modulo

问题 I want to find a way to know if an integer is divided by 3 or 7 without using division, because it is very slow in MIPS assembly. I have done a lot of research but found nothing. 回答1: There's a method described by Granlund & Montgomery that requires the modular / multiplicative inverse of the (odd) divisor modulo 2**b . (Some parts of this paper have been improved recently) The divisors: (d) = 3, 7 (odd numbers) are an easy case. Assuming 32 bit (unsigned) arithmetic, the inverses modulo 2*

问题 I am trying to find the sum of the first r binomial coefficients for a fixed n. (nC1 + nC2 + nC3 + ... + nCr) % M where r < = n. Is there an efficient algorithm to solve this problem ? 回答1: Note that the "first" binomial coefficient for fixed n is nC0 . Let f(n) = nC0 + nC1 + ... + nC(r-1) . Using the "Pascal's triangle" identity, nCk = (n-1)C(k-1) + (n-1)Ck we have nC0 + nC1 + nC2 + ... + nC(r-1) = (n-1)C(-1) + (n-1)C0 + (n-1)C0 + (n-1)C1 + (n-1)C1 + (n-1)C2 + ... + (n-1)C(r-2) + (n-1)C(r-1)

问题 This works fine, but I want to make it prettier - and accommodate all values that are divisible by 4: if i==4 || i==8 || i==12 || i==16 || i==20 || i==24 || i==28 || i==32 # ... end Any clever, short method to do this? 回答1: Try this: if i % 4 == 0 This is called the "modulo operator". 回答2: There's also modulo , which allows you to do 420.modulo(4).zero? There's nothing stopping you doing that with % , but it looks weird: 420.%(4).zero? 回答3: This is always a good conversation starter: if (i &

问题 This works fine, but I want to make it prettier - and accommodate all values that are divisible by 4: if i==4 || i==8 || i==12 || i==16 || i==20 || i==24 || i==28 || i==32 # ... end Any clever, short method to do this? 回答1: Try this: if i % 4 == 0 This is called the "modulo operator". 回答2: There's also modulo , which allows you to do 420.modulo(4).zero? There's nothing stopping you doing that with % , but it looks weird: 420.%(4).zero? 回答3: This is always a good conversation starter: if (i &

问题 When I put in Python interpreter a ** b % c with large a (20 figures) b (4 figures) c (20 figures) I saw that Python calculates it pretty fast, almost like pow (a,b,c). I expect another behavior that Python first calculate a ** b then get the modulo (%) of result and such calculation will take significantly more time. Where is the magic behind the scene? 回答1: If you are typing into the Python interpreter something like: 20937505974095709374 ** 3438 Then seeing a couple of seconds wait. Then

问题 I am trying to make a simple objective-C height converter. The input is a (float) variable of feet, and I want to convert to (int) feet and (float) inches: float totalHeight = 5.122222; float myFeet = (int) totalHeight; //returns 5 feet float myInches = (totalHeight % 12)*12; //should return 0.1222ft, which becomes 1.46in However, I keep getting an error from xcode, and I realized that the modulo operator only works with (int) and (long). Can someone please recommend an alternative method?

问题 I am trying to make a simple objective-C height converter. The input is a (float) variable of feet, and I want to convert to (int) feet and (float) inches: float totalHeight = 5.122222; float myFeet = (int) totalHeight; //returns 5 feet float myInches = (totalHeight % 12)*12; //should return 0.1222ft, which becomes 1.46in However, I keep getting an error from xcode, and I realized that the modulo operator only works with (int) and (long). Can someone please recommend an alternative method?

问题 I have a question. I have this array of the alphabet. And every even letter needs to be !$letter, so it needs to echo out !b, !d, !f I'm unsure how to do this. I've been told to use the modulo, % for this. But reading a lot about it on the internet and things haven't gotten any clearer for me. I appreciate anyone that can help me on this matter! Thanks in advance! 回答1: foreach ($alphabet as $i => $letter) { echo (($i % 2) == 1 ? '!' : '') . $letter; } $i is the position of the letter in the

问题 I'm currently new to learning python and stumbled upon this problem: Exercise 2-7 Get the Time Write an algorithm that reads the amount of time in seconds and then displays the equivalent hours, minutes and remaining seconds. • One hour corresponds to 60 minutes. • One minute corresponds to 60 seconds. Here's how I wrote it: def amount_time(t): h = t//3600 m = int((t/3600 - h)*60) s = int(((t/3600 - h)*60 - m)*60) print("Number of hours:",h) print("Number of minutes:",m) print("Number of

问题 I am currently writing a program that requires the calculation of: (x^2) mod y Both these numbers are integers. Both can be large numbers, though they never exceed 10^9. That is still enough to overflow integer for x squared. Speed is crucial for this code so gradual multiplication is not usable. Thank you. 回答1: You can take a reference from this post This should be a relatively easy example. 4^23=2^46. Now, since 2^5=32≡1(mod31), 2^46=(2^5)9×2=32^9×2≡1×2≡2(mod31) 回答2: The solution was really