例题

一个模型

一件事情成功的概率为p，则期望的第一次成功的次数是多少？

答：1/p

第一次成功的概率为$p$

第二次成功的概率为$(1-p)\times p$

第三次成功的概率为 $(1-p)^2 \times p$

…

期望次数 $E(x) = p + 2\times (1-p)\times p$ + $3 \times (1-p)^2 \times p + \ldots = \sum_{k\ge 1}{k(1-p)^{k-1}}p$

设$f(x) = \sum_{k\ge 1}{x^k}=\frac x{1-x}(\lvert x \rvert <1)$

求导得，$f’(x) =\sum_{k\ge 1}{kx^{k-1}}=\frac{(1-x)-x\times (-1)}{(1-x)^2}=\frac 1 {(1-x)^2}$

$\Rightarrow f’(x-1)=\sum_{k\ge 1}{k(x - 1)^{k-1}}=\frac 1 {x^2}$

$\Rightarrow E(x) = f’(1-p) \times p = \frac p {p^2}=\frac 1 p$

Kevin的抽奖黑幕

这里终点不唯一，但起点唯一，所以从后往前推，最后的状态固定。

令$dp[i][j]$为当前是第$i$轮，已经$j$轮未获奖，到$n$轮抽奖结束获得奖品数量的期望。

状态转移：

最后的答案即$dp[0][0] \times n$(因为$dp[0][0]$表示一个人的期望，有n个人所以乘上n)

参考代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
#include<bits/stdc++.h>
#define endl '\n'
#define int long long
using namespace std;
using ll = long long;

bool multi = 1;

template<class T>
constexpr T power(T a, long long b) {
    T res = 1;
    for (; b; b /= 2, a *= a) {
        if (b % 2) {
            res *= a;
        }
    }
    return res;
}

constexpr long long mul(long long a, long long b, long long p) {
    long long res = a * b - (long long)(1.L * a * b / p) * p;
    res %= p;
    if (res < 0) {
        res += p;
    }
    return res;
}
template<long long P>
struct MLong {
    long long x;
    constexpr MLong() : x{} {}
    constexpr MLong(long long x) : x{norm(x % getMod())} {}

    static long long Mod;
    constexpr static long long getMod() {
        if (P > 0) {
            return P;
        } else {
            return Mod;
        }
    }
    constexpr static void setMod(long long Mod_) {
        Mod = Mod_;
    }
    constexpr long long norm(long long x) const {
        if (x < 0) {
            x += getMod();
        }
        if (x >= getMod()) {
            x -= getMod();
        }
        return x;
    }
    constexpr long long val() const {
        return x;
    }
    explicit constexpr operator long long() const {
        return x;
    }
    constexpr MLong operator-() const {
        MLong res;
        res.x = norm(getMod() - x);
        return res;
    }
    constexpr MLong inv() const {
        assert(x != 0);
        return power(*this, getMod() - 2);
    }
    constexpr MLong &operator*=(MLong rhs) & {
        x = mul(x, rhs.x, getMod());
        return *this;
    }
    constexpr MLong &operator+=(MLong rhs) & {
        x = norm(x + rhs.x);
        return *this;
    }
    constexpr MLong &operator-=(MLong rhs) & {
        x = norm(x - rhs.x);
        return *this;
    }
    constexpr MLong &operator/=(MLong rhs) & {
        return *this *= rhs.inv();
    }
    friend constexpr MLong operator*(MLong lhs, MLong rhs) {
        MLong res = lhs;
        res *= rhs;
        return res;
    }
    friend constexpr MLong operator+(MLong lhs, MLong rhs) {
        MLong res = lhs;
        res += rhs;
        return res;
    }
    friend constexpr MLong operator-(MLong lhs, MLong rhs) {
        MLong res = lhs;
        res -= rhs;
        return res;
    }
    friend constexpr MLong operator/(MLong lhs, MLong rhs) {
        MLong res = lhs;
        res /= rhs;
        return res;
    }
    friend constexpr std::istream &operator>>(std::istream &is, MLong &a) {
        long long v;
        is >> v;
        a = MLong(v);
        return is;
    }
    friend constexpr std::ostream &operator<<(std::ostream &os, const MLong &a) {
        return os << a.val();
    }
    friend constexpr bool operator==(MLong lhs, MLong rhs) {
        return lhs.val() == rhs.val();
    }
    friend constexpr bool operator!=(MLong lhs, MLong rhs) {
        return lhs.val() != rhs.val();
    }
};

