随 机 数 生 成 器

利用泊松分布实现随机数生成器

不多说，直接上代码，这是在华师大算法课上做的实验代码，C++可运行。

#includeiostream

#includetime.h

#includecmath

using namespace std;

class Random {

Random(bool pseudo = true);

double random_real();

int random_integer(int low, int high);

int poisson(double mean);

void randomByAvg(double avg,int num);

private:

int reseed(); -- Re-randomize the seed.

int seed,multiplier,add_on;

-- constants for use in arithmetic operations

Random::Random(bool pseudo)

Post: The values of seed, add_on, and multiplier are

initialized. The seed is initialized randomly only if pseudo == false.

if (pseudo) seed = 1;

else seed = time(NULL) % INT_MAX;

multiplier = 2743;

add_on = 5923;

int Random::reseed()

Post: The seed is replaced by a pseudorandom successor.

seed = seed * multiplier + add_on;

return seed;

double Random::random_real()

Post: A random real number between 0 and 1 is returned.

double max = INT_MAX + 1.0; --INT_MAX = (2)31 -1

double temp = reseed();

if (temp 0) temp = temp + max;

return temp - max;

int Random::random_integer(int low, int high)

Post: A random integer between low and high is returned. if (low high) return random_integer(high, low);

else return ((int) ((high - low) * random_real())) + low; int Random::poisson(double mean)

Post: A random integer, reflecting a Poisson distribution with parameter mean, is returned.

double limit = exp(-mean);

double product = random_real();

int count = 0;

while (product limit) {

count++;

product *= random_real();

return count;

void Random::randomByAvg(double avg,int num) { double p = 1 - 1.0 - (avg + 1);

int t;double sum=0,ave;

for(int i=0;inum;i++){

t=poisson(avg);

coutt" ";

coutendl;

ave=sum-num*1.0;

cout"随机整数序列的平均值为："aveendl;

int main(){

cout"请输入概率均值:"endl;

double rand;

cinrand;

cout"请输入随机整数的个数："endl;

--产生随机序列

Random random;int t;double sum=0;

random.randomByAvg(rand,num);

return 0;

运行结果：

int t; -- t为n最大倍数，且满足 t = m * m

printf ("It is %lf.-n", random ()) ;

std::uniform_real_distributiondouble dis2(0.0, 1.0);

SplittableRandom random = new SplittableRandom();

生成10000个数然后计算分布比例应该可以，只要在90%上下就算通过。

nRF51822_Product Specification_v3.1.pdf中介绍nrf自带真随机生成器：

sum+=Integer.parseInt(list.get(i).get("value").toString());

下面主要介绍的是线性同余法，如C的rand函数和JAVA的java.util.Random类就是使用该方法实现的，其公式为：

code.append(N60_CHARS[i]);

为了生成伪随机数，伪随机数生成器需要称为种子（seed）的信息。伪随机数的种子是用来对伪随机生成器内部状态进行初始化的。