multiplication

问题 If I use a calculator, 2/3 is 0.6666666667 which is about 67% . However if I try to do the same thing with css calc I get an error. width: calc(2 / 3); Is there a working way for this? I don't think it looks that good writing it like 0.666666666667 . Any ideas are welcome. 回答1: The problem is with calc(2 / 3) you will just get a number without an unit. CSS can't display just a number as width. This would be like if you set width: 3 which obviously doesn't work. If you want the percentage you

问题 As you see in the screen shot above, I have the following expression in my Matlab m-file code: K = P * D * D' * P; Where, P is an nxn matrix, and D is a nx1 column vector (n=4, if it matters). Why am I getting this warning message? What changes if I use or don't use parenthesis there? 回答1: Floating-point arithmetic is not associative. So in general, a * (b * c) won't necessarily give the same result as (a * b) * c . Your statement as written is equivalent to ((P * D) * D') * P , so the

问题 In my case, product of two INT_MAX numbers is 296447233 , which is incorrect. long long int product = 0; product = 2137483647 * 2137483647; printf("product: %lli\n", product); What I am doing wrong, and how to correct it ?? Thanks ! 回答1: Both of your 2137483647 are of type int . So they stay that type and overflow. Make them long long s: product = 2137483647LL * 2137483647LL; or cast: product = (long long)2137483647 * 2137483647; 回答2: Try product = 2137483647LL * 2137483647LL; to ensure that

问题 First off, this is homework. I'm trying to read a 5 digit number into the register bx. The number is assumed to be no greater than 65535 (16 bits). Below is how I am attempting to do so. However, when I attempt to print the number, I am only printing the very last digit that was entered. Which leads me to guess that when I add another number to bx it is overwriting the previous number, but I am unable to see the problem. Any help would be appreciated, I'm almost certain that it is something

问题 I want to be able to compute g^x = g * g * g * ... * g (x times) where g is in a finite field GF(2^m). Here m is rather large, m = 256, 384, 512, etc. so lookup tables are not the solution. I know that there are really fast algorithms for a similar idea, modpow for Z/nZ (see page 619-620 of HAC). What is a fast, non-table based way to compute cycles (i.e. g^x)? This is definitely a wishful question but here it comes: Can the idea of montgomery multiplication/exponentiation be 'recycled' to

问题 I am having trouble getting divide and conquer matrix multiplication to work. From what I understand, you split the matrices of size nxn into quadrants (each quadrant is n/2) and then you do: C11 = A11⋅ B11 + A12 ⋅ B21 C12 = A11⋅ B12 + A12 ⋅ B22 C21 = A21 ⋅ B11 + A22 ⋅ B21 C22 = A21 ⋅ B12 + A22 ⋅ B22 My output for divide and conquer is really large and I'm having trouble figuring out the problem as I am not very good with recursion. example output: Original Matrix A: 4 0 4 3 5 4 0 4 4 0 4 0 4

问题 Does the compiler optimize out any multiplications by 1? That is, consider: int a = 1; int b = 5 * a; Will the expression 5 * a be optimized into just 5? If not, will it if a is defined as: const int a = 1; 回答1: It will pre-calculate any constant expressions when it compiles, including string concatenation. Without the const it will be left alone. Your first example compiles to this IL: .maxstack 2 .locals init ([0] int32, [1] int32) ldc.i4.1 //load 1 stloc.0 //store in 1st local variable ldc

问题 I'm writing a Fixedpoint class, but have ran into bit of a snag... The multiplication, division portions, I am not sure how to emulate. I took a very rough stab at the division operator but I am sure it's wrong. Here's what it looks like so far: class Fixed { Fixed(short int _value, short int _part) : value(long(_value + (_part >> 8))), part(long(_part & 0x0000FFFF)) {}; ... inline Fixed operator -() const // example of some of the bitwise it's doing { return Fixed(-value - 1, (~part)

问题 This question already has answers here : Is floating point math broken? (31 answers) Closed 4 years ago . I am executing the following code in java but i got two different answers for what should be the same number mathematically. public class TestClass { public static void main(String[] args) { double a=0.01; double b=4.5; double c=789; System.out.println("Value1---->"+(a*b*c)); System.out.println("Value2---->"+(b*c*a)); } } Output: Value1---->35.504999999999995 Value2---->35.505 Please tell

问题 I read in a paper that division and multiplication of a number by a power of 2 is a trivial process. I have searched a lot of internet for the explanation but doesn't get it. Can any one explain in easy words what does this actually meant to say. 回答1: It is trivial from the bit operations perspective. Multiplying by 2 is equivalent to a shift left by 1 bit, division is a right shift. similarly it is the same trivial to multiply and divide by any power of 2. int a = 123; // or in binary format