hdu 6377 度度熊看球赛 (dp)

大意: $n$对情侣, $2n$个座位, 对于一个方案, 若$k$对情侣相邻, 则喧闹值增加$D^k$, 求喧闹值期望.

跟CF 840C一样, 设$dp[i][j]$为$i$个人, 有$j$对情侣相邻, 枚举每个人转移即可.

#include <iostream>
#include <cstdio>
#define REP(i,a,n) for(int i=a;i<=n;++i)
using namespace std;
typedef long long ll;
const int N = 2e3+10, P = 998244353;
int n, d, dp[N][N], fac[N][N];

int main() {
	dp[0][0] = 1;
	REP(i,0,2000) REP(j,0,i/2) if (dp[i][j]) {
		int &r = dp[i][j];
		if (i&1) {
			dp[i+1][j] = (dp[i+1][j]+(ll)(i+1-j-2)*r)%P;
			dp[i+1][j+1] = (dp[i+1][j+1]+2ll*r)%P;
			if (j) dp[i+1][j-1] = (dp[i+1][j-1]+(ll)j*r)%P;
		}
		else {
			dp[i+1][j] = (dp[i+1][j]+(ll)(i+1-j)*r)%P;
			if (j) dp[i+1][j-1] = (dp[i+1][j-1]+(ll)j*r)%P;
		}
	}
	REP(i,1,1000) {
		fac[i][0] = 1;
		REP(j,1,1000) fac[i][j] = (ll)fac[i][j-1]*i%P;
	}
	while (~scanf("%d%d", &n, &d)) {
		int ans = 0;
		REP(i,0,n) { 
			ans = (ans+(ll)dp[2*n][i]*fac[d][i])%P;
		}
		printf("%d\n", ans);
	}
}