牛客网暑期ACM多校训练营（第七场）C：Bit Compression（思维）

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

时间限制：C/C++ 2秒，其他语言4秒
空间限制：C/C++ 262144K，其他语言524288K
Special Judge, 64bit IO Format: %lld
题目描述
A binary string s of length $N = 2^n$ is given. You will perform the following operation n times :

• Choose one of the operators AND (&), OR (|) or XOR (^). Suppose the current string is $S = s_1s_2...s_k$. Then, for all $1\le i \le \frac{k}{2}$ , replace $s_{2i-1}s_{2i}$ with the result obtained by applying the operator to $s_{2i-1}$ and $s_{2i}$. For example, if we apply XOR to {1101} we get {01}.

After n operations, the string will have length 1.

There are $3^n$ ways to choose the n operations in total. How many of these ways will give 1 as the only character of the final string.
输入描述:
The first line of input contains a single integer $n (1 ≤ n ≤ 18)$.

The next line of input contains a single binary string $s (|s| = 2^n)$. All characters of $s$ are either 0 or 1.
输出描述:
Output a single integer, the answer to the problem.
示例1
输入
2
1001
输出
4
说明
The sequences (XOR, OR), (XOR, AND), (OR, OR), (OR, AND) works.

题解：暴力的复杂度是$3^n$的，只需要再优化一点点！
• 如果剩下了16个变量，可能性只有65536个。
• 预处理他们，得到以个$3^{n-4}$的做法。
• 事实上，直接暴力，常数优化即可

#include<bits/stdc++.h>
using namespace std;
const int MAX=1e6+10;
const int MOD=1e9+7;
typedef long long ll;
char s[20][MAX];
ll d[5][MAX];
ll cal(int x,int n)
{
    if(d[n][x]!=-1)return d[n][x];
    if(n==0)return d[0][x]=x;
    ll tot=0,y=0;
    for(int i=0,j=0;i<(1<<n);i+=2,j++)y+=(1<<j)*(((x>>i)&1)^(x>>(i+1)&1));tot+=cal(y,n-1);y=0;
    for(int i=0,j=0;i<(1<<n);i+=2,j++)y+=(1<<j)*(((x>>i)&1)&(x>>(i+1)&1));tot+=cal(y,n-1);y=0;
    for(int i=0,j=0;i<(1<<n);i+=2,j++)y+=(1<<j)*(((x>>i)&1)|(x>>(i+1)&1));tot+=cal(y,n-1);y=0;
    return d[n][x]=tot;
}
ll dfs(int n)
{
    if(n<=4)
    {
        int x=0;
        for(int i=0;i<(1<<n);i++)x+=(1<<i)*(s[n+1][i]-'0');
        return d[n][x];
    }
    ll tot=0;
    for(int i=0,j=0;i<(1<<n);i+=2,j++)s[n][j]=((s[n+1][i]-'0')^(s[n+1][i+1]-'0'))+'0';tot+=dfs(n-1);
    for(int i=0,j=0;i<(1<<n);i+=2,j++)s[n][j]=((s[n+1][i]-'0')|(s[n+1][i+1]-'0'))+'0';tot+=dfs(n-1);
    for(int i=0,j=0;i<(1<<n);i+=2,j++)s[n][j]=((s[n+1][i]-'0')&(s[n+1][i+1]-'0'))+'0';tot+=dfs(n-1);
    return tot;
}
int main()
{
    memset(d,-1,sizeof d);
    for(int i=0;i<(1<<16);i++)cal(i,4);//预处理
    int n;
    cin>>n;
    scanf("%s",s[n+1]);
    cout<<dfs(n)<<endl;
    return 0;
}