I didn't come up with a positive solution during the exam, and finally I scored a violent cheat. The violence was good... I searched every state, and then I enumerated the violence. If I have time, I still have to play.

It's true, but I'm too konjaku to fight...

And then the positive solution is the search.

First understand a property

No matter what the operation is, as long as each operation is determined, no matter how the order is changed, the result will not be changed.

On the proof, you can think of each position as a block, and each moving block moves with it, regardless of the order result.

Then the question can be divided into four situations

Set the interval length in the search to 2^k, so the current interval length of 2^k-1 should be incremented;

When there is a 2^k interval that does not meet the increasing condition at this time, it will be interchanged internally

When there are two, it is necessary to cross call in four situations (at this time, it is better not to use array to store the number of blocks to be operated, because it will be backtracked after searching)

Search directly for the next when there is no time

When there are too many return s, they can't be changed.

Code:

  1 #include<iostream>
  2 #include<algorithm>
  3 #include<cstdio>
  4 #include<cmath>
  5 #include<cstring>
  6 #include<vector>
  7 #include<string>
  8 #define MAXN 21000
  9 #define pt printf("---------\n");
 10 using namespace std;
 11 int tot;
 12 bool bian[MAXN];
 13 int base[MAXN];int n;
 14 int a[MAXN];
 15 int jie[MAXN];
 16 void qian(int k,int x,int y)//2^k Long paragraph x Paragraph 1 y paragraph
 17 {
 18     int star=(x-1)*base[k]+1;
 19     int la=x*base[k];
 20     for(int i=star;i<=la;++i)
 21     {
 22        swap(a[i],a[i+(y-x)*base[k]]);
 23     }
 24 }
 25 int sum;
 26 vector<int>wr;vector<int>ci;
 27 int work(int kx,int xx,int yy,int cc1,int cc2)
 28 {
 29         qian(kx-1,xx,yy);
 30         int star_1=(cc1-1)*base[kx]+base[kx-1];
 31         int star_2=(cc2-1)*base[kx]+base[kx-1];
 32         //printf("2--search a[%d]=%d a[%d]=%d wr0%d wr1=%d %d\n",star_1,a[star_1],star_2,a[star_2],xx,yy,kx+1);
 33         if(a[star_1]+1!=a[star_1+1]||a[star_2]+1!=a[star_2+1])
 34         {
 35                 qian(kx-1,xx,yy);
 36                 return 0;
 37         }
 38         return 1;
 39 }
 40 void DFS(int kx,int cnt)//Search length is 2^k The interval is then used cnt One operation
 41 {
 42     if(kx>n)
 43     {
 44         for(int i=1;i<=base[n];++i)
 45         {if(a[i-1]+1!=a[i]) return ;}
 46         sum+=jie[cnt];
 47        // printf("sum=%d cnt=%d\n",sum,cnt);
 48         return ;
 49     }
 50    // printf("kx=%d cnt=%d\n",kx,cnt);
 51     int duan=base[n-kx];
 52     wr.clear();ci.clear();
 53     for(int k=1;k<=duan;++k)
 54     {
 55         int star=(k-1)*base[kx]+base[kx-1];//   ([][*]|[*][])
 56         if(a[star]+1!=a[star+1])
 57         {
 58     //        printf("The first%d paragraph a[%d]=%d a[%d]=%d duan=%d %d\n",k,star,a[star],star+1,a[star+1],k*2-1,k*2);
 59             wr.push_back(k*2-1);
 60             wr.push_back(k*2);
 61             ci.push_back(k);
 62         }
 63     }
 64     if(wr.size()>4)
 65     {
 66        return ;
 67     }
 68     if(wr.size()==2)
 69     {
 70         qian(kx-1,wr[0],wr[1]);
 71         int star=(ci[0]-1)*base[kx]+base[kx-1];
 72         if(a[star]+1!=a[star+1])
 73         {  
 74             qian(kx-1,wr[0],wr[1]);
 75             return ;
 76         }
 77         else
 78         {
 79             int xx=wr[0],yy=wr[1];
 80      //       printf("1--search a[%d]=%d a[%d]=%d wr0%d wr1=%d %d\n",star,a[star],star+1,a[star+1],wr[0],wr[1],kx+1);
 81             DFS(kx+1,cnt+1);
 82             qian(kx-1,xx,yy);
 83         }
 84     }
 85     if(wr.size()==4)
 86     {
 87          int xx1=wr[0],xx2=wr[1],yy1=wr[2],yy2=wr[3];
 88          int cc1=ci[0];int cc2=ci[1];
 89          if(work(kx,xx1,yy1,cc1,cc2)==1)
 90          {
 91              DFS(kx+1,cnt+1);
 92              qian(kx-1,xx1,yy1);
 93     //         printf("2----%d %d\n",xx1,yy1);
 94          }
 95          if(work(kx,xx1,yy2,cc1,cc2)==1)
 96          {
 97              DFS(kx+1,cnt+1);
 98              qian(kx-1,xx1,yy2);
 99    //          printf("2----%d %d\n",xx1,yy2);        
100          }
101          if(work(kx,xx2,yy1,cc1,cc2)==1)
102          {
103             DFS(kx+1,cnt+1);
104             qian(kx-1,xx2,yy1);
105     //        printf("2----%d %d\n",xx2,yy1);   
106          }
107          if(work(kx,xx2,yy2,cc1,cc2)==1)
108          {
109             DFS(kx+1,cnt+1);
110             qian(kx-1,xx2,yy2);
111    //         printf("2----%d %d\n",xx2,yy2);   
112          }
113     }
114     if(wr.size()==0)
115     {
116         DFS(kx+1,cnt);
117     }
118     return ;
119 }
120 int main()
121 {
122      scanf("%d",&n);
123      base[0]=1;jie[0]=1;
124      for(int i=1;i<=n;++i)
125      {
126         base[i]=base[i-1]*2;jie[i]=jie[i-1]*i;
127      } 
128      for(int i=1;i<=base[n];++i)
129      {
130         scanf("%d",&a[i]);
131      }
132      DFS(1,0);
133      printf("%d\n",sum);
134 }