division

问题 For the following program: #include <iostream> #include <iomanip> using namespace std; int main() { for (float a = 1.0; a < 10; a++) cout << std::setprecision(30) << 1.0/a << endl; return 0; } I recieve the following output: 1 0.5 0.333333333333333314829616256247 0.25 0.200000000000000011102230246252 0.166666666666666657414808128124 0.142857142857142849212692681249 0.125 0.111111111111111104943205418749 Which is definitely not right right for the lower place digits, particularly with respect

问题 System.out.println((26.55f/3f)); or System.out.println((float)( (float)26.55 / (float)3.0 )); etc. returns the result 8.849999. not 8.85 as it should. Can anyone explain this or should we all avoid using floats? 回答1: What Every Programmer Should Know About Floating-Point Arithmetic: Q: Why don’t my numbers, like 0.1 + 0.2 add up to a nice round 0.3, and instead I get a weird result like 0.30000000000000004? A: Because internally, computers use a format (binary floating-point) that cannot

问题 I have a few divs on my code all of them positioned correctly how i want it. The only problem is that they do not re size when i shrink the page. I want to be able to re size them according to the web browsers size. I do not know exactly how to do it and im worried it might mess up their correct top & left position. All of my divs are currently set as absolute positioning. 回答1: Set their sizes to percent values rather than pixels (I asume that's what you're using) 回答2: Set all divs with

问题 Not terribly familiar with Xcode or Objective-C but trying to learn. Hopefully someone can help me out with a problem I'm having. I have two fields, one called price and one called units and I'm trying divide the inputs of the cells by each other and then display the result with the correct currency of the 'nationality' of the device when a button is pressed. So far I have of the action of the button I have; - (IBAction)calculate:(id)sender { int x = [price.text floatValue]; int y = [units

问题 i have an equation , ((a*b*c*d)/(e*f*g*h))%m My question is , Can i first apply multiplication property (a*b) mod(n) = (a*mod(n)) * (b*mod(n) ) mod(n) to numerator and then denominator , so that numerator and denominator becomes a single value , and then solve the division operation? (a/b) mod(n) = (a*inv(b)) mod(n) 回答1: Let N = a*b*c*d and D = e*f*g*h . We want to calculate: (N/D) mod n = (N * inv(D)) mod n We can use the multiplication property here in the following way: (N * inv(D)) mod n

问题 So of course I know there are simple solutions to this such as using the GMP library or numerous other arbitrary-precision libraries. This is for class work so I am not allowed to take any of these routes. After we build up all our operations we are leading up to be able to do a RSA encryption scheme. I am using vectors to store n-bit numbers represented in binary. I have conversions to decimal later but I must operate on the binary and only convert for display. I have successfully

问题 I'm having an issue I want to learn more about, and how to avoid. I've got this code len :: (Num r ) => [a] -> r len [] = 0 len xs = 1 + len ( tail xs ) avg :: (Num t) => [t] -> Double avg xs = ( sum xs ) / ( len xs ) Which renders the following error len.hs:6:9: Couldn't match expected type `Double' against inferred type `t' `t' is a rigid type variable bound by the type signature for `avg' at len.hs:5:12 In the expression: (sum xs) / (len xs) In the definition of `avg': avg xs = (sum xs) /

问题 I have 2 different query which will return values : 1502.00 and 6 SELECT replace(CURRENT_VALUE,'$','') curVal FROM form_attributes_values WHERE TEST_ID=2 AND ATTRIBUTE_ID = ( SELECT ATTRIBUTE_ID FROM form_attributes WHERE FORM_ID=6 AND FORM_FIELD_NAME='REGRINDABLECUTTERCOST' ) and SELECT replace(CURRENT_VALUE,'$','') curVal FROM form_attributes_values WHERE TEST_ID=2 AND ATTRIBUTE_ID = ( SELECT ATTRIBUTE_ID FROM form_attributes WHERE FORM_ID=6 AND FORM_FIELD_NAME='REGRINDSPOSSIBLE' ) I am

问题 Given range x, y. I need to count all numbers in between and are divisible by n. I know the simple way to do this is by looping over the whole range for(int i=x;i<=y;i++) if(i%n ==0) counter++; The counter holds the answer. But this works so slow for big ranges. for example x=0 , and y=3,000,000,000. I'm sure that there is some relation that I can use to reduce the number of iteration and optimizing this code for speed. I searched but i could not find it out. Can any one help me with that

问题 I really don't think this is a precision problem, the answer should be about 0.226. Here's the exact code: val = I(i,j) bucketSize pos = val / bucketSize I is just a matrix I'm taking values from. Here is the output from MATLAB: val = 29 bucketSize = 128 pos = 0 What am I missing? 回答1: My guess would be that your matrix I is pixel data loaded from an image file, which will have values that are typically unsigned 8-bit integers. As already mentioned, converting both integer values to a double