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;
}

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