luogu P4859 已经没有什么好害怕的了

嘟嘟嘟


题中给的\(k\)有点别扭，我们转换成\(a > b\)的对数是多少，这个用二元一次方程解出来是\(\frac{n + k}{2}\)。


然后考虑dp，令\(dp[i][j]\)表示前\(i\)个数中，有\(j\)对满足\(a > b\)的方案数，转移的时候考虑这一组是否满足\(a > b\)即可：\(dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1] * (num[i] - (j - 1))\)。其中\(num[i]\)表示比\(a[i]\)小的\(b[i]\)的个数。


求完这个还没有完事，因为我们只保证了有\(j\)个满足\(a > b\)，而剩下的位置并不清楚。
于是令\(g[i] = dp[n][i] * (n - i)!\)，表示\(n\)组匹配中，至少有\(i\)组满足\(a > b\)的方案数，因为剩下的\(n - i\)个位置是瞎排的，所以不知道是否会出现\(a > b\)。


令\(f[i]\)表示恰好有\(i\)个匹配满足\(a > b\)，那么能列出\(g[k] = \sum _ {i = k} ^ {n} C_{i} ^ {k} f[i]\)（其实自己并不是十分懂这一步），然后通过二项式反演就可以求出\(f[k] = \sum _ {i = k} ^ {n} (-1) ^ {i - k} C_{i} ^ {k} g[i]\)。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
#include<assert.h>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 2e3 + 5;
const ll mod = 1e9 + 9;
In ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
In void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
In void MYFILE()
{
#ifndef mrclr
  freopen("ha.in", "r", stdin);
  freopen("ha.out", "w", stdout);
#endif
}

int n, K, a[maxn], b[maxn], num[maxn];

In ll inc(ll a, ll b) {return a + b < mod ? a + b : a + b - mod;}

ll fac[maxn], C[maxn][maxn];
In void init()
{
  fac[0] = 1;
  for(int i = 1; i <= n; ++i) fac[i] = fac[i - 1] * i % mod;
  C[0][0] = 1;
  for(int i = 1; i <= n; ++i)
    {
      C[i][0] = 1;
      for(int j = 1; j <= i; ++j) C[i][j] = inc(C[i - 1][j - 1], C[i - 1][j]);
    }
}

ll dp[maxn][maxn];

int main()
{
  MYFILE();
  n = read(), K = (n + read()) >> 1;
  init();
  for(int i = 1; i <= n; ++i) a[i] = read();
  for(int i = 1; i <= n; ++i) b[i] = read();
  sort(a + 1, a + n + 1), sort(b + 1, b + n + 1);
  for(int i = 1; i <= n; ++i) num[i] = lower_bound(b + 1, b + n + 1, a[i]) - b - 1;
  dp[0][0] = 1;
  for(int i = 1; i <= n; ++i)
    {
      dp[i][0] = dp[i - 1][0];
      for(int j = 1; j <= i; ++j)
    dp[i][j] = inc(dp[i - 1][j] % mod, dp[i - 1][j - 1] * (num[i] - j + 1) % mod);
    }
  ll ans = 0;
  for(int i = K; i <= n; ++i)
    {
      int flg = (i - K) & 1;
      ll tp = C[i][K] * fac[n - i] % mod * dp[n][i] % mod;
      ans = inc(ans, flg ? mod - tp : tp);
    }
  write(ans), enter;
  return 0;
}

原文地址：https://www.cnblogs.com/mrclr/p/10981378.html