一道概率问题的编程解

问题：投掷21次硬币，至少一次连续5次或5次以上正面朝上的概率是多少？
我的解答：（brute函数是水木上另一个人给出的解）
#include <cstdio>
#include <cstdlib>
#include <memory.h>
#include <cmath>
#include <vector>
using namespace std;

template<int n, int m> int brute()
{
int c = 0;
for(unsigned int i = 0; i < (1 << n); i++)
{
for(unsigned int t = i; t != 0; t >>= 1)
{
if( (t & ((1 << m) - 1)) == ((1 << m) - 1))
{
c++;
break;
}
}
}
return c;
}

template<int n, int m> double probability()
{
int c = brute<n, m>();
return (double)c / (1 << n);
}
// continuous M from N element
float getPossibility(int N, int M) {
int n = M-1;
vector< vector<int> > f(2, vector<int>(M,0));

for (int i = 0; i < M-1; ++i) {
f[0][i] = 1<<(M-i-2);
}
f[0][M-1] = 1;

int cur = 0;

while (n < N) {
n++;
int last = cur;
cur = (cur+1)%2;

int sum = f[last][4];
for (int cnt = 1; cnt < M; ++cnt) {
f[cur][cnt] = f[last][cnt-1];
sum += f[last][cnt-1];
}
f[cur][0] = sum;
}
int ret = 0;
for (int cnt = 0; cnt < M; ++cnt) {
ret += f[cur][cnt];
}
return 1.0-((float)ret)/(1<<N);
}
int main()
{
float p = getPossibility(5, 5);
printf("%d, %f, %f/n", brute<5, 5>(), probability<5, 5>(), p);
p = getPossibility(6, 5);
printf("%d, %f, %f/n", brute<6, 5>(), probability<6, 5>(), p );
p = getPossibility(21, 5);
printf("%d, %f, %f/n", brute<21, 5>(), probability<21, 5>(), p );
return 0;
}
简单的说明：计算所有5次以下的事件的个数。计算过程可用动态规划进行。