HDU 4945 2048 DP 组合

思路：

　　这个题写了一个背包的解法，超时了。搜了下题解才发现我根本不会做。

　　思路参见这个：

　　其实我们可以这样来考虑，求补集，用全集减掉不能组成2048的集合就是答案了。

　　因为只要达到2048就可以了，所以求补集会大大减小枚举的次数。

代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <cmath>
 6 #include <algorithm>
 7 #include <string>
 8 #include <queue>
 9 #include <stack>
10 #include <vector>
11 #include <map>
12 #include <set>
13 #include <functional>
14 #include <cctype>
15 #include <time.h>
16
17 using namespace std;
18
19 typedef __int64 ll;
20
21 const int INF = 1<<30;
22 const int MAXN = 1e5+55;
23 const int MAXM = 2055;
24 const ll MOD = 998244353;
25
26 int cnt[MAXM];
27 ll dp[20][MAXM];
28 ll fact[MAXN], inv[MAXN];
29 int n;
30
31 ll myPow(ll x, int d) {
32     ll res = 1;
33     while (d>0) {
34         if (d&1) res = (res*x)%MOD;
35         x = (x*x)%MOD;
36         d >>= 1;
37     }
38     return res;
39 }
40
41 inline ll C(int x, int y) {
42     return ((fact[x]*inv[y])%MOD * inv[x-y])%MOD;
43 }
44
45 inline int lowBit(int x) {
46     return x&(-x);
47 }
48
49 void solve() {
50     int sum = n;
51     for (int i = 0; i < 12; i++) sum -= cnt[1<<i];
52
53     memset(dp, 0, sizeof(dp));
54
55     for (int j = min(2048, cnt[1]); j >= 0; j--) dp[0][j] = C(cnt[1], j);
56
57     for (int i = 1; i < 12; i++) {
58         int cc = 2048>>i;
59         for (int k = min(cc, cnt[1<<i]); k >= 0; k--) { //选k个
60             ll CC = C(cnt[1<<i], k);
61             for (int j = k; j <= cc; j++) { //
62                 dp[i][j] = (dp[i][j] + (((dp[i-1][(j-k)<<1]+dp[i-1][(j-k)<<1|1])%MOD)*CC)%MOD )%MOD;
63             }
64         }
65     }
66
67     ll ans = ((myPow(2, n-sum)-dp[11][0])%MOD+MOD)%MOD;
68     ans = (ans*myPow(2, sum))%MOD;
69     printf("%I64d\n", ans);
70 }
71
72 int main() {
73     #ifdef Phantom01
74         freopen("HDU4945.txt", "r", stdin);
75     #endif //Phantom01
76
77     fact[0] = 1;
78     for (int i = 1; i < MAXN; i++) fact[i] = (fact[i-1]*i)%MOD;
79     inv[MAXN-1] = myPow(fact[MAXN-1], MOD-2);
80     for (int i = MAXN-1; i > 0; i--) inv[i-1] = (inv[i]*i)%MOD;
81
82     int T = 1;
83
84     while (scanf("%d", &n)!=EOF && n!=0) {
85         printf("Case #%d: ", T++);
86         memset(cnt, 0, sizeof(cnt));
87         int x;
88         for (int i = 0; i < n; i++) {
89             scanf("%d", &x);
90             cnt[x]++;
91         }
92         solve();
93     }
94
95     return 0;
96 }