题面：Luogu
题解：期望dp+gauss消元先吐槽一下这个题面，根本看不懂，看讨论区半天才发现，另外m个就是用来降低概率的
设\(prob_i\)表示一轮降了\(i\)点的概率
\[Prob_i=\frac{\binom{k}{i}m^{k-i}}{(m+1)^k}\]
总共\((m+1)^{k}\)种方案，选\(i\)次降一点，剩下的不降
由于\(k\)很大，所以\(\binom{k}{i}\)不能通过预处理阶乘算出，而要递推（否则会#6~#10RE）
递推方法很多，我的是
\[prob_i=\frac{1}{i!}*m^{k-i}*\frac{k!}{(k-i)!}*\frac{1}{(m+1)^k}\]
设\(Q_{x,y}\)表示一轮操作从\(x\)点变成\(y\)点
\[Q_{x,y}=\begin{split}
\left\{
\begin{array}{lr}
0 & x=n \&\& y=n+1 &\\
Prob_{x-y} & x=n\\
\frac{1}{m+1}Prob_0 & y=x+1\\
\frac{m}{m+1}Prob_{x-y}+\frac{1}{m+1}Prob_{x-y+1} & \text{otherwise}
\end{array}
\right.
\end{split}
\]
设\(dp_i\)表示从\(i\)点变成\(0\)的期望步数
\[
\begin{split}
& dp_i=1+\sum_{j=1}^{i+1} Q_{i,j}dp_j & (1\le i<n) \\
& dp_n=1+\sum_{j=1}^nQ_{n,j}dp_j
\end{split}
\]
于是可以推一下式子
\[dp_{i+1}=\frac{dp_i-1-\sum_{j=1}^{i}{Q_{i,j}dp_{j}}}{Q_{i,i+1}} (1\le i<n)\]
可以发现，我们只需求出\(dp_1\)即可求出其它
然后是两种办法
1.高斯消元
矩阵大概是这样
\[
\left[
\begin{matrix}
1-Q_{1,1} & -Q_{1,2} & 0 &\cdots & 0\\
-Q_{2,1} & 1-Q_{2,2} & -Q_{2,3} &\cdots & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots \\
-Q_{n-1,1} & -Q_{n-1,2} & -Q_{n-1,3} &\cdots & -Q_{n-1,n} \\
-Q_{n,1} & -Q_{n,2} & -Q_{n,3} &\cdots & 1-Q_{n,n}
\end{matrix}
\right]*
\left[
\begin{matrix}
dp_{1}&\\
dp_{2}&\\
\vdots&\\
dp_{n-1}&\\
dp_{n}&
\end{matrix}
\right]=
\left[
\begin{matrix}
1&\\
1&\\
\vdots&\\
1&\\
1&
\end{matrix}
\right]
\]
然后可以通过一些手法把它搞成只要\(n^2\)的
2.设\(dp_1=X\)
则每一项都可以表示为\(A[i]*X+B[i]\)
然后通过\(dp_n\)的递推式也能得到一个\(dp_n=An*X+Bn\)
所以
\[A[n]*X+B[n]=An*X+Bn \\
X=\frac{Bn-B[n]}{A[n]-An}
\]
然后带回去就可以了
我也是写的这种方法
还有一件令我迷惑的事：prob每次不清零会WA#1~#5，但是我写的每次应该都会重新算一遍，在没被改变的地方应该也没有被调用
#include<bits/stdc++.h>
using namespace std;
template<typename T>
inline void read(T& x)
{
x = 0; char c = getchar();
while (!isdigit(c)) c = getchar();
while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
}
#define maxn 1505
#define P 1000000007
int fac[maxn], inv[maxn], prob[maxn];
int A[maxn], B[maxn], An, Bn, X;
int n, p, m, k, invM;
inline int mul(int x, int y) { return 1ll * x * y % P; }
inline int qpow(int x, int y)
{
int ans = 1;
for (; y; x = mul(x, x), y >>= 1) if (y & 1) ans = mul(ans, x);
return ans;
}
inline int getp(int x, int y)//Q_{x,y}
{
if (x == n && y == n + 1) return 0;
if (x == n) return prob[x - y];
if (y == x + 1) return mul(invM, prob[0]);
return (0ll + mul(mul(m, prob[x - y]), invM) + mul(invM, prob[x - y + 1])) % P;
}
inline void exgcd(int a, int b, int& x, int& y)
{
if (!b) { x = 1, y = 0; return; }
exgcd(b, a % b, y, x);
y -= a / b * x;
}
inline int getinv(int a)
{
int x, y;
exgcd(a, P, x, y);
x %= P;
return (x < 0) ? x + P : x;
}
inline void init(int maxnum = 1500)
{
fac[0] = inv[0] = 1;
for (int i = 1; i <= maxnum; ++i) fac[i] = 1ll * fac[i - 1] * i % P;
inv[maxnum] = getinv(fac[maxnum]);
for (int i = maxnum - 1; i; --i) inv[i] = 1ll * inv[i + 1] * (i + 1) % P;
}
inline void getprob()
{
invM = getinv(m + 1);
int inv1 = qpow(invM, k);
for (int i = 0, tp = qpow(m, k), tp2 = getinv(m), tpm = 1; i <= min(n, k); ++i)
prob[i] = mul(mul(mul(inv[i], tpm), tp), inv1), tp = mul(tp, tp2), tpm = mul(tpm, k - i);
//tp=m^{k-i},inv[i]=\frac{1}{i!},tpm=\frac{k!}{(k-i)!},inv1=\frac{1}{(m+1)^k}
}
inline void solve()
{
A[1] = 1, B[1] = 0; An = 0, Bn = 1;
for (int i = 2; i <= n; ++i)
{
A[i] = A[i - 1], B[i] = B[i - 1] - 1;
for (int j = 1, tp; j < i; ++j)
{
tp = getp(i - 1, j);
A[i] = (0ll + A[i] - mul(tp, A[j])) % P;
B[i] = (0ll + B[i] - mul(tp, B[j])) % P;
}
if (A[i] < 0) A[i] += P;
if (B[i] < 0) B[i] += P;
int tp = getinv(getp(i - 1, i));
A[i] = mul(A[i], tp), B[i] = mul(B[i], tp);
}
for (int i = 1, tp; i <= n; ++i)
{
tp = getp(n, i);
An = (0ll + An + mul(tp, A[i])) % P;
Bn = (0ll + Bn + mul(tp, B[i])) % P;
}
X = mul((Bn - B[n] + P) % P, getinv((A[n] - An + P) % P));
printf("%d\n", (mul(A[p], X) + B[p]) % P);
memset(prob, 0, sizeof(prob));
}
int main()
{
int T; read(T);
init();
while (T--)
{
read(n), read(p), read(m), read(k);
if (!k || (!m && k == 1)) { puts("-1"); continue; }//特判
if (!m)
{
int ans = 0;
while (p > 0)
{
if (p < n) ++p;
p -= k, ++ans;
}
printf("%d\n", ans);
continue;
}
getprob();
solve();
}
return 0;
}
来源：https://www.cnblogs.com/123789456ye/p/12536762.html