Question

这应该掷6个骰子，并记录有多少是唯一的。例如：1 2 3 4 5 6 = 6个唯一编号，1 1 1 1 1 1 = 1个唯一编号，1 2 3 3 3 3 = 3个唯一编号。我每掷一个骰子的比例都很高，约为16.7％。

#include <iostream>
#include <cstdlib>
#include <ctime>
using namespace std;

int numberGenerator()
{
int x = (rand() % 6) + 1;
return x;
}

int diceCounter()
{

int counter[6] = {0,0,0,0,0,0};

for (int i = 0; i < 6; i++)
    {
    int k = numberGenerator();
        if (k == 1)
           counter[0] = 1;
        if (k == 2)
           counter[1] = 1;
        if (k == 3)
           counter[2] = 1;
        if (k == 4)
           counter[3] = 1;
        if (k == 5)
           counter[4] = 1;
        if (k == 6)
           counter[5] = 1;
     }
return counter[0]+counter[1]+counter[2]+counter[3]+counter[4]+counter[5];
}                 // returns how many unique numbers were rolled


int main()
{
srand(time(NULL));
cout << diceCounter() << endl;

int a1 = 0;
int a2 = 0;
int a3 = 0;
int a4 = 0;
int a5 = 0;
int a6 = 0;
int p;


for (int j = 0; j < 1000000; j++)
{
p = numberGenerator();        // for number of unique numbers, it adds +1 to counter
if (p == 1)
a1 += 1;
if (p == 2)
a2 += 1;
if (p == 3)
a3 += 1;
if (p == 4)
a4 += 1;
if (p == 5)
a5 += 1;
if (p == 6)
a6 += 1;
}
cout << "1 ==> " << (a1 / 1000000.0)*100 << "%" << endl;
cout << "2 ==> " << (a2 / 1000000.0)*100 << "%" << endl;
cout << "3 ==> " << (a3 / 1000000.0)*100 << "%" << endl;
cout << "4 ==> " << (a4 / 1000000.0)*100 << "%" << endl;
cout << "5 ==> " << (a5 / 1000000.0)*100 << "%" << endl;
cout << "6 ==> " << (a6 / 1000000.0)*100 << "%" << endl;

                // this results in uniform 16.7% percentages
}

Answer 1

在您的循环中，您正在调用numberGenerator而不是diceCounter。

因此，不是计算滚动6个骰子的唯一结果数，而是计算单个掷骰的每个数字的计数。正如你所期望的那样，每个数字都会出现1/6的时间。

Answer 2

添加@ MichaelAnderson的回答：

随机数实际上都是伪随机数。

他们处理seed值。因此，每次拨打rand()时，请先调用srand()重新设置随机数生成器：像这样：

srand(time(NULL));
int r = rand();

为什么这些骰子结果如此均匀？

2 个答案: