modulo

问题 Someone does 20 Hours 42 Minutes & 16 Seconds in one shift totaling 74536 seconds. How do I get the hours from number of seconds the person has done for that shift? 20 * 60 * 60 = 72000 42 * 60 = 2520 16 = 16 + ----- Total = 74536 ____________________________ Total % 60 = Seconds (16) Total % ? = Minutes (42) Total % ? = Hours (20) Tried 84600 already; turns out when a number is lower the modulus, it really is not very helpful, and something I am going to have to catch should someone only

问题 I need to do modulo 256 arithmetic in C. So can I simply do unsigned char i; i++; instead of int i; i=(i+1)%256; 回答1: No. There is nothing that guarantees that unsigned char has eight bits. Use uint8_t from <stdint.h> , and you'll be perfectly fine. This requires an implementation which supports stdint.h : any C99 compliant compiler does, but older compilers may not provide it. Note: unsigned arithmetic never overflows, and behaves as "modulo 2^n". Signed arithmetic overflows with undefined

问题 Currently I write some ugly code like def div(dividend: Int, divisor: Int) = { val q = dividend / divisor val mod = dividend % divisor (q, mod) } Is it specified in standard library? 回答1: No (except for BigInt , as mentioned in other answers), but you can add it: implicit class QuotRem[T: Integral](x: T) { def /%(y: T) = (x / y, x % y) } will work for all integral types. You can improve performance by making separate classes for each type such as implicit class QuotRemInt(x: Int) extends

问题 How do I find the modulo (%) of two long values in Java? My code says 'Integer number too large' followed by the number I'm trying to mod. I tried casting it to a long but it didn't work. Do I have to convert it to a BigInteger and use the remainder method? Thanks. 回答1: The % operator does work for longs. It sounds like you may have forgotten to stick L at the end of a numeric literal, as in 123456789L . Can we see your code? 回答2: You can only have an integer up to 2 147 483 647. If you want

问题 What would be an elegant, efficient and Pythonic way to perform a h/m/s rounding operation on time related types in Python with control over the rounding resolution? My guess is that it would require a time modulo operation. Illustrative examples: 20:11:13 % (10 seconds) => (3 seconds) 20:11:13 % (10 minutes) => (1 minutes and 13 seconds) Relevant time related types I can think of: datetime.datetime \ datetime.time struct_time 回答1: For a datetime.datetime rounding, see this function: https:/

问题 I need to calculate the factorial of a big number (<=1.000.000) and I need the result modulo 1000000007. I wrote the following but it generates an error when run (test.exe has stopped working). It works only for small numbers. long long unsigned modulo(long long unsigned nr){ return nr % 1000000007; } long long unsigned fact(long long unsigned nr){ if(nr)return modulo(nr * fact(nr - 1)); else return 1; } UPDATE 1: long long unsigned fact(long long unsigned nr){ long long unsigned r = nr;

问题 Hi I am writing a code to calculate P^Q where P, Q are positive integers which can have number of digits upto 100000 I want the result as result = (P^Q)modulo(10^9+7) Example: P = 34534985349875439875439875349875 Q = 93475349759384754395743975349573495 Answer = 735851262 I tried using the trick: (P^Q)modulo(10^9+7) = (P*P*...(Q times))modulo(10^9+7) (P*P*...(Q times))modulo(10^9+7) = ((Pmodulo(10^9+7))*(Pmodulo(10^9+7))...(Q times))modulo(10^9+7) Since both P and Q are very large, I should

问题 Hi I am writing a code to calculate P^Q where P, Q are positive integers which can have number of digits upto 100000 I want the result as result = (P^Q)modulo(10^9+7) Example: P = 34534985349875439875439875349875 Q = 93475349759384754395743975349573495 Answer = 735851262 I tried using the trick: (P^Q)modulo(10^9+7) = (P*P*...(Q times))modulo(10^9+7) (P*P*...(Q times))modulo(10^9+7) = ((Pmodulo(10^9+7))*(Pmodulo(10^9+7))...(Q times))modulo(10^9+7) Since both P and Q are very large, I should

问题 I am learning to work with some math like PHP query and just got to the modulo, I am not quite sure in what situations to use this because of something i stumbled on and yes I did already read one of the posts here about the modulo : Understanding The Modulus Operator % (This explanation is only for positive numbers since it depends on the language otherwise) The quote above is in the top answer there. But if I focus on PHP only and i use the modulo like this: $x = 8; $y = 10; $z = $x % $y;

问题 Recently I discovered that C#'s operator % is applicable to double. Tried some things out, and after all came up with this test: class Program { static void test(double a, double b) { if (a % b != a - b * Math.Truncate(a / b)) { Console.WriteLine(a + ", " + b); } } static void Main(string[] args) { test(2.5, 7); test(-6.7, -3); test(8.7, 4); //... } } Everything in this test works. Is a % b always equivalent to a - b*Math.Round(a/b) ? If not, please explain to me how this operator really