template<>
long long MLong<0LL>::Mod = (long long)(1E18) + 9;

template<int P>
struct MInt {
    int x;
    constexpr MInt() : x{} {}
    constexpr MInt(long long x) : x{norm(x % getMod())} {}
  
    static int Mod;
    constexpr static int getMod() {
        if (P > 0) {
            return P;
        } else {
            return Mod;
        }
    }
    constexpr static void setMod(int Mod_) {
        Mod = Mod_;
    }
    constexpr int norm(int x) const {
        if (x < 0) {
            x += getMod();
        }
        if (x >= getMod()) {
            x -= getMod();
        }
        return x;
    }
    constexpr int val() const {
        return x;
    }
    explicit constexpr operator int() const {
        return x;
    }
    constexpr MInt operator-() const {
        MInt res;
        res.x = norm(getMod() - x);
        return res;
    }
    constexpr MInt inv() const {
        assert(x != 0);
        return power(*this, getMod() - 2);
    }
    constexpr MInt &operator*=(MInt rhs) & {
        x = 1LL * x * rhs.x % getMod();
        return *this;
    }
    constexpr MInt &operator+=(MInt rhs) & {
        x = norm(x + rhs.x);
        return *this;
    }
    constexpr MInt &operator-=(MInt rhs) & {
        x = norm(x - rhs.x);
        return *this;
    }
    constexpr MInt &operator/=(MInt rhs) & {
        return *this *= rhs.inv();
    }
    friend constexpr MInt operator*(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res *= rhs;
        return res;
    }
    friend constexpr MInt operator+(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res += rhs;
        return res;
    }
    friend constexpr MInt operator-(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res -= rhs;
        return res;
    }
    friend constexpr MInt operator/(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res /= rhs;
        return res;
    }
    friend constexpr std::istream &operator>>(std::istream &is, MInt &a) {
        long long v;
        is >> v;
        a = MInt(v);
        return is;
    }
    friend constexpr std::ostream &operator<<(std::ostream &os, const MInt &a) {
        return os << a.val();
    }
    friend constexpr bool operator==(MInt lhs, MInt rhs) {
        return lhs.val() == rhs.val();
    }
    friend constexpr bool operator!=(MInt lhs, MInt rhs) {
        return lhs.val() != rhs.val();
    }
};

template<>
int MInt<0>::Mod = 998244353;

template<int V, int P>
constexpr MInt<P> CInv = MInt<P>(V).inv();

constexpr int P = 998244353;//模数
using Z = MInt<P>;

//a ^ b mod p
int qpow(int a,int b,int p){
    long long res=1%p;
    while(b){
        if(b&1) res=res*a%p;
        a=(long long)a*a%p;
        b>>=1;
    }
    return res;
}
const int N = 2010;
int n, m, k, d;
Z dp[N][N];
void solve() {
    cin >> n >> m >> k >> d;
    for(int i = 0; i <= d; i++) {
        dp[m][i] = 0;
    }
    Z p1 = k / (Z)n, p2 = 1 - p1;
    for(int i = m - 1; i >= 0; i--) {
        for(int j = d - 1; j >= 0; j--) {
            if(j == d - 1) dp[i][j] = dp[i + 1][0] + 1;
            else dp[i][j] = (dp[i + 1][0] + 1) * p1 + dp[i + 1][j + 1] * p2;
        }
    }
    
    cout << dp[0][0] * n << '\n';
}

signed main() {
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int T = 1;
    if(multi) cin >> T;
    while(T--) {
        solve();
    }

    return 0;
}