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Found this bit of code...
if(fabs([tempPlacemark getCoordinate].latitude-latitude) < latDelta && fabs([tempPlacemark getCoordinate].longitude-longitude)<longDelta )
...
refers to this in Math.h:
extern float fabsf(float);
extern double fabs(double);
extern long double fabsl(long double);
So what am I looking at?
double fabs( double ) - returns the absolute value of the argument
NSLog(#"res: %.f", fabs(10)); //result 10
NSLog(#"res: %.f", fabs(-10)); //result 10
found here.
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This may have been asked before but none of the answers I saw worked for me. I tried Lucas Theorem,Fermat's theorem but none of them worked. Is there an efficient way to find the value of:
nCr mod 10^9+7 where n<=10^9 and r<=1000
Any help will be very useful
n is large while r is small, you are better off compute nCr by n(n-1)...(n-r+1)/(1*2*...*r)
You may need to find multiplicate inverse of 1, 2, ... r mod 10^9+7
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I'm trying to generate a random number between 0 and 3 by saying
int i = arc4Random() % 3;
but it keeps giving me the warning "implicit declaration of function 'arc4Random' is invalid in c99
Try it without the capital r
int i = arc4random() % 3;
You have a capital "R" in arc4Random. Should be: int i = arc4random % 3;.
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Can anyone help me to understand that Is 2^(n^2 )=Θ(2^(n^3 )) ? it will be great if also provide the proof for this. As per my view this does not need to be equal.
The given assumption is not true:
First of all, 2^(n^2) is a function and Theta(2^(n^3)) is a set of functions, so it would be correct to say that 2^(n^2) ∈ Theta(2^(n^3)). The = is just a common abuse of notation, but it actually means ∈. To find out whether that statement is true, solve the following limit:
lim (n->infinity) of (2^(n^2)) / (2^(n^3))
If the result is 0 or infinite then the function does not belong to that particular Theta class. If it is some other value, it does belong to that class.
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If you evaluate the following in VBA, the output is True:
(2 And 3) = 2
Can someone explain this to me? Thanks!
You have:
2 (base 10) == 10 (base 2)
3 (base 10) == 11 (base 2)
Performing a bitwise AND gives
10 (base 2) == 2 (base 10)
0 = false, everything else is true in VB
I tried to find some link for MS that explained this but I could not.
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int RandomSource_next(int bits, double* seed) {
*seed = (((long long) *seed * 0x5DEECE66DLL) + 0xBLL) & ((1LL << 48) - 1);
return (int)((signed long long) *seed >> (48 - bits));
}
I think it got something to do with the address.
Most probably you are passing incorrect address as seed. Maybe you're passing not an address but a value?
The following should work
double seed = 0;
RandomSource_next(48, &seed);
The following should crash
RandomSource_next(48, 0);