Because calculus is introductory college mathematics for the quantitative sciences and engineering. It's a basic literacy in those fields. Logic and discrete math are so special case and niche that they benefit no one but the mathematician and computer scientist. They are also harder for a lot of people than calculus is.
Maybe (I'm not even sure)
You would have to replace int, class, var and other things with emoticons, first.
In many situation, verbosity decrease the readability of formula, though.
It wasn't totally rhetorical, but it's just a shape associated to a quantity. Would a different shape be better? I doubt it, and the benefit would be marginal at best, with a huge cost.
You're welcome to try devising a better system, but I can't see how you'll succeed.
I think it would end up increasing cognitive load. Reading an equation that's spelled out in words makes it really hard for me to grasp the underlying math. On the other hand, an equation using common symbols is a lot easier to understand, to sort into components, and to work with.
The number of divisorsof k?
CountDivisors(int a, int b, int d)
{
int amod = a % d;
if (amod != 0)
{
if (a < 0)
{
a -= (amod);
}
else
{
a += (d - amod);
}
}
int bmod = b % d;
if (b < 0)
{
b -= (d + bmod);
}
else
{
b -= bmod;
}
return (b - a) / d + 1;
}This is what my placeholder did but if I have a very large range and small k.. then it might take a while. Anyway hopefully the solution I came up with is correct.
The former if possible.
I see...
( abs(b-a) - (abs(a) % abs(d)) ) / abs(d) + 1Mouse is the name of the argument and sweatdrops is the value. He's calling the poop-poop-poop method of the poop var.
First argument doesn't need a name. This is removed in Swift 3.x.
what it does is that it copies the designated folder and its contained files to another designated folder. Then it renamed the folder and the containing files to something else I wanted. I was pretty happy about that and I want to make it even better. I work in a Windows environment so I wrote a code to check if the folder and files already exist in the designated area. If so, I want to delete it and copy it from source again so it works like a replace. However, I get the error that says files already in use so I os.remove() won't work. Any help you guys can provide on that? Thanks thanks!
I'm deeply sorry, I was tired yesterday, after looking at news channel till late in the night, and I've written something stupid. It's far worse than that: the formula simply doesn't work.edit: it doesn't handle a == b though but that's trivial enough to fix.
int Count(int a, int b, int d) {
if (a<b) {
int n = (a-1)%d;
if (n<0) n+=d; // To deal with different % in C
return (b-a+n+1)//d;
} else {
int n = (b-1)%d;
if (n<0) n+=d;
return (a-b+n+1)//d;
}
}int Count(int a, int b, int d) {
if (a>b) return Count(b, a, d);
int n = (a-1)%d;
return (b-a+n+1)/d + (n<0?1:0);
}int Count(int a, int b, int d) {
int n = (a-1)%d;
return (b-a+n+1)/d + (n<0?1:0);
}| ?- likes(Prolog, weird).
Do you have an idea about who could use the file? How do you check whether the files exists?However, I get the error that says files already in use so I os.remove() won't work. Any help you guys can provide on that? Thanks
Soubds like you need functors and monads.
float f = 7.15f;
float f2 = truncf(f * 10.0f) / 10.0f;
printf("%0.1f", f2);float truncf_n(float f, unsigned places) {
float factor = pow(10.0f, places);
return truncf(f * factor) / factor;
}
int main(int argc, char **argv) {
printf("%0.1f", truncf_n(7.15f, 1));
}What you're asking for isn't going to be simple because it's incorrect and unexpected behavior. Truncating places demands rounding, otherwise you are misrepresenting the number. So that's why the typical formatting options round.
Not sure it's even a proper way to do this, depending on what you're after...
printf("%.4f", truncf_n(0.2047, 4));So the Python script I made is supposed to do the follow:
anyways, hope this is more clear with what's happening. Any help would be appreciated! Thanks I see. But the issue is that Windows will forbid you to remove the file as long as there's someone "using" it. It can be an unclosed file descriptor somewhere, or something like this. There's no easy way around it, the first task would be toHowever, it is being used by another process so the remove operation cannot be performed
Those rounding issues keeps bothering me. I keep getting issues with floats, or troubles trying to prove an algorithm work, and last time I wanted to write an algorithm that wouldn't work because of floats rounding issues, I discovered that the "banker's rounding" made the task impossible: the algorithm was working in all cases!
printf("The percentage is %.1f", (value-0.05) )Python 3.2.5 [GCC 4.8.3] on cygwin
>>> 1.0 + (2.0**-53 + 2.0**-64) == 1.0
TruePython 3.4.1 |Continuum Analytics, Inc.| on Windows (32 bits).
In [1]: 1.0 + (2.0**-53 + 2.0**-64) == 1.0
Out[1]: False