概率期望

$dp[i][j]$表示从$i,j$到$r,c$的期望，有$dp[i][j]=p_0\times dp[i][j]+p_1\times dp[i][j+1]+p_2\times dp[i+1][j]+2$

有$dp[i][j]=(p_1\times dp[i][j+1]+p_2\times dp[i+1][j]+2)\times \frac{1}{1-p_0}$，从后面开始转移。

#include <bits/stdc++.h>using namespace std;const double eps = 1e-7;const int N = 1e3 + 10;double dp[N][N], p[N][N][3];int n, m;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);while (scanf("%d %d", &n, &m) != EOF) {for (int i = 1; i <= n; i++) {for (int j = 1; j <= m; j++) {for (int k = 0; k < 3; k++) {scanf("%lf", &p[i][j][k]);}}}memset(dp, 0, sizeof dp);for (int i = n; i >= 1; i--) {for (int j = m; j >= 1; j--) {dp[i][j] = (p[i][j][1] * dp[i][j + 1] + p[i][j][2] * dp[i + 1][j] + 2) / (1 - p[i][j][0]);}}printf("%.3f\n", dp[1][1]);}return 0;
}

类似上一题，但是就是要每次碰到能跳的位置要一直跳转到不能再跳为止。

#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 10;double dp[N];int a[N], n, m;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);while (scanf("%d %d", &n, &m) && (n + m)) {memset(dp, 0, sizeof dp), memset(a, 0, sizeof a);for (int i = 1; i <= m; i++) {int x, y;scanf("%d %d", &x, &y);a[x] = y;}for (int i = n - 1; i >= 0; i--) {for (int j = 1; j <= 6; j++) {int cur = i + j;while (a[cur]) {cur = a[cur];}dp[i] += (dp[cur] + 1) / 6.0;}}printf("%.4f\n", dp[0]);}return 0;
}

$$\begin{gathered} 容易推出：E_i=\sum p_k{E}_{i+k}+p_0{E}_0+1 \\ 接下来用待定系数法 \\ {E}_0为我们要求的，假设E_i=a_i{E}_0+b_i \\ {E}_i=\sum p_k(a_{i+k}{E}_0+b_{i+k})+p_0{E}_0+1 \\ {E}_i=(\sum p_k{a}_{i+k}+p_0)E_0+\sum p_k{b}_{i+k}+1 \\ {a}_i=\sum p_k{a}_{i+k}+p_0,b_i=\sum p_k{b}_{i+k}+1 \end{gathered}$$

#include <bits/stdc++.h>using namespace std;const int N = 510;double A[N], B[N], p[20];int n, k1, k2, k3, a, b, c;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T;scanf("%d", &T);while (T--) {scanf("%d %d %d %d %d %d %d", &n, &k1, &k2, &k3, &a, &b, &c);memset(A, 0, sizeof A), memset(B, 0, sizeof B), memset(p, 0, sizeof p);p[0] = 1.0 / (k1 * k2 * k3);for (int i = 1; i <= k1; i++) {for (int j = 1; j <= k2; j++) {for (int k = 1; k <= k3; k++) {if (i == a && j == b && k == c) {continue;}p[i + j + k] += p[0];}}}for (int i = n; i >= 0; i--) {for (int j = 3; j <= k1 + k2 + k3; j++) {A[i] += p[j] * A[i + j];B[i] += p[j] * B[i + j];}A[i] += p[0];B[i] += 1;}printf("%.15f\n", B[0] / (1 - A[0]));}return 0;
}

每次共有四种情况发生：

• 这个$bug$出现在已有的子系统里，并且在已有的$bug$里，概率为$\frac{i}{s}\times \frac{i}{n}$
• 这个$bug$出现在已有的子系统里，但是是一个从未出现的新$bug$，概率为$\frac{i}{s}\times \frac{n-i}{n}$
• 这个$bug$出现在新的子系统里，但是是一个已有的$bug$，概率为$\frac{s-i}{s}\times \frac{i}{n}$
• 这个$bug$出现在新的子系统里，并且是一个新的$bug$，概率为$\frac{s-i}{s}\times \frac{n-i}{n}$

综上：我们定义$dp[i][j]$表示已经在$i$个子系统里出现过$j$个$bug$，到在$s$个子系统里出现过$n$个$bug$的步数期望，

$dp[i][j]=\frac{i\times j}{s\times n}dp[i][j]+\frac{i\times (n-j)}{s\times n}dp[i][j+1]+\frac{(s-i)\times j}{s\times n}dp[i+1][j]+\frac{(s-i)\times (s-j)}{s\times n}dp[i+1][j+1]$

