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 }