彩色圆环:
题目大意:
一个环上有 (n) 个点,每个点随机染为 (m) 种颜色之一。求环上同色连续段长度之积的期望值。
思路:
破环为链,就有 (f_{i,[0,1]}) 表示到第 (i) 个数,环首尾是否同种颜色的期望值。则有:
[egin{aligned}
f_{i,1}&=sum_{j=0}^{i-1} frac{p_{i-j}cdot f_{j,0} cdot (i - j)}{m}\
f_{i,0}&=sum_{j=0}^{i-1} p_{i-j}cdot (i - j)left(frac{(m-2)f_{i,0}}{m} + frac{(m-1)f_{i,1}}{m}
ight)
end{aligned}]
其中 (p_i=m^{-i}),即连续 (i) 个相同颜色的概率。
代码:
const int N = 210;
inline ll Read()
{
ll x = 0, f = 1;
char c = getchar();
while (c != '-' && (c < '0' || c > '9')) c = getchar();
if (c == '-') f = -f, c = getchar();
while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
return x * f;
}
double f[N][2], p[N], ans, m;
int n;
int main()
{
n = Read(); scanf ("%lf", &m);
p[1] = 1.0;
m = 1.0 / m;
for (int i = 2; i <= n; i++)
p[i] = p[i - 1] * m;
f[0][1] = 1.0;
for (int i = 1; i <= n; i++)
for (int j = 0; j < i; j++)
f[i][1] += p[i - j] * (i - j) * f[j][0] * m,
f[i][0] += p[i - j] * (i - j) * (f[j][0] * (1 - 2 * m) + f[j][1] * (1 - m));
double ans = p[n] * n;
for (int i = 1; i < n; i++)
ans += 1.0 * i * i * f[n - i][0] * p[i]; //处理首尾不相接
printf ("%.10lf
", ans);
return 0;
}