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I'm writing a c++ implementation of RSA algorithm. It work but It takes hours to calcultate d key. Any help to make it work faster would be awesome.
unsigned __int64 calcolo_d(const unsigned __int64 eulero, const unsigned __int64 e) {
register unsigned __int64 d = 0;
while (!((e*d) % eulero == 1))
{
++d;
}
return d; }
What you are computing is the modular inverse of e mod eulero. This can be done efficiently via the extended euclidean algorithm.
There are many, many implementations out there to choose from.
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using concept of function overloading making a program to find maximum of 3 integer no.
I don't understand how do i show overload a function that is taking 3 integers only
You can use the fact that the maximum of three numbers is the bigger of 1 of the numbers and the maximum of the other two.
int maximum(int a,int b) { return /* ... */ }
int maximum(int a,int b, int c) { return /* ... */ }
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I came Across a line of code in an example that I can not understand how is it being utilized and what benefit on using this type of declaration.
following in the part of code extracted from that example
const uint16_t cycles = 8;
typedef uint8_t invariant(value < cycles) TIndex_t;
struct TData
{
uint16_t readings[cycles];
volatile uint16_t sum;
TIndex_t index;
void addReading(uint16_t arg)
writes(*this; volatile)
pre(invar())
pre(arg <= 1023)
post(invar())
{
sum = sum - readings[index] + arg;
readings[index] = arg;
index = static_cast<TIndex_t>((index + 1u) % cycles);
}
void init()
writes(*this; volatile)
post(invar());
ghost(
bool invar() const
returns((forall r in readings :- r <= 1023) && sum == + over readings);
)
This is GNU C++ the struct is a kind of class definition as I can tell
This is not pure C++.
It looks like somebody has written either a compiler extension or a series of macros to allow the specification of pre- and post-conditions for a block of code.
You're going to have to ask that person.
(As an aside, this is why just "extracting code" is often insufficient for understanding; you need to observe the full context in order to know what's going on. In this case, the context includes definitions and/or documentation found elsewhere and not included in your question.)
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I have a little problem.
Who knows how we can calculate the log
base n with Shift_L or Shift_R?
for example: for n=2 we had this solution:
int log(int n){
int res = 0;
while((n>>=1))
res++;
return res;
}
You don't seem to want the logarithm for a base b, but the largest integer n so that n <= log_b(x). If that's the case, the following function should serve your needs:
int intlog(double base, double x) {
return (int)(log(x) / log(base));
}
well this is rather a math problem instead of an actuall programming problem, if i understand your problem correctly:
log_2 (x) = log_a (x) / log_a (2) where a can be any base.
Therefore you could use the math.h's function log(double)
double res = log(x)/log(2);
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#include <iostream>
using namespace std;
int main()
{
int a = 101;
return 0;
}
Question : How do I know that the number (1) is repeated twice in the variable
If you look at the code, you will see that the number 101 is assigned to the variable a and that number has the digit 1 twice in its decimal representation. So direct inspection is the way to go. I wouldn't even write the code for such a trivial requirement.
Use modulus 10 and division 10 to find it. Rough idea is,
while( a > 0 )
{
if( a % 10 == 1 )count_one++;
a=a/10;
}
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I have written a Sign function and am wondering whether it is correct or not (Stupid question to ask, I know!) I'm just interested in knowing whether this is the best method to solve this particular task:
template<typename T>
T sign(T n)
{
if(n < 0) return -1;
if(n > 0) return 1;
return 0;
}
Would this give accurate enough results for large datasets? Can anyone see a problem, that I haven't come across that may arise when putting this into a real-life context?
Thanks
I would change return 0; to return n;. If n is NaN, sign should return NaN, not 0.