P4859 已经没有什么好害怕的了(二项式反演)

Posted 2021-11-11 吃花椒的妙酱

传送门
题目大意：给定长度为n的两个数组a,b，没有两个相同的数，求使得ai > bi 对数比ai < bi对数多k对的求方案数，对1e9+9取余

思路：二项式反演

显然(n+k)为奇数时对数为0

设ai > bi 的对数为 h = (n+k)/2

设dp[i][j]为a数组中前i个数，有j个已经匹配的方案数，

则有转移方程dp [i] [j] = dp[i-1] [j] + dp[i-1] [j-1] * (last[i] - (j-1))，其中last[i]表示b数组中有多少个数比a[i]小

dp[n] [h]则表示n个数匹配h个的方案数，则至少有h个匹配的方案数为 dp[n] [h] * (n-h)!，剩下n-h个随便排

设g(i)为只是有i个匹配的方案数，f(i)表示恰好有i个匹配的方案数,则有
$$g(i)=\sum _{i=h}^n{C}_i^h\ast f(i)$$
我们要求的是f(k),二项式反演一波
$$f(k)=\sum _{i=h}^n(-1{)}^{i-h}\ast C_i^h\ast g(i)$$
即为所求

#define _CRT_SECURE_NO_WARNINGS
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <list>
#include <queue>
#include <vector>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <deque>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define _for(i,a,b) for(int i=(a) ;i<=(b) ;i++)
#define scd(v) scanf("%d",&v)
#define scdd(a,b) scanf("%d %d",&a,&b)
#define endl "\\n"
#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define pb push_back
#define all(v) v.begin(),v.end()
#define mst(v,a) memset(v,a,sizeof(v))
#define ls p<<1
#define rs p<<1|1
#define int long long
#define inf 0x7f7f7f7f
#define fi first
#define se second
#define pii pair<int , int >
#define ls p<<1
#define rs p<<1|1
#define lson p<<1,l,mid
#define rson p<<1|1,mid+1,r
#define AC return 0
const int N = 2e3  + 100;
const int mod = 1e9+9;
int n, m,k;
int a[N],b[N];
int dp[2010][2010];
int F[N], g[N],f[N],ni[N];
void de()
{
    _for(i, 1, 10)
    {
        cout << " i" <<i<< endl;
        _for(j, 1, i) cout << dp[i][j] << " ";
        cout << endl;
    }
}
ll qsm(int a, int b)
{
    int ans = 1, temp = a;
    while (b )
    { 
        if (b & 1) ans = (ans * temp) % mod;
        temp = (temp * temp) % mod;
        b >>= 1;
    }
    return ans;
}
int C(int n, int m)
{
    return f[n] * qsm(f[n - m], mod - 2) % mod * qsm(f[m], mod - 2) % mod;
}
void solve()
{
    dp[0][0] = 1;
    _for(i, 1, n)
    {
        int temp = lower_bound(b + 1, b + 1 + n, a[i]) - b - 1;
     //   cout << " temp " << temp << endl;
        dp[i][0] = 1;
        _for(j, 1, i)
        {
            dp[i][j] =( dp[i - 1][j] + dp[i - 1][j - 1] * (temp - (j - 1))%mod ) %mod;
        }
    }
  //  de();
   // _for(i, 1, n) cout << dp[i][k] << " ";
   // cout << endl;
    _for(i, k, n) g[i] = dp[n][i] * f[n - i] % mod;
    int ans = 0;
    _for(i, k, n)
    {
        int flag = 1;
        if ((i - k) & 1) flag = -1;
        ans += flag * g[i]%mod * C(i,k)%mod;
        ans = (ans % mod + mod) % mod;
    }
    cout << ans << endl;
}
signed main()
{
    //freopen("data.txt","r",stdin);
    IOS;
    f[0] = 1;
    _for(i, 1, 2000) f[i] = (f[i - 1] * i) % mod;
    cin >> n >> k;
    _for(i, 1, n) cin >> a[i];
    _for(i, 1, n) cin >> b[i];
    if ((n + k) & 1)
    {
        cout << 0 << endl;
        AC;
    }
    k = (n + k) / 2;
    sort(a + 1, a + 1 + n);
    sort(b + 1, b + 1 + n);
    solve();
    AC;
}