modulo

I have a table software and columns in it as dev_cost , sell_cost . If dev_cost is 16000 and sell_cost is 7500. How do I find the quantity of software to be sold in order to recover the dev_cost ? I have queried as below: select dev_cost / sell_cost from software ; It is returning 2 as the answer. But we need to get 3, right? What would be the query for that? Thanks in advance. Ilmari Karonen Your columns have integer types, and integer division truncates the result towards zero . To get an accurate result, you'll need to cast at least one of the values to float or decimal : select cast(dev

My problem is to compute (g^x) mod p quickly in JavaScript, where ^ is exponentiation, mod is the modulo operation. All inputs are nonnegative integers, x has about 256 bits, and p is a prime number of 2048 bits, and g may have up to 2048 bits. Most of the software I've found that can do this in JavaScript seems to use the JavaScript BigInt library ( http://www.leemon.com/crypto/BigInt.html ). Doing a single exponentiation of such size with this library takes about 9 seconds on my slow browser (Firefox 3.0 with SpiderMonkey). I'm looking for a solution which is at least 10 times faster. The

Most of us are familiar with the maximum sum subarray problem . I came across a variant of this problem which asks the programmer to output the maximum of all subarray sums modulo some number M. The naive approach to solve this variant would be to find all possible subarray sums (which would be of the order of N^2 where N is the size of the array). Of course, this is not good enough. The question is - how can we do better? Example: Let us consider the following array: 6 6 11 15 12 1 Let M = 13. In this case, subarray 6 6 (or 12 or 6 6 11 15 or 11 15 12) will yield maximum sum ( = 12 ). We can

I'm trying a line like this: for i in {1..600}; do wget http://example.com/search/link $i % 5; done; What I'm trying to get as output is: wget http://example.com/search/link0 wget http://example.com/search/link1 wget http://example.com/search/link2 wget http://example.com/search/link3 wget http://example.com/search/link4 wget http://example.com/search/link0 But what I'm actually getting is just: wget http://example.com/search/link Mark Longair Try the following: for i in {1..600}; do echo wget http://example.com/search/link$(($i % 5)); done The $(( )) syntax does an arithmetic evaluation of

I have a function f(x)=(1^1)*(2^2)*(3^3)*.....(x^x) i have to calculate (f(x)/(f(x-r)*f(r)))modulo c i can calculate f(x) and (f(x-r)*f(r)). assume f(x) is a and f(x-r)*f(r) is b. c is some number that is very larger. `` so i how can calculate (a/b)%c your f(x) is just ᴨ (PI cumulative multiplication) squared it is hard to write it in here so i will deifine g(x0,x1) instead g(x0,x1)=x0*(x0+1)*(x0+2)*...*x1 so: f(x)=g(1,x)^2 computing h(x,r,c)=f(x)/(f(x-r)*f(r))%c when you rewrite it to g() you get: h(x,r,c)=((g(1,x)/(g(1,x-r)*g(1,r)))^2)%c now simplify (and lower magnitude) as much as you can

Preamble This question is not about the behavior of (P)RNG and rand() . It's about using power of two values uniformly distributed against modulo. Introduction I knew that one should not use modulo % to convert a value from a range to another, for example to get a value between 0 and 5 from the rand() function: there will be a bias. It's explained here https://bitbucket.org/haypo/hasard/src/ebf5870a1a54/doc/common_errors.rst?at=default and in this answer Why do people say there is modulo bias when using a random number generator? But today after investigating some code which was looking wrong,

long long x; for (int i = 1; i <= n; i++) { x = (x * i) % m; } cout << x; This is the trick to calculate (n!) mod m (assume m > n). However, I don't know why it's true. Can you explain the math mechanism behind this? The basic idea here is that you can take the modulus before, during, or after multiplication and get the same value after taking the modulus of the final result. As @Peter points out, (a * b) % m == ((a % m) * (b % m)) % m For the factorial, n! = 1 * 2 * 3 * ... * (n-1) * n so we have n! % m = (((((((1 * 2) % m) * 3) % m) * ... * n-1) % m) * n) % m taking the modulus after each