1. 题目描述
对于m面的骰子。有两种查询，查询0表示求最后n次摇骰子点数相同的期望；查询1表示最后n次摇骰子点数均不相同的期望。
2. 基本思路
由期望DP推导，求得最终表达式。
(1) 查询0
不妨设$dp[k]$表示当前已经有k次相同而最终实现n次相同的期望。
\begin{align}
dp[0] &= 1 + dp[1] \notag \\
dp[1] &= 1 + \frac{1}{m}dp[2] + \frac{m-1}{m}dp[1] \notag \\
dp[2] &= 1 + \frac{1}{m}dp[3] + \frac{m-1}{m}dp[1] \notag \\
&\cdots \notag \\
dp[n-1] &= 1 + \frac{1}{m}dp[n] + \frac{m-1}{m}dp[1] \notag \\
dp[n] &= 0
\end{align}
两两相减可得。
\begin{align}
dp[0]-dp[1] &= \frac{1}{m}(dp[1]-dp[2]) \notag \\
dp[1]-dp[2] &= \frac{1}{m}(dp[2]-dp[3]) \notag \\
dp[2]-dp[3] &= \frac{1}{m}(dp[3]-dp[4]) \notag \\
&\cdots \notag \\
dp[n-2]-dp[n-1] &= \frac{1}{m}(dp[n-1]-dp[n])
\end{align}
由$dp[0]=1+dp[1], dp[0]-dp[1]=1$代入可得。
\begin{align}
dp[0]-dp[1] &= 1 \notag \\
dp[1]-dp[2] &= m \notag \\
dp[2]-dp[3] &= m^2 \notag \\
&\cdots \notag \\
dp[n-1]-dp[n] &= m^{n-1}
\end{align}
显然是一个等比数列，累加后可得$dp[0]-dp[n]=dp[0]$.
\begin{align}
dp[0] = \frac{m^n-1}{m-1}
\end{align}
（2）查询1
不妨设$dp[k]$表示当前已经有k个不同而最终实现n个不同的期望。
\begin{align}
dp[0] &= 1 + dp[1] \notag \\
dp[1] &= 1 + \frac{1}{m}dp[1] + \frac{m-1}{m}dp[2] \notag \\
dp[2] &= 1 + \frac{1}{m}dp[1] + \frac{1}{m}dp[2] + \frac{m-2}{m}dp[3] \notag \\
&\cdots \notag \\
dp[n-1] &= 1 + \Sigma_{i=1}^{n-1}{\frac{1}{m}dp[i]} + \frac{m-(n-1)}{m}dp[n] \notag \\
dp[n] &= 0
\end{align}
两两相减可得。
\begin{align}
dp[0]-dp[1] &= \frac{m-1}{m}(dp[1]-dp[2]) \notag \\
dp[1]-dp[2] &= \frac{m-2}{m}(dp[2]-dp[3]) \notag \\
&\cdots \notag \\
dp[n-2]-dp[n-1] &= \frac{m-(n-1)}{m}(dp[n-1]-dp[n])
\end{align}
由$dp[0]=1+dp[1], dp[0]-dp[1]=1$代入累加后可得。
\begin{align}
dp[0] = \Sigma_{i=1}^{n} \frac{m^i}{\prod_{j=0}^{i-1}(m-j)}
\end{align}
3. 代码
1 /* 4652 */
2 #include <iostream>
3 #include <sstream>
4 #include <string>
5 #include <map>
6 #include <queue>
7 #include <set>
8 #include <stack>
9 #include <vector>
10 #include <deque>
11 #include <bitset>
12 #include <algorithm>
13 #include <cstdio>
14 #include <cmath>
15 #include <ctime>
16 #include <cstring>
17 #include <climits>
18 #include <cctype>
19 #include <cassert>
20 #include <functional>
21 #include <iterator>
22 #include <iomanip>
23 using namespace std;
24 //#pragma comment(linker,"/STACK:102400000,1024000")
25
26 #define sti set<int>
27 #define stpii set<pair<int, int> >
28 #define mpii map<int,int>
29 #define vi vector<int>
30 #define pii pair<int,int>
31 #define vpii vector<pair<int,int> >
32 #define rep(i, a, n) for (int i=a;i<n;++i)
33 #define per(i, a, n) for (int i=n-1;i>=a;--i)
34 #define clr clear
35 #define pb push_back
36 #define mp make_pair
37 #define fir first
38 #define sec second
39 #define all(x) (x).begin(),(x).end()
40 #define SZ(x) ((int)(x).size())
41 #define lson l, mid, rt<<1
42 #define rson mid+1, r, rt<<1|1
43
44 #define LL __int64
45
46 int t;
47 int op, n, m;
48
49 LL Pow(LL base, int n) {
50 LL ret = 1;
51
52 while (n) {
53 if (n & 1)
54 ret *= base;
55 n >>= 1;
56 base *= base;
57 }
58
59 return ret;
60 }
61
62 void solve() {
63 double ans = 0.0;
64
65 if (op) {
66 double tmp = 1.0;
67 rep(i, 0, n) {
68 tmp = tmp * m / (m-i);
69 ans += tmp;
70 }
71 } else {
72 LL fz = Pow(m, n) - 1;
73 ans = fz / (m-1);
74 }
75 printf("%.9lf\n", ans);
76 }
77
78 int main() {
79 ios::sync_with_stdio(false);
80 #ifndef ONLINE_JUDGE
81 freopen("data.in", "r", stdin);
82 freopen("data.out", "w", stdout);
83 #endif
84
85 while (scanf("%d", &t)!=EOF) {
86 while (t--) {
87 scanf("%d%d%d", &op, &m, &n);
88 solve();
89 }
90 }
91
92 #ifndef ONLINE_JUDGE
93 printf("time = %d.\n", (int)clock());
94 #endif
95
96 return 0;
97 }