#include <bits/stdc++.h>using namespace std;const int N = 1e3 + 10;double dp[N][N];int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int n, s;while (scanf("%d %d", &n, &s) != EOF) {memset(dp, 0, sizeof dp);for (int i = n; i >= 0; i--){for (int j = s; j >= 0; j--){if(i == n && j == s) {continue;}double factor = n * s - i * j;dp[i][j] = dp[i + 1][j] * j * (n - i) + dp[i][j + 1] * (s - j) * i + dp[i + 1][j + 1] * (n - i) *(s - j) + n * s;dp[i][j] /= factor;}}printf("%.4f\n", dp[0][0]);}return 0; 
}

容易想到$dp$转移方程$dp[i]$表示已经收集了$i$张卡到收集$n$张卡所需地步数期望，

$dp[i]=\sum p_k\times dp[i+k]+p_0\times dp[i]+1$，

$dp[i]=\frac{\sum p_k\times dp[i+k]+1}{1-p_0}$

$p_0+\sum p_k=1$

然后逆向进行$dp$即可。

#include <bits/stdc++.h>using namespace std;const int N = 25;double dp[1 << 21], p[N];int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int n;while (scanf("%d", &n) != EOF) {for (int i = 0; i < n; i++) {scanf("%lf", &p[i]);}memset(dp, 0, sizeof dp);for (int i = (1 << n) - 2; i >= 0; i--) {double delt = 0;for (int j = 0; j < n; j++) {if (i >> j & 1) {continue;}delt += p[j];dp[i] += p[j] * dp[i | (1 << j)];}dp[i] += 1;dp[i] /= delt;}printf("%.4f\n", dp[0]);}return 0; 
}

在每一层我们的期望应该是$a_0=\frac{n\times (n-1)}{2}$，我们每一次有$q=\frac{n-m}{n}$的概率继续游戏，

所以有如下$a_0,a_0\times q,a_0\times q^2,a_0\times a^3,\dots ,a_{n-1}\times q^{n-1},a_n\times a^n$。

$a_0\times \frac{1-q^n}{1-q},n$趋于无穷大时，$q^n=0$，有$\frac{a_0}{1-q}=\frac{a_0}{\frac{m}{n}}=\frac{n^2\times (n-1)}{2\times m}$。

所以期望为$\frac{n\times (n-1)}{2\times m}$。

#include <bits/stdc++.h>using namespace std;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);double n, m;scanf("%lf %lf", &n, &m);printf("%.6f\n", n * (n - 1) / 2 / m);return 0;
}

$$\begin{gathered} \sum _{i=1}^n=\frac{n^2+n}{2} \\ 所以我们分别求出n^2,n的期望即可，f[i]表示从i到n,n的期望,g[i]表示从i到n,n^2的期望。 \\ n的期望: \\ f[i]=\frac{i}{n}(f[i]+1)+\frac{n-i}{n}(f[i+1]+1) \\ {n}^2的期望 \\ g[i]=\frac{i}{n}(f[i]+1{)}^2+\frac{n-i}{n}(f[i+1]+1{)}^2 \\ f[i{]}^2=g[i],f[i+1{]}^2=g[i+1] \\ 化简有:f[i]=f[i+1]+\frac{n}{n-i} \\ g[i]=g[i+1]+2f[i+1]+2\frac{i}{n-i}f[i]+\frac{n}{n-i} \end{gathered}$$

#include <bits/stdc++.h>using namespace std;const int N = 1e4 + 10;double f[N], g[N];int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int n;scanf("%d", &n);for (int i = n - 1; i >= 0; i--) {f[i] = f[i + 1] + 1.0 * n / (n - i);g[i] = g[i + 1] + 2 * f[i + 1] + 2.0 * i / (n - i) * f[i] + 1.0 * n / (n - i);}printf("%.2f\n", (f[0] + g[0]) / 2);return 0;
}

#include <bits/stdc++.h>using namespace std;int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);double a[7], sum = 0;for (int i = 0; i < 7; i++) {cin >> a[i];sum += a[i];}double ans = 1;for (int i = 0; i < 6; i++) {ans = ans * (i + 1) * a[i] / (sum - i);}ans *= a[6];printf("%.3f\n", ans * 7.0);return 0;
}