map--Bit Compression

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

Bit Compression

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 = 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-1 and s2i. 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| = 2n). 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：给出一个01串，每次从&、|、^里面选一个，对这个01串进行相应的操作，问最后得到1有多少中方法。

方法：就是思维。记录得到某个串有几种方法，再由这个串出发，得到的方法累加。

不懂的话可以手动运行3 11001100这个样例，再结合我上面说的记录得到某个串几种方法，递推下去。

int main()
{
    map<int,string>m;//第一个是关键字，first,键。第二个是键值。
    m[0]="休闲无心";
    m[1]="大好河山";
    m[2]="木青子";
    m[3]="红颜祸水";
    map<int,string>::iterator  Map_iter;
    for(Map_iter=m.begin();Map_iter!=m.end();Map_iter++)
    {
        cout<<"key:"<<Map_iter->first<<" "<<"map:"<<Map_iter->second<<endl;
    }
}

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#include<bits/stdc++.h>
using namespace std;
map<string,int> mp[20];//string是关键字it->first.
map<string,int>::iterator it;
int main()
{
    int n;
    string s,sa,sb,sc;
    cin>>n>>s;
    mp[n][s]=1;
    for(int i=n;i>=1;i--)
    {
        for(it=mp[i].begin();it!=mp[i].end();it++)
        {
            s=it->first;
            int s1=it->second;
            int len=1<<i;
            sa=sb=sc="";
            for(int j=0;j<len;j+=2)
            {
                sa+=((s[j]-'0')&(s[j+1]-'0'))+'0';
                sb+=((s[j]-'0')^(s[j+1]-'0'))+'0';
                sc+=((s[j]-'0')|(s[j+1]-'0'))+'0';
            }
            mp[i-1][sa]+=s1;
            mp[i-1][sb]+=s1;
            mp[i-1][sc]+=s1;
        }
    }
    cout<<mp[0]["1"]<<endl;
    return 0;
}