【数论】[逆元，错排]P4071排列计数

题目描述

求有多少种长度为n的系列A，满足以下条件：

1~n这n个数在序列中各出现一次；若第i个数a[i]的值为i，则称i是稳定的。序列恰有m个数是稳定的。

输出序列个数对1e9+7取模的结果。

Solution

显然是从N个数中选m个数稳定，剩下的错排。答案即为：(C^m_n * d[n - m])

(C^m_n = frac{n!}{m!(n - m)!} = n! * (m!)^{p - 2} * ((n - m)!)^{p - 2}，p = 1e9 +7)

(d[n] = (n - 1) (d[n - 1] + d[n - 2]))

#include <iostream>
#include <cstdio>
using namespace std;
inline long long read() {
  long long x = 0; int f = 0; char c = getchar();
  while (c < '0' || c > '9') f |= c == '-', c = getchar();
  while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  return f? -x : x;
}

const int mod = 1e9 + 7;
const int maxn = 1000006;
int T, n, m;
long long d[maxn], c[maxn];
inline long long pow(long long a, long long k) {//快速幂
  long long x = 1;
  while (k) {
    if (k & 1) x = x * a % mod;
    a = a * a % mod; k >>= 1;
  }
  return x;
}
int main() {
  T = read();
  c[0] = 1;//预处理
  for (int i = 1; i <= maxn; ++i) c[i] = c[i - 1] * i % mod;
  d[1] = 0; d[0] = 1;
  for (int i = 2; i <= maxn; ++i)
    d[i] = (d[i - 1] + d[i - 2]) % mod * (i - 1) % mod;
  while (T--) {
    n = read(); m = read();
    printf("%lld
",d[n - m] % mod * 
      c[n] * pow(c[m], mod - 2) % mod * pow(c[n - m], mod - 2) % mod);
  }
  return 0;
}