牛客多校第七场 C Bit Compression 思维

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

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...s_k. Then, for all , replace s_2i-1s_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 3ⁿ 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

复制

2
1001

输出

复制

4

说明

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


题意：给你2^n个数，每次可以将前2^(n-1)个数和后2^(n-1)个数进行二进制与、或、异或操作中的一种，问有多少种不同的操作顺序使得最后的结果为一个一？
分析：考虑直接暴力的时间复杂度是3^18，这个数与题目给出的时间限制已经很接近，所以我们只需要在直接暴力的基础上做一点点优化就可以了
　　　考虑如果最后要得到一个一，则数列中一定要还存在一，否则无论与、或、异或操作都无法再得到一
　　　所以我们只需要在暴力的时候判断下剩下的数中是否还有一，没有一就没必要继续操作
AC代码：

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e6 + 10;
const double eps = 1e-8;
const ll mod = 1e9 + 7;
const ll inf = 1e9;
const double pi = acos(-1.0);
string s;
ll t[maxn*2]; //开始将数字存在1<<n后，每次数字合并后产生的1<<(n-1)往前面放，方便合并运算
ll res = 0;
void dfs( ll n ) {
    if( n == -1 ) {
        if( t[2] == 1 ) { //只剩一个一
            res ++;
        }
        return ;
    }
    ll k = 1<<n;
    for( ll j = 0; j <= 2; j ++ ) {
        ll cnt = 0;
        for( ll i = 1; i <= k; i ++ ) {
            ll tmp = i+k;
            if( j == 0 ) {
                t[tmp] = t[tmp*2-1]^t[tmp*2];
            } else if( j == 1 ) {
                t[tmp] = t[tmp*2-1]|t[tmp*2];
            } else {
                t[tmp] = t[tmp*2-1]&t[tmp*2];
            }
            if( t[tmp] == 0 ) {
                cnt ++;
            }
        }
        if( cnt == k ) { //如果都是零就不可能再出现一
            continue;
        }
        dfs(n-1);
    }
}
int main() {
    ll n;
    cin >> n;
    cin >> s;
    ll k = 1<<n;
    for( ll i = 1; i <= k; i ++ ) {
        t[i+k] = s[i-1] - '0';
    }
    dfs(n-1);
    cout << res << endl;
    return 0;
}