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-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.
#include<bits/stdc++.h>
using namespace std;const int maxn=20;map<string,int> mp[maxn];
map<string,int>::iterator it;string s,sa,sb,sc;
int n;
/*
题目大意:给定二进制串，和三种对应的放缩方式，
每次长度递减一倍，然后计数最终压缩成串1的情况组合数。暴力大法好啊。。。。
但这道题参照了别人的代码也可以用状压写，用map作为
状压计数的容器，不断在新的长度中插入上一个长度压缩来的字符串，并累加其计数。另外还有DFS做法，比赛时找数学规律找崩了。*/int main()
{cin>>n>>s;mp[n][s]=1;for(int i=n;i>=1;i--){for(it=mp[i].begin();it!=mp[i].end();it++){int cnt=it->second;string ss=it->first;sa=sb=sc="";///置空int len=1<<i;for(int j=0;j<len;j+=2){sa+=((ss[j]-'0') | (ss[j+1]-'0'))+'0';sb+=((ss[j]-'0') ^ (ss[j+1]-'0'))+'0';sc+=((ss[j]-'0') & (ss[j+1]-'0'))+'0';}mp[i-1][sa]+=cnt;mp[i-1][sb]+=cnt;mp[i-1][sc]+=cnt;}}cout<<mp[0]["1"]<<endl;return 0;
}