使用线性全等生成器我能够创建两个均匀分布的独立伪随机数序列。我试图改变我的程序，允许它使用这些序列并执行Box-Muller变换将它们更改为正态分布集。
然而，我遇到的问题是，无论两个统一序列的输入种子值如何，我的新“正态分布随机数”（Z）总是等于零。
任何提示都将不胜感激。 非常感谢
#define _USE_MATH_DEFINES
#include <iostream>
#include <cmath>
#include <math.h>
using namespace std;
#define M 4294967295
unsigned long get_rand(unsigned long x) //establishing function to generate random numbers
{
unsigned long a = 2269477;
unsigned long b = 1; //Values taken from wikipedia for Linear Congruence Method
unsigned long m = M;
unsigned long y;
y = (a * x + b) % m;
return y;
}
unsigned long get_normal(unsigned long x1, unsigned long x2)
{
unsigned long R;
unsigned long phi;
unsigned long u;
R = sqrt(-2 * log(x1)); //Box-Muller Transform
phi = (2 * M_PI*x2);
u = R*cos(phi);
return u;
}
double u1, u2, Z;
double bin0 = 0;
double bin1 = 0;
double bin2 = 0; //Variables used to store frequency of each number range
double bin3 = 0;
double bin4 = 0;
double bin5 = 0;
double bin6 = 0;
double bin7 = 0;
double bin8 = 0;
double bin9 = 0;
int main() {
double seed1,seed2;
cout << "Please enter seed values " << endl;
cin >> seed1;
cout << "\n";
cin >> seed2;
double x;
cout << "Please enter how many random numbers you want " << endl;
cin >> x;
cout << endl;
cout << "Random Numbers generated shown below: " << endl;
for (int i = 0; i < x; i++) //generate as many random numbers as the user has asked for
{
seed1 = get_rand(seed1);
seed2 = get_rand(seed2);
u1 = (double(seed1) / M); //changing to double and dividing by 'M' gets all values between 0 and 1
cout <<"U1 = " << u1 << endl; //type conversion to prevent integer rounding in division
u2 = (double(seed2) / M);
cout << "U2 = " << u2 << endl;
Z = get_normal(u1, u2);
cout <<"Z = " << Z << endl;
if (Z >= 0.0 && Z <= 0.1)
{ //checking for which intervals each random number falls into
bin0++; //if a number falls into this interval, increment the counter by 1 each time
}
else if (Z > 0.1 && Z <= 0.2) //if it doesnt fall into first interval, it will check the next interval, and so on...
{
bin1++;
}
else if (Z > 0.2 && Z <= 0.3)
{
bin2++;
}
else if (Z > 0.3 && Z <= 0.4)
{
bin3++;
}
else if (Z > 0.4 && Z <= 0.5)
{
bin4++;
}
else if (Z > 0.5 && Z <= 0.6)
{
bin5++;
}
else if (Z > 0.6 && Z <= 0.7)
{
bin6++;
}
else if (Z > 0.7 && Z <= 0.8)
{
bin7++;
}
else if (Z > 0.8 && Z <= 0.9)
{
bin8++;
}
else if (Z > 0.9 && Z <= 1.0)
{
bin9++;
}
}
double binTotal = bin0 + bin1 + bin2 + bin3 + bin4 + bin5 + bin6 + bin7 + bin8 + bin9;
cout << endl;
int bin0Percent = (bin0 / binTotal) * 100; //working out a percentage
cout << " Number of values in range 0.0-0.1: " << bin0 << endl; //output screen for each interval
cout << " Percentage of values in this interval: " << bin0Percent << "%" << endl;
cout << endl;
int bin1Percent = (bin1 / binTotal) * 100;
cout << " Number of values in range 0.1-0.2: " << bin1 << endl;
cout << " Percentage of values in this interval: " << bin1Percent << "%" << endl;
cout << endl;
int bin2Percent = (bin2 / binTotal) * 100;
cout << " Number of values in range 0.2-0.3: " << bin2 << endl;
cout << " Percentage of values in this interval: " << bin2Percent << "%" << endl;
cout << endl;
int bin3Percent = (bin3 / binTotal) * 100;
cout << " Number of values in range 0.3-0.4: " << bin3 << endl;
cout << " Percentage of values in this interval: " << bin3Percent << "%" << endl;
cout << endl;
int bin4Percent = (bin4 / binTotal) * 100;
cout << " Number of values in range 0.4-0.5: " << bin4 << endl;
cout << " Percentage of values in this interval: " << bin4Percent << "%" << endl;
cout << endl;
int bin5Percent = (bin5 / binTotal) * 100;
cout << " Number of values in range 0.5-0.6: " << bin5 << endl;
cout << " Percentage of values in this interval: " << bin5Percent << "%" << endl;
cout << endl;
int bin6Percent = (bin6 / binTotal) * 100;
cout << " Number of values in range 0.6-0.7: " << bin6 << endl;
cout << " Percentage of values in this interval: " << bin6Percent << "%" << endl;
cout << endl;
int bin7Percent = (bin7 / binTotal) * 100;
cout << " Number of values in range 0.7-0.8: " << bin7 << endl;
cout << " Percentage of values in this interval: " << bin7Percent << "%" << endl;
cout << endl;
int bin8Percent = (bin8 / binTotal) * 100;
cout << " Number of values in range 0.8-0.9: " << bin8 << endl;
cout << " Percentage of values in this interval: " << bin8Percent << "%" << endl;
cout << endl;
int bin9Percent = (bin9 / binTotal) * 100;
cout << " Number of values in range 0.9-1.0: " << bin9 << endl;
cout << " Percentage of values in this interval: " << bin9Percent << "%" << endl;
cout << endl;
}
答案 0 :(得分：1)
M太大了，在long的限制范围内。因此，此M的任何long除以（或模数）将导致0. Perphaps您应该使用unsigned long long。
同时
而不是R = sqrt(-2 * log(x1));尝试R= sqrt(fabs(2 * log(x1)));
另外
phi = (2 * M_PI*x2);
u = R*cos(phi);
phi总是2 * PI的倍数，因此cos（phi）= 1。
答案 1 :(得分：1)
get_normal返回long，它不能介于0和1之间，因为它是一个整数。将函数返回的整数存储到double（Z）并不能神奇地恢复丢弃的小数部分。
我认为你应该在double中使用浮点运算（即get_normal s），并且还要更改返回类型。
顺便说一下，C ++标准库有很多随机数分布。您可能想要使用它而不是尝试编写自己的。