牛客网暑期ACM多校训练营（第七场）Bit Compression

链接：https://www.nowcoder.com/acm/contest/145/C
来源：牛客网

题目描述
A binary string s of length N = 2n is given. You will perform the following operation n times :

- Choose one of the operators AND (&), OR (|) or XOR (^). Suppose the current string is S = s1s2...sk. Then, for all , replace s2i-1s2i with the result obtained by applying the operator to s2i- and s2i. For example, if we apply XOR to {} we get {}.

After n operations, the string will have length .

There are 3n ways to choose the n operations in total. How many of these ways will give  as the only character of the final string.
输入描述:
The first line of input contains a single integer n ( ≤ n ≤ ).

The next line of input contains a single binary string s (|s| = 2n). All characters of s are either  or .
输出描述:
Output a single integer, the answer to the problem.
示例1
输入

复制

输出

复制

说明

The sequences (XOR, OR), (XOR, AND), (OR, OR), (OR, AND) works.

官方题解把n<=4打印出来。这样就跑到4就结束了。不能只开一个数组，因为等于0不代表没跑到过。比如0000.跑到了ans还是0

#include<math.h>
#include<stdio.h>
#include<algorithm>
#include<string>
#include<iostream>
using namespace std;
const int maxn=<<;
int num[][maxn];
int vis[][maxn];
int val[][maxn];
int solve(int x)
{
 if(x==)
 return num[x][];
 int tmp=;
  if(x<=)
  {
      for(int i=;i<=<<x;i++)
      {
          tmp=(tmp<<)+num[x][i];
      }
      if(vis[x][tmp])
          return val[x][tmp];
  }
    int ans=;
    for(int i=,j=;i<=<<x;i+=){
        num[x-][++j]=num[x][i]^num[x][i+];
    }
    ans+=solve(x-);
    for(int i=,j=;i<=<<x;i+=){
        num[x-][++j]=num[x][i]&num[x][i+];
    }
    ans+=solve(x-);
    for(int i=,j=;i<=<<x;i+=){
        num[x-][++j]=num[x][i]|num[x][i+];
    }
    ans+=solve(x-);
    if(x<=){
        val[x][tmp]=ans;
        vis[x][tmp]=;
    }
    return ans;

}
int main()
{
    int n;
    string s;
    cin>>n>>s;
    for(int i=;i<<<n;i++)
    {
        num[n][i+]=s[i]-'';
    }
    printf("%d\n",solve(n));

    return ;
}