题解 嗑瓜子

传送门

按题意DP即可

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 2010
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n;
ll inv[N*3];
const ll mod=998244353;

namespace task1{
	ll dp[N][N<<1];
	void solve() {
		for (int i=1; i<=n; ++i) {
			for (int j=0; j<=2*(n-i); ++j) {
				dp[i][j]=(i*inv[i+j]%mod*dp[i-1][j+2]%mod+j*inv[i+j]%mod*dp[i][j-1]%mod+1)%mod;
			}
		}
		printf("%lld\n", dp[n][0]);
		exit(0);
	}
}

signed main()
{
	freopen("eat.in", "r", stdin);
	freopen("eat.out", "w", stdout);

	n=read();
	inv[0]=inv[1]=1;
	for (int i=2; i<N*3; ++i) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
	task1::solve();

	return 0;
}