I'm implementing universal hashing and using the following universal hash function :
h(k)=((A*k)mod 2^64) rsh 64-r
where A is a random number between
2^61 and 2^62.
The rand()
function in C++ has return type integer and it can't generate that big numbers. So how can i generate random numbers in this range? (numbers should be very random i.e. every number should have equal probability to be selected)
Note:
long long int random=rand();
doesn't work as the number returned by rand
is int.
In C++11 you can use the random header and std::uniform_int_distribution along with a 64-bit instance of std::mersenne_twister_engine this should do what you want (see it live):
#include <iostream>
#include <random>
#include <cmath>
int main()
{
std::random_device rd;
std::mt19937_64 e2(rd());
std::uniform_int_distribution<long long int> dist(std::llround(std::pow(2,61)), std::llround(std::pow(2,62)));
std::cout << std::llround(std::pow(2,61)) << std::endl;
std::cout << std::llround(std::pow(2,62)) << std::endl;
for (int n = 0; n < 10; ++n) {
std::cout << dist(e2)<< ", " ;
}
std::cout << std::endl ;
}
If C++11 is not an option then it seems there is source code available for several 64-bit Mersenne Twister implementations.
((long long)rand() << 32) | rand()
EDIT: that's assuming that rand() produces 32 random bits, which it might not.
来源:https://stackoverflow.com/questions/21096015/how-to-generate-64-bit-random-numbers