Codeforces 1151F Sonya and Informatics
题意：
有一个数列，只有\(0\)和\(1\)，每一次操作选择两个数交换位置，求\(k\)次操作之后这个数列为非递减数列的概率是多少
思路：
假设一共有\(m\)个\(0\)，和\(n - m\)个\(1\)
\(f[i][j]\)表示到第\(i\)个操作，前\(m\)个数字中有\(j\)个\(1\)的方案数
有以下转移：
- 前m个数字随便取两个交换，或者后(n - m)个数字随便取两个交换，不会改变j
\[\begin{eqnarray*}
f[i][j] += f[i - 1][j] * ({m \choose 2} + {n - m \choose 2})
\end{eqnarray*}
\]
- 前m个数字中的1和后(n - m)个数字中的1进行交换
\[\begin{eqnarray*}
f[i][j] += f[i - 1][j] * j * (n - m - j)
\end{eqnarray*}
\]
- 前m个数字中的0和后(n - m)个数字中的0进行交换
\[\begin{eqnarray*}
f[i][j] += f[i - 1][j] * (m - j) * j
\end{eqnarray*}
\]
- 前m个数字中取一个0和后(n - m)个数字中的1进行交换
\[\begin{eqnarray*}
f[i][j] += f[i - 1][j - 1] * (m - j + 1) * (n - m - j + 1)
\end{eqnarray*}
\]
- 前m个数字中取一个1和后(n - m)个数字中的0进行交换
\[\begin{eqnarray*}
f[i][j] += f[i - 1][j + 1] * (j + 1) * (j + 1)
\end{eqnarray*}
\]
然后矩阵快速幂加速递推。
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define N 110
const ll p = (ll)1e9 + 7;
int n, m, k, a[N];
ll f[N][N], C[N][N];
ll qmod(ll base, ll n) {
ll res = 1;
while (n) {
if (n & 1) {
res = res * base % p;
}
base = base * base % p;
n >>= 1;
}
return res;
}
struct node {
int len;
ll a[N][N];
node () {}
node (int len) {
this->len = len;
memset(a, 0, sizeof a);
}
node operator * (const node &other) const {
node res = node(len);
for (int i = 0; i <= len; ++i) {
for (int j = 0; j <= len; ++j) {
for (int k = 0; k <= len; ++k) {
(res.a[i][j] += a[i][k] * other.a[k][j] % p) %= p;
}
}
}
return res;
}
}base, res;
void qmod(node &res, node &base, ll n) {
while (n) {
if (n & 1) {
res = res * base;
}
base = base * base;
n >>= 1;
}
}
int main() {
memset(C, 0, sizeof C);
C[1][0] = C[1][1] = 1;
for (int i = 1; i <= 100; ++i) {
for (int j = 0; j <= i; ++j) {
(C[i + 1][j + 1] += C[i][j]) %= p;
(C[i + 1][j] += C[i][j]) %= p;
}
}
while (scanf("%d%d", &n, &k) != EOF) {
m = 0;
for (int i = 1; i <= n; ++i) {
scanf("%d", a + i);
m += (a[i] == 0);
}
base = node(m); res = node(m);
if (m == 0) {
res.a[0][0] = 1;
}
for (int i = 1, j = 0; i <= m; ++i) {
j += (a[i] == 1);
if (i == m) {
res.a[0][j] = 1;
}
}
for (int i = 0; i <= m; ++i) {
(base.a[i][i] += C[m][2] + C[n - m][2]) %= p;
(base.a[i][i] += i * (n - m - i) % p) %= p;
(base.a[i][i] += (m - i) * i % p) %= p;
if (i != 0) {
(base.a[i - 1][i] += (m - i + 1) * (n - m - i + 1) % p) %= p;
}
if (i != m) {
(base.a[i + 1][i] += (i + 1) * (i + 1) % p) %= p;
}
}
qmod(res, base, k);
ll a = res.a[0][0], b = 0;
for (int i = 0; i <= m; ++i) {
(b += res.a[0][i]) %= p;
}
printf("%lld\n", a * qmod(b, p - 2) % p);
/*
memset(f, 0, sizeof f);
for (int i = 1, j = 0; i <= m; ++i) {
j += (a[i] == 1);
if (i == m) {
f[0][j] = 1;
}
}
for (int i = 1; i <= k; ++i) {
for (int j = 0; j <= m; ++j) {
(f[i][j] += f[i - 1][j] * (C[m][2] + C[n - m][2]) % p) %= p;
(f[i][j] += f[i - 1][j] * j % p * (n - m - j) % p) %= p;
(f[i][j] += f[i - 1][j] * (m - j) % p * j % p) %= p;
if (j != 0) {
(f[i][j] += f[i - 1][j - 1] * (m - j + 1) % p * (n - m - j + 1) % p) %= p;
}
if (j != m) {
(f[i][j] += f[i - 1][j + 1] * (j + 1) % p * (j + 1) % p) %= p;
}
}
}
ll a = 0, b = f[k][0];
for (int i = 0; i <= m; ++i) {
(a += f[k][i]) %= p;
}
printf("%lld\n", b * qmod(a, p - 2) % p);
*/
}
return 0;
}
