HDU4945 2048(dp)

先是看错题意。。然后知道题意之后写了发dp..无限TLE..实在是不知道怎么优化了，跑了遍数据是对的，就当作理论AC掉好了。。

#pragma warning(disable:4996)

#include <iostream>

#include <cstring>

#include <string>

#include <vector>

#include <cstdio>

#include <queue>

#include <algorithm>

#include <cmath>

#include <ctime>

using namespace std;



#define ll long long

#define maxn 120000

#define mod 998244353



ll mod_pow(ll a, ll n){

	ll ret = 1;

	while (n){

		if (n & 1) ret = ret*a%mod;

		a = a*a%mod;

		n >>= 1;

	}

	return ret;

}



ll fac[maxn];

ll fac_inv[maxn];



int cnt[2500];

int dp[13][2500];

int two[13] = { 0, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048 };

int two_com[13] = { 0, 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1 };

int n;



inline int getint() {

	int ret = 0; bool ok = 0;

	for (;;) {

		int c = getchar();

		if (c >= '0'&&c <= '9')ret = (ret << 3) + ret + ret + c - '0', ok = 1;

		else if (ok)return ret;

	}

}



inline ll comb(int n, int m){

	return fac[n] * fac_inv[m] % mod*fac_inv[n - m] % mod;

}

inline void add(int &a, int b){

	a += b;

	if (a >= mod) a -= mod;

}



int main()

{

	//freopen("1001.in", "r", stdin);

	//freopen("out.txt", "w", stdout);

	//double t1 = clock();

	fac[0] = fac_inv[0] = 1;

	for (int i = 1; i <= 100000; ++i){

		fac[i] = fac[i - 1] * i%mod;

	}

	fac_inv[100000] = mod_pow(fac[100000], mod - 2);

	for (int i = 99999; i >= 0; --i){

		fac_inv[i] = fac_inv[i + 1] * (i + 1) % mod;

	}

	int ca = 0;

	while (~scanf("%d", &n) && n){

		for (int i = 1; i <= 12; ++i) cnt[two[i]] = 0;

		int tmp;

		for (int i = 0; i < n; ++i) {

			tmp = getint();

			cnt[tmp]++;

		}

		int pn = 0;

		for (int i = 1; i <= 12; ++i) pn += cnt[two[i]];

		memset(dp, 0, sizeof(dp)); dp[0][0] = 1;

		ll sum,cb;

		for (int i = 1; i <= 12; ++i){

			int num = cnt[two[i]];

			for (int j = 0; j <= two_com[i - 1]; ++j){

				sum = 0;

				int k;

				for (k = 0; (j >> 1) + k <= two_com[i] && k <= num; ++k){

					cb = comb(num, k);

					sum = sum + cb; if (sum >= mod) sum -= mod;

					add(dp[i][(j >> 1) + k], dp[i - 1][j] * cb%mod);

				}

				if ((j >> 1) + num > two_com[i]){

					ll res = ((mod_pow(2, num) - sum) % mod + mod) % mod;

					add(dp[i][two_com[i]], res*dp[i - 1][j] % mod);

				}

			}

		}

		ll ans = dp[12][1] * mod_pow(2, n - pn) % mod;

		printf("Case #%d: %I64d\n", ++ca, ans);

	}

	//double t2 = clock();

	//cout << t2 - t1 << endl;

	return 0;

}