modulo

问题 I try to calculate with JS' modulo function, but don't get the right result (which should be 1). Here is a hardcoded piece of code. var checkSum = 210501700012345678131468; alert(checkSum % 97); Result: 66 Whats the problem here? Regards, Benedikt 回答1: A bunch of improvements to Benedikt's version: "cRest += '' + cDivident;" is a bugfix; parseInt(divisor) makes it possible to pass both arguments as strings; check for empty string at the end makes it always return numerical values; added var

问题 While writing a shader program today, I encountered a situation where I have to use % to find the remainder. GLSL gave me an error saying that it is not available in the current version. I've tried several problems. GLSL doesn't support recursive function and while loops, which is needed if I want to create a function that can give me the result of (a % b) . So, I'm currently stuck. Can someone help me with this? Edit . I was trying to emulate a CRT screen using some shader code from this

问题 This question already has answers here : Python modulo on floats (3 answers) Closed 5 years ago . I just want to show you the results of the operations in python. I cannot explain. >>> 1.0%1.0 0.0 (OK) >>> 1.0%0.1 0.09999.... >>> 1.0%0.001 0.00999.... >>> 1.0 %0.0001 0.000999... ... and so on. I need something that allows me to understand whether the remainder of 'x%y' is 0.0, namely 'y' divides 'x' exactly N times, where N is an integer. Due to the previous behavior I don't know how to set a

问题 How can i check number is divisible by particular number or not both numbers are decimal. with decimal point value of two digits. i had tried with below (((dn / size) % 1) == 0) but in some cases it not provide proper out put. how can i resolve it. here i put some example values like double dn = 369.35,369.55.370.55 and size may be 0.05,0.10,0.5 etc... if(((dn / size) % 1) == 0) { Log.d(TAG,"OK"); } else { Log.d(TAG,"OK"); } please help me to short out it. 回答1: (dn / size) % 1 == 0 whilst

问题 After reading a few errors of the type 'int not callable' on stackoverflow, I see that most errors of the type involve treating an int like a function. I am getting this error on the following program and I'm not sure what's going on: find the power of n that satisfies the equation for n in range(100): if ((2^n // 3) % 2) == 1: print n The error traceback reads: File "<stdin>", line 1, in <module> TypeError: 'int' object is not callable 回答1: You have a variable named range , which you're

问题 All, How can I calculate 2^301 mod 77? I did check out the link StackOverflow. But did not understand the step wherein 625 mod 221 = 183 mod 221. How did the conversion take place? 回答1: Take a look at the question here for an answer to your question. Basically, (X * Y) % Z == ((X % Z) * (Y % Z)) % Z . So, as a starting point, 2^301 % 77 == ((2^150 % 77) * (2^151 % 77)) % 77 . Keep splitting until you have reasonable numbers, then recombine. You will be able to keep your numbers at a

问题 I was working with a toroidal 2D grid in c++ (ie. it wraps around at the sides), and wrote an obvious neighbor / relative-point function: point neighbor(point p0, int dx, int dy) { point p=p0; p.x += dx; p.y += dy; p.x %= width; if(p.x<0) p.x += width; p.y %= height; if(p.y<0) p.y += height; return p; } I was totally clueless why my program wasn't working, since the implementation of this function seemed trivial. I thought I understood the % operator, I even remembered to check for negative

问题 Hi I have an algorithm in which I need to apply operations to BigInt's. I understand that BigInt's can be manipulated using the Maths class such as: import java.math.*; BigInteger a; BigInteger b = BigInteger.ZERO; BigInteger c = BigInteger.ONE; BigInteger d = new BigInteger ("3"); BigInteger e = BigInteger.valueOf(5); a.multiply(b); a.add(b); a.substract(b); a.divide(b); I need to be able to apply greater than for a while condition e.g. while (a > 0) { Which gives me a syntax error saying

问题 Edited: I have a big number that C does not have a type for it natively. I have to use a char array to hold it. As an example, I create a 32-byte array. It represents a large number up to 2 ^ 256. unsigned char num[32]; // The size could be any number for this question. I want to apply modulo operation on it, for example, I want to mod the big number by a small divisor and get an integer type result. int divisor = 1234; // Note that the divisor is much smaller than the big number int result;

问题 According to Google Calculator (-13) % 64 is 51 . According to Javascript (see this JSBin) it is -13 . How do I fix this? 回答1: Number.prototype.mod = function(n) { return ((this%n)+n)%n; }; Taken from this article: The JavaScript Modulo Bug 回答2: Using Number.prototype is SLOW, because each time you use the prototype method your number is wrapped in an Object . Instead of this: Number.prototype.mod = function(n) { return ((this % n) + n) % n; } Use: function mod(n, m) { return ((n % m) + m) %