牛客多校第七场C Bit Compression （dfs+剪枝）

 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 1 s 2 ...s k . Then,
for all , 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.

题意 给定字符串 求出进行xor & | 操作后，有多少种能是最后的结果是1。

题解 如果是暴力的话，需要3的18次，会超时，稍微剪下枝，就能过了 ，于是发现剩下的数字全为零时 三个操作都无法得到1，dfs+剪枝就过了。

#include<bits/stdc++.h>
using namespace std;
const double PI=acos(-1.0);
typedef long long ll;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define clr(x) memset(x,0,sizeof(x))
const int INF=0x3f3f3f3f;
const int maxn=1e6+5;
char s[maxn];
bool a[20][maxn];
int n;
int ans;
void dfs(int pos,int len,int op){
	int k=len>>1;
	int sum0=0;
	if(op==1)
	rep(i,1,k+1){
			a[pos+1][i]=a[pos][i<<1]|a[pos][i<<1-1];
			if(a[pos+1][i]==0){
				sum0++;
			}
	}
	if(op==2)
	rep(i,1,k+1){
			a[pos+1][i]=a[pos][i<<1]&a[pos][i<<1-1];
			if(a[pos+1][i]==0){
				sum0++;
			}
	}
	if(op==3)
	rep(i,1,k+1){
			a[pos+1][i]=a[pos][i<<1]^a[pos][i<<1-1];
			if(a[pos+1][i]==0){
				sum0++;
			}
	}
	if(len==2){
		ans+=a[pos+1][1];
		return ;
	}
	if(sum0==k)
	return;
	dfs(pos+1,k,1);
	dfs(pos+1,k,2);
	dfs(pos+1,k,3);
}
int main(){
	//本地测试
	#ifdef ONLINE_JUDGE
	#else
    freopen("C://Users//yan//Desktop//in.txt","r",stdin);
	#endif
	ios::sync_with_stdio(false);//取消同步
	std::cin.tie(0);//解除cin与cout的绑定，进一步加快执行效率。
	cin>>n;
	n=pow(2,n);
	cin>>s;
	rep(i,1,n+1){
		a[0][i]=s[i-1]-'0';
	}
	ans=0;
	dfs(0,n,1);
	dfs(0,n,2);
	dfs(0,n,3); 
	cout<<ans;
	
}