codeforces 1156F Card Bag

題解： dp[i][j] 表示第i次使用j個數的方案數， 在求總數的時候不要忘記補充win後的方案

代碼如下：

#include<bits/stdc++.h>
using namespace std;
const int MAX_N = 5e3 + 5;
typedef long long ll;
ll a[MAX_N], dp[MAX_N][MAX_N], r[MAX_N], rf[MAX_N], f[MAX_N];
const ll mod = 998244353;
ll mod_pow(int x, int n) {
	ll res = 1;
	while (n) {
		if (n & 1) res = 1ll * res*x%mod;
		x = 1ll * x*x%mod; n >>= 1;
	}
	return res;
}

int main() {
	r[1] = 1;
	for (int i = 2; i < MAX_N; i++) r[i] = r[mod%i] * (mod - mod / i)%mod;
	rf[0] = 1; f[0] = 1;
	for (int i = 1; i < MAX_N; i++) {
		rf[i] = r[i] * rf[i - 1] % mod;
		f[i] = i * f[i - 1] % mod;
	}
	int n; cin >> n;
	for (int x, i = 0; i < n; i++) {
		cin >> x; a[x]++;
	}
	ll win_times = 0;
	dp[0][0] = 1;
	for (int i = 1; i <= n; i++) {
		for (int j = 0; j <= i; j++) {
			if(j>=1)
				dp[i][j] = dp[i - 1][j - 1] * a[i]%mod;
			if (a[i] > 1) win_times = (win_times + dp[i][j] * (a[i] - 1)%mod*f[n-j-1]%mod)% mod;
			dp[i][j] = (dp[i][j] + dp[i - 1][j]) % mod;
		}
	}
	cout << win_times*mod_pow(f[n],mod-2)%mod << endl;
	//system("pause");
}