二项分布是从N条伯努利路径中获得n次成功的离散概率分布Pp(n | N)(具有x = 0和x = 1标记的两个可能结果)。x = 1是成功,x = 0是成功发生的概率为p,失败发生的概率为q = 1 – p。)因此,二项分布可以写为
$$P_ {p} \ lgroup n \:\ arrowvert \ N \ rgroup = \ left(\ begin {array} {c} N \\ n \ end {array} \ right)p ^ {n} \ lgroup1-p \ rgroup ^ {Nn} $$
#include <iostream>
#include <random>
using namespace std;
int main(){
const int nrolls = 10000; // number of rolls
const int nstars = 100; // maximum number of stars to distribute
default_random_engine generator;
binomial_distribution<int> distribution(9,0.5);
int p[10]={};
for (int i=0; i<nrolls; ++i) {
int number = distribution(generator);
p[number]++;
}
cout << "binomial_distribution (9,0.5):" << endl;
for (int i=0; i<10; ++i)
cout << i << ": " << string(p[i]*nstars/nrolls,'*') << endl;
}输出结果
0: 1: * 2: ****** 3: *************** 4: ************************* 5: ************************ 6: **************** 7: ******* 8: * 9: