Codeforces #564div2 E1（记忆化搜索）

虽然不是正解但毕竟自己做出来了还是记下来吧～
对每个人分别dfs得到其期望，某两维的组合情况有限所以Hash一下避免了MLE。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;

const int maxn = 51, mod = 998244353;
int n, m, like[maxn], w[maxn];
int sum[2], cnt[2], val[maxn * maxn][maxn * maxn], dp[maxn][maxn << 1][5000][2];//in fact, 2500 is ok
unordered_map<long long, int> mp;
int hashcnt;

int ksm(int a, int b) {
	int res = 1;
	for (; b; b >>= 1) {
		if (b & 1)	res = 1LL * res * a % mod;
		a = 1LL * a * a % mod;
	}
	return res;
}

int calc(int wi, int sum) {//wi / sum
	if (val[wi][sum] != -1)	return val[wi][sum];
	return val[wi][sum] = 1LL * wi * ksm(sum, mod - 2) % mod;	
}

int Hash(int x, int y) {//[suma][sumb] is MLE, so hash it
	long long t = 10000LL * x + y;
	return mp.count(t) ? mp[t] : mp[t] = hashcnt++;
}

int dfs(int depth, int wi, int suma, int sumb, int like) {
	if (depth == m + 1) {
		return wi;
	}
	int &x = dp[depth][wi][Hash(suma, sumb)][like];
	if (x != -1)	return x;
	int add = like ? 1 : -1, ts = 0;
	//select this
	if (wi > 0)	ts = (1LL * calc(wi, suma + sumb) * dfs(depth + 1, wi + add, suma + add, sumb, like) + ts) % mod;
	//select one the same as it
	if (suma > wi)	ts = (1LL * calc(suma - wi, suma + sumb) * dfs(depth + 1, wi, suma + add, sumb, like) + ts) % mod;
	//select one from opposite
	if (sumb > 0)	ts = (1LL * calc(sumb, suma + sumb) * dfs(depth + 1, wi, suma, sumb - add, like) + ts) % mod;
	return x = ts;
}

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 0; i < n; i++) {
		scanf("%d", &like[i]);
		cnt[like[i]]++;
	}
	for (int i = 0; i < n; i++) {
		scanf("%d", &w[i]);
		sum[like[i]] += w[i];//sum[0] sum[1]
	}
	memset(val, -1, sizeof val);
	memset(dp, -1, sizeof dp);
	for (int i = 0; i < n; i++) {
		printf("%d\n", dfs(1, w[i], sum[like[i]], sum[like[i] ^ 1], like[i]));
	}
}