loop,bit-manipulation,modulus

well I do not see any point in your code ... but there are many ways how to create repetitive series you need here few examples example1: int tab[4]={0,1,0,2}; for (int i=0;i<8;i=(i+1)&3) output(tab[i]); use a table example2: for (int i=0;i<8;i+=4) { output(0); output(1); output(0); output(2); } direct output ...

It shouldn't matter much, although being a multiple of N (or close to one) will help around the edges, especially for smaller M. The more important thing is that your random numbers are evenly distributed between 0 and M.

12345, Here 5 comes first time. Then you are passing actually 4-1, 3-1, 2-1, 1-1 Use return temp + sumDigits(x) instead of return temp + sumDigits(x - 1) You should also change here: def sumDigits(x): if x <= 0: return 0 Otherwise, it will add extra one...

It's impossible to reverse this multiplication. First, let's take the two's complement of the first number. 3904635256 = 2147483648 - -1757151608 Next, we'll multiply the second number 369018468323820520 = 3904635256 * 94507795 Now, here's the tricky part. We convert the product to hex, and drop the digits larger than a...

I misunderstood your question because I focused on the assumption you had that the output should be binary [0 or 1], which is wrong. To reproduce the output of the dyadic transformation as in the link you provided, your code works fine (for 1 value), and you can use this...

Since 36 decimal digits is too much for a typical long long, you need to perform your modulus operation during the conversion: int a[36]={1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9}; long long int temp=0; for(int i=35 ; i >= 0 ; i--) { temp = 10*temp + a[i]; temp %= 1000000007; } printf("%lld",temp); I made two...

Mod() operates on integer values. If the value you're passing isn't an integer value, VBScript will convert it to one by rounding it to the nearest integer. That's what's happening with your value of 1705. After three iterations, you're left with 1.705, which Mod() rounds to 2. So now the...

DBL_EPSILON is the smallest quantity that can be added to 1.0. If your number is larger than 1.0, the epsilon for that number will be proportionately larger as well. Simply multiply DBL_EPSILON by the number to get a threshold. double a = remainder(92.66, 0.01); if (a <= DBL_EPSILON * 92.66)...

MySQL isn't giving you a bogus result, it's simply using a different implementation of modulus than you are expecting. Unfortunately the term modulus seems to have been defined somewhat ambiguously, and it's implementation varies from language to language. From what I can tell in the Wikipedia on Modulo, MySQL's implementation...

@Kevin is right I just need to add: Linear interpolation for range change so if you have number x on interval <x0,x1> and want to change it to y on interval <y0,y1> then use this formula: y=y0+((x-x0)*(y1-y0)/(x1-x0)); it is the formula for 2D line and also base for DDA algorithms...

The reason is that Matlab uses double floating-point arithmetic by default. A number as large as 688^79 can't be represented accurately as a double. (The largest integer than can be accurately represented as a double is of the order of 2^53). To obtain the right result you can use symbolic...

python,for-loop,enumeration,modulus

Since Python 2.6, enumerate() takes an optional start parameter to indicate where to start the enumeration. See the documentation for enumerate.

Let's look at a small example: mb = (k*b + k*(k-1)*a*a)%MOD; Here, k*b, k*(k-1)*a*a can overflow, so can the sum, taking into account (x + y) mod m = (x mod m + y mod m) mod m we can rewrite this (x= k*b, y=k*(k-1)*a*a and m=MOD) mb = ((k*b)...

modulus,number-theory,modular-arithmetic

There two important theories that would help you to solve this problem :- Modular Exponentiation Fermat's little theorem Modular Exponentiation :- This simple recursive formula to make you understand :- (a^n)%p = ((a^(n-1))%p*a)%p The above formula can help you prevent the overflow which occurs if a^n is large. Moreover fast...

python,lambda,primes,modulus,sieve-of-eratosthenes

It looks like a compact (but somewhat obscure) implementation of the Sieve of Eratosthenes [EDIT: as pointed out in the comments, this is in fact an "unfaithful sieve" as the trial division causes worse time complexity than the actual Sieve of Eratosthenes]. The first line is just an arbitrary search...

algorithm,math,modulus,integer-division

I just want to use some equation to evaluate this thing rather than a modulus operator because the program needs to produce results within 1 sec. I have tried this counter=(((int)(N/A))+((int)(N/B)))-((int)(N/(A*B))); but it fails for input N=200 A=20 B=8 You are already on the right track, your formula indicates...

