【折半枚举】 String Coloring

Problem Statement

You are given a string $\mathrm{SS}$ of length $\mathrm{2N2N}$ consisting of lowercase English letters.

There are ${\mathrm{22N22N}}^{}$ ways to color each character in $\mathrm{SS}$ red or blue. Among these ways, how many satisfy the following condition?

• The string obtained by reading the characters painted red from left to right is equal to the string obtained by reading the characters painted blue from right to left.

Constraints

• $\mathrm{1\le N\le 181\le N\le 18}$
• The length of $\mathrm{SS}$ is $\mathrm{2N2N}$.
• $\mathrm{SS}$ consists of lowercase English letters.

---

Input

Input is given from Standard Input in the following format:

$\mathrm{NN}$
$\mathrm{SS}$

Output

Print the number of ways to paint the string that satisfy the condition.

---

Sample Input 1 Copy

Copy

4
cabaacba

Sample Output 1 Copy

Copy

4

There are four ways to paint the string, as follows:

• cabaacba
• cabaacba
• cabaacba
• cabaacba

---

Sample Input 2 Copy

Copy

11
mippiisssisssiipsspiim

Sample Output 2 Copy

Copy

504

---

Sample Input 3 Copy

Copy

4
abcdefgh

Sample Output 3 Copy

Copy

0

---

Sample Input 4 Copy

Copy

18
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Sample Output 4 Copy

Copy

9075135300

The answer may not be representable as a $3232$-bit integer.

代码如下：

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
bool vis[50];
int n,m;
string s;
pair<string ,string > ps[1<<18];
map<pair<string,string >,int > mp;
typedef long long ll;
ll cnt=0;
void solve()
{
    for(register int i=0;i< 1<<n;i++)
    {
        string t,tt;
        for(register int j=0;j<n;j++)
        {
            if(i>>j&1)
            {
                t+=s[j];
                //vis[j]=1;
            }
            else
                tt+=s[j];
        }
        ps[i]=make_pair(t,tt);
        mp[ps[i]]++;
    }
    for(register int i=0;i<1<<n;i++)
    {
        string t,tt;
        for(register int j=0;j<n;j++)
        {
            if(i>>j&1)
            {
                t+=s[n+j];
            }
            else
                tt+=s[n+j];
        }
        reverse(t.begin(),t.end());
        reverse(tt.begin(),tt.end());
        cnt+=mp[make_pair(t,tt)];
    }
}
inline int read()                                //读入优化
{
    int x=0,f=1;
    char c=getchar();
    while (!isdigit(c))
        f=c=='-'?-1:1,c=getchar();
    while (isdigit(c))
        x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int main()
{
    n=read();
    cin>>s;
    solve();
    cout << cnt << endl;
    //cout << "Hello world!" << endl;
    return 0;
}
/*
4
cabaacba
*/