bzoj3622

容斥原理

看见恰好k个就要容斥

g[i]表示有几个b比a小

dp[i][j]表示前i个数至少有j个大的方案数，dp[i][j]=dp[i-1][j]+dp[i-1][j-1]*(g[i]-j+1),就是可以不匹配，或者在剩下的g[i]-j+1选一个

然后就是容斥了，那个系数搞的不是很清楚，和spring一样

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = , P = 1e9 + ;
int n, k;
ll ans;
ll a[N], b[N], inv[N], facinv[N], fac[N], dp[N][N], g[N];
ll C(int n, int m)
{
    return fac[n] * facinv[m] % P * facinv[n - m] % P;
}
int main()
{
    scanf("%d%d", &n, &k);
    for(int i = ; i <= n; ++i) scanf("%d", &a[i]);
    for(int i = ; i <= n; ++i) scanf("%d", &b[i]);
    if((n + k) & )
    {
        puts("");
        return ;
    }
    k += (n - k) >> ;
    sort(a + , a + n + );
    sort(b + , b + n + );
    inv[] = ; fac[] = facinv[] = ;
    for(int i = ; i <= n; ++i)
    {
        if(i != ) inv[i] = (P - P / i) * inv[P % i] % P;
        facinv[i] = facinv[i - ] * inv[i] % P;
        fac[i] = fac[i - ] * i % P;
    }
    for(int i = , j = ; i <= n; ++i)
    {
        while(a[i] > b[j + ] && j +  <= n) ++j;
        g[i] = j;
    }
    for(int i = ; i <= n; ++i) dp[i][] = ;
    for(int i = ; i <= n; ++i)
        for(int j = ; j <= g[i]; ++j)
            dp[i][j] = (dp[i - ][j] + dp[i - ][j - ] * (g[i] - j + ) % P) % P;
    for(int i = k; i <= n; ++i) ans = ((ans + (((i - k) & ) ? - : ) * dp[n][i] * fac[n - i] % P * C(i, k) % P) % P + P) % P;
    printf("%lld\n", ans);
    return ;
}