Luogu3067 平衡的奶牛群 Meet in the middle

题意：给出$N$个范围在$[1,10^8]$内的整数，问有多少种取数方案使得取出来的数能够分成两个和相等的集合。$N \leq 20$

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发现爆搜是$O(3^N)$的，所以考虑双向搜索。

先把前$3^\frac{N}{2}$搜完，然后每一次搜出后$3^\frac{N}{2}$的时候，枚举前面的$2^\frac{N}{2}$，每一个对应一下看有没有和为$0$的方案即可。复杂度为$O(6^\frac{N}{2})$，虽然不开O2过不去qwq

 #include<bits/stdc++.h>
 using namespace std;

 inline int read(){
     ;
     char c = getchar();
     while(!isdigit(c))
     c = getchar();
     while(isdigit(c)){
     a = (a << ) + (a << ) + (c ^ ');
     c = getchar();
     }
     return a;
 }

 struct HashTable{
 #define MOD 103
     struct node{
         int num;
         node* nxt;
     }*begin[MOD] , *last[MOD];
     void insert(int num){
         int t = num % MOD;
         )
             t += MOD;
         if(last[t] == NULL){
             begin[t] = new node;
             begin[t]->num = num;
             begin[t]->nxt = NULL;
             last[t] = begin[t];
         }
         else{
             node* now = new node;
             now->num = num;
             now->nxt = NULL;
             last[t]->nxt = now;
             last[t] = now;
         }
     }

     bool count(int num){
         int t = num % MOD;
         )
             t += MOD;
         for(node* i = begin[t] ; i != NULL ; i = i->nxt)
             if(i->num == num)
                 ;
         ;
     }
 }zt[ << ];
 ] , N , ans;
  << ][ << ];

 void init(int now , int end , int cnt , int sum){
     if(now > end){
         zt[cnt].insert(sum);
         return;
     }
     init(now +  , end , cnt , sum);
     init(now +  , end , cnt | ( << now) , sum + M[now]);
     init(now +  , end , cnt | ( << now) , sum - M[now]);
 }

 void getAns(int now , int end , int cnt , int sum){
     if(now > end){
      ; i <  << (N >> ) ; i++)
         if(!is[cnt][i] && (zt[i].count(sum) || zt[i].count(-sum))){
             ;
             ans++;
         }
     return;
     }
     getAns(now +  , end , cnt , sum);
     getAns(now +  , end , cnt | ( << now - (N >> )) , sum + M[now]);
     getAns(now +  , end , cnt | ( << now - (N >> )) , sum - M[now]);
 }

 int main(){
     N = read();
      ; i < N ; i++)
     M[i] = read();
     init( , (N >> ) -  ,  , );
     getAns(N >>  , N -  ,  , );
     cout << ans - ;
     ;
 }

再放一个复杂度似乎不对但是很快的方法

 #include<bits/stdc++.h>
 using namespace std;

 struct node{
     int zt , sum;
 }num1[] , num2[];
 ];
 ];

 void dfs(node* num , int& cnt , int now , int end , int sum , int zt){
     if(now > end){
         num[++cnt].sum = sum;
         num[cnt].zt = zt;
         return;
     }
     dfs(num , cnt , now +  , end , sum , zt);
     dfs(num , cnt , now +  , end , sum + M[now] , zt | ( << now));
     dfs(num , cnt , now +  , end , sum - M[now] , zt | ( << now));
 }

 bool cmp(node a , node b){
     return a.sum < b.sum;
 }

 bool operator == (node a , node b){
     return a.zt == b.zt && a.sum == b.sum;
 }

 int main(){
     cin >> N;
      ; i < N ; i++)
         cin >> M[i];
     dfs(num1 , cnt1 ,  , (N - ) >>  ,  , );
     dfs(num2 , cnt2 , N >>  , N -  ,  , );
     sort(num1 +  , num1 + cnt1 +  , cmp);
     sort(num2 +  , num2 + cnt2 +  , cmp);
     cnt1 = unique(num1 +  , num1 + cnt1 + ) - num1 - ;
     cnt2 = unique(num2 +  , num2 + cnt2 + ) - num2 - ;
     int p1 = cnt2 , p2 = cnt2;
      ; i <= cnt1 ; i++){
         )
             p1--;
         p2 = p1;
         )
             p2--;
         while(++p2 <= p1)
             vis[num1[i].zt | num2[p2].zt] = ;
     }
     ;
      ; i <  << N ; i++)
         ans += vis[i];
     cout << ans;
     ;
 }