luogu P4859 已经没有什么好害怕的了

Posted 2021-12-22 mrclr

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\\(\\fracn + k2\\)。


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;