Your result will always be false because gets.chomp returns a String: number = gets.to_i The above code will work but you must always type an integer. There are better ways to manage checking to see if the input is valid. So you could check first like this: number = gets...

asp.net,ms-access,dataset,warnings,modulus

I managed to work around the warnings being thrown by .NET by generating the query's columns in my DataSet (taking the query with Mod in it, removing it then executing the query). This way the columns are properly filled. ASP.NET was not letting me generate the columns due to the...

actionscript-3,division,flash-cs6,modulus

I think the way you've framed the question may not actually fit well with your problem. If you want to know each time your score passes a 100 increment, you could simply do this: // represents current hundreds (1=100, 2=200, etc) private var currentLevel:int = 0; // each time score...

I'm not sure why it was using modulus on the count to determine which div to scroll to, but I'm certain your problem has something to do with that. Try this code: $(function() { $("div.box").swipe( { swipeDown:function(event, direction, distance, duration, fingerCount) { var scrollTarget = $(window).scrollTop() > $(this).offset().top ? $(this).offset().top...

With modulus, long long int a = fib(k + 1); long long int b = fib(k); a >= b is not necessary true, (2 * a >= b neither). so return (b * ((2 * a - b) % MOD)) % MOD; may return negative number....

c++,algorithm,do-while,modulus

while ( ( a || b ) != 0 ); this is wrong. It should be while ( a != 0 && b != 0 ); On a side note, the Euclid's algorithm can be concisely implemented recursively: int gcd(int a, int b) { return b == 0 ? a...

Becuase you can multiply by 0: c = 100*0 + 50; It's the + 50 that is returned as modulo....

Excel columns aren't a normal number system. It's not just base 26. The first two-digit column is "AA". In any normal number system, the first two digit number is composed of two different digits. Basically, in excel column numbering, there is no "zero" digit. To account for this difference, 1...

operators,bit-manipulation,modulus,and-operator

Compilers are pretty smart; if either is actually faster in practice, both &1 and %2 will quite possibly compile to the same - faster - code. With a stupid compiler, I'd place a bet on &1 being at least as fast as the alternative since it maps to bit operations...

The . after the string is an operator used in some languages like PHP to append strings, but it does not work in Python, so a code like this: days = "Mon" print "Here are the days: %s ". % days Produces the following output: File "h.py", line 3 print...

python,performance,division,modulus,divmod

To measure is to know: >>> import timeit >>> timeit.timeit('divmod(n, d)', 'n, d = 42, 7') 0.22105097770690918 >>> timeit.timeit('n // d, n % d', 'n, d = 42, 7') 0.14434599876403809 The divmod() function is at a disadvantage here because you need to look up the global each time. Binding it...

c,cryptography,modulus,algebra

You can utilize the fact, that (N-1) % N == -1. Thus, (65536 * a) % 65537 == -a % 65537. Also, -a % 65537 == -a + 1 (mod 65536), when 0 is interpreted as 65536 uint16_t fastmod65537(uint16_t a, uint16_t b) { uint32_t c; uint16_t hi, lo; if (a...

Don't use modulus. i++; if (i >= 10) i = -10; ...

The problem is that your loop sets cNumberV[nLength - n], but then prints out cNumberV[n]. So the first half of the loop prints uninitialized array entries, and the second half of the loop prints the result of the first half's calculation in reverse order (but due to an off-by-one error...

The problem you are facing results from the behavior of the methods BigInteger.toString(int radix) and BigInteger.toByteArray(). When you call the BigInteger.toString(int radix) method, it returns only the significant digits of the number. So if the value is supposed to be, for example, 05ABFF, it returns only 5ABFF. This is natural...

Operator '%' is for element-wise matrix multiplication. You have to create your own function: /** * Extend division reminder to vectors * * @param a Dividend * @param n Divisor */ template<typename T> T mod(T a, int n) { return a - floor(a/n)*n; } ...

You can add a column (in sample B) with the formula: B2 -> =IF(A2=A1;"";COUNTIF($A:$A;A2)) and autofomplete. In a cell you make the check: =MOD(SUM(B2:B9999);70) ...

Move if ( r == 0){...} to the top if your else ifs When that is evaluated first, the remaining else blocks won't be....

swift,modulo,floating-point-precision,modulus,integer-division

52 goes into 66 1 time evenly with 14 left over. When you divide 66 / 52, you get 1.2692, or 52 / 52 + 14 / 52. So, if you're only after the decimals here, they can be acquired like this (66 % 52) / 52

php,loops,foreach,division,modulus

I got this to work. My problem was that my array was multidimensional, so I converted it to a single array. Afterwards, I used array_chunks. $chunks = array_chunk($l, 2); $i=0; foreach($chunks as $mychunk){ if($i%2== 0){ echo "<tr class=\"r0\">"; } else { echo "<tr class=\"r1\">"; } $i++; foreach($mychunk as $newchunk) {...