Hdu4945 2048 (DP)

First, read the wrong question .. After knowing the meaning of the question, I wrote DP... unlimited tle... but I really don't know how to optimize it. It's right to run the data all over, and it's just as a theoretical AC ..

#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 998244353ll 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;}