【HDOJ】4652 Dice

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
+ \frac{m-1}{m}dp[1] \notag \\
dp
&= 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
)
\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
&= m^{n-1}
\end{align}
显然是一个等比数列，累加后可得$dp[0]-dp
=dp[0]$.
\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
\notag \\
dp
&= 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
)
\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

/* 4652 */
#include <iostream>
#include <sstream>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <deque>
#include <bitset>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <climits>
#include <cctype>
#include <cassert>
#include <functional>
#include <iterator>
#include <iomanip>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,1024000")

#define sti                set<int>
#define stpii            set<pair<int, int> >
#define mpii            map<int,int>
#define vi                vector<int>
#define pii                pair<int,int>
#define vpii            vector<pair<int,int> >
#define rep(i, a, n)     for (int i=a;i<n;++i)
#define per(i, a, n)     for (int i=n-1;i>=a;--i)
#define clr                clear
#define pb                 push_back
#define mp                 make_pair
#define fir                first
#define sec                second
#define all(x)             (x).begin(),(x).end()
#define SZ(x)             ((int)(x).size())
#define lson            l, mid, rt<<1
#define rson            mid+1, r, rt<<1|1

#define LL __int64

int t;
int op, n, m;

LL Pow(LL base, int n) {
LL ret = 1;

while (n) {
if (n & 1)
ret *= base;
n >>= 1;
base *= base;
}

return ret;
}

void solve() {
double ans = 0.0;

if (op) {
double tmp = 1.0;
rep(i, 0, n) {
tmp = tmp * m / (m-i);
ans += tmp;
}
} else {
LL fz = Pow(m, n) - 1;
ans = fz / (m-1);
}
printf("%.9lf\n", ans);
}

int main() {
ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
freopen("data.in", "r", stdin);
freopen("data.out", "w", stdout);
#endif

while (scanf("%d", &t)!=EOF) {
while (t--) {
scanf("%d%d%d", &op, &m, &n);
solve();
}
}

#ifndef ONLINE_JUDGE
printf("time = %d.\n", (int)clock());
#endif

return 0;
}