multiplication

问题 I'm having a lot of trouble getting some simple math done in VHDL. I'm terrible at this language so if my syntax is stupid or something, I have an excuse :P. I'm trying to implement a very simple random number generator that calculates a pseudo-random number by this formula: seed = (seed*1103515245) + 12345 How I'm trying to do it: Signalss here signal seed: std_logic_vector(31 downto 0) := x"2B4C96B9"; signal multiply: std_logic_vector(31 downto 0) := x"41C64E6D"; signal add: std_logic

问题 i am trying to do the following in Scheme: List<int> list = new List<int>(); List<int> list1 = new List<int>(); List<int> list2 = new List<int>(); list.Add(1); list.Add(2); list.Add(3); list.Add(4); list1.Add(2); list1.Add(4); list1.Add(6); list1.Add(8); for (int i = 0; i < list.Count; i++) { for (int p = 0; p < list1.Count; p++) { list2.Add(list[i] * list1[p]); } } as seen in the code above, I am trying to multiply each element of the first list with every element in the second list. So 1*2,

问题 I've been working on a Batch polygon area calculator and I got a problem. I need to multiply 2 variables, but sometimes it return a negative number if the two positive variables are large. Here's an example: 999999*999999 returns -729379967 . Code goes below: REM Calc square area :PolySqu Cls Echo Polygon Area Calculator For /L %%P In (1,1,57) Do Echo. Set /P "InputPolygonCalSqu=Enter one of the line's length in cm :" Set /A SquArea=InputPolygonCalSqu * InputPolygonCalSqu Cls Echo Polygon

问题 So I'm trying, for all intents and purposes, to realize for myself a java version of C++'s STL algorithm inner_product for vector parameters. So far, my code(which is probably fundamentally wrong) looks like this: public static<T,K> double inner_product(Vector<T> v1, Vector<K> v2) { double isum = 0; for(int i=0;i<v1.size()&&i<v2.size();i++) { isum+=v1.elementAt(i)*v2.elementAt(i); } return isum; } The problem is that the operator * is undefined for the types T, K. However my knowledge so far

问题 I have tried to implement karatsuba algorithm using the below code. The problem starts when the number of digits in x and y(parameters) mismatch as the recursive call doesn't work in that case with the below logic. As of now, am getting the correct output when the number of digits in x and y is the same. To be more precise, I think the problem starts with the calculations of z1 and z3 because that is where the number of digits for x and y mismatches frequently. I also am a bit confused about

问题 I have the following problem Under what circumstances can multiplication be regarded as a unit time operation? But I thought multiplication is always considered to be taking unit time. Was I wrong? 回答1: It depends on what N is. If N is the number of bits in an arbitrarily large number, then as the number of bits increases, it takes longer to compute the product. However, in most programming languages and applications, the size of numbers are capped to some reasonable number of bits (usually

问题 How can I make multiplication table to look like this: http://i.imgur.com/rR6JSua.png ? With my code it only has one column. #include<stdio.h> int main() { int i, j; for(i = 1;i <= 9;i++) { for(j = 1;j <= 9;j++) { printf("%d * %d = %d\n",i , j,i*j); } printf("%d * %d = %d\n",i , 10,i*10); printf("\n"); } return 0; } 回答1: Try This: #include<stdio.h> int main() { int i, j; for(i = 1;i <= 9;i+=3) { for(j = 1;j <= 10;j++) { printf("%2d * %2d = %2d ",i , j,(i)*j); printf("%2d * %2d = %2d ",i+1 , j

问题 Execution of: #define HIGH32(V64) ((uint32_t)((V64 >> 32)&0xffFFffFF)) #define LOW32(V64) ((uint32_t)(V64&0xffFFffFF)) uint32_t a = 0xffFFffFF; uint32_t b = 0xffFFffFF; uint64_t res = a * b; printf("res = %08X %08X\n", HIGH32(res), LOW32(res)); Gives: "res = 00000000 00000001" But I expect: fffffffe00000001. What did I do wrong? Single assignment: res = 0x0123456789ABCDEF; printf("res = %08X %08X\n", HIGH32(res), LOW32(res)); Gives res = 01234567 89ABCDEF Environment: $gcc --version gcc (GCC)

问题 I'm searching how to tell SymPy to use a multiplication of exponentials rather than an exponential of a sum. That is, it currently gives me exp(a + b) and I would want to get exp(a)*exp(b). There must be a fairly easy way but I can't seem to find it. 回答1: You could use the expand() function to show the expression with multiplication of bases rather than the sum of exponents: >>> from sympy import * >>> a, b = symbols('a b') >>> expr = exp(a + b) >>> expr exp(a + b) >>> expr.expand() exp(a)

问题 I have a tuple of integers such as (1, 2, 3, 4, 5) and I want to produce the tuple (1*2, 2*3, 3*4, 4*5) by multiplying adjacent elements. Is it possible to do this with a one-liner? 回答1: Short and sweet. Remember that zip only runs as long as the shortest input. print tuple(x*y for x,y in zip(t,t[1:])) 回答2: >>> t = (1, 2, 3, 4, 5) >>> print tuple(t[i]*t[i+1] for i in range(len(t)-1)) (2, 6, 12, 20) Not the most pythonic of solutions though. 回答3: If t is your tuple: >>> tuple(t[x]*t[x+1] for x