【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.

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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

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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：

 1 #include <iostream>
 2 #include <bits/stdc++.h>
 3 using namespace std;
 4 bool vis[50];
 5 int n,m;
 6 string s;
 7 pair<string ,string > ps[1<<18];
 8 map<pair<string,string >,int > mp;
 9 typedef long long ll;
10 ll cnt=0;
11 void solve()
12 {
13     for(register int i=0;i< 1<<n;i++)
14     {
15         string t,tt;
16         for(register int j=0;j<n;j++)
17         {
18             if(i>>j&1)
19             {
20                 t+=s[j];
21                 //vis[j]=1;
22             }
23             else
24                 tt+=s[j];
25         }
26         ps[i]=make_pair(t,tt);
27         mp[ps[i]]++;
28     }
29     for(register int i=0;i<1<<n;i++)
30     {
31         string t,tt;
32         for(register int j=0;j<n;j++)
33         {
34             if(i>>j&1)
35             {
36                 t+=s[n+j];
37             }
38             else
39                 tt+=s[n+j];
40         }
41         reverse(t.begin(),t.end());
42         reverse(tt.begin(),tt.end());
43         cnt+=mp[make_pair(t,tt)];
44     }
45 }
46 inline int read()                                //读入优化
47 {
48     int x=0,f=1;
49     char c=getchar();
50     while (!isdigit(c))
51         f=c=='-'?-1:1,c=getchar();
52     while (isdigit(c))
53         x=(x<<1)+(x<<3)+(c^48),c=getchar();
54     return x*f;
55 }
56 int main()
57 {
58     n=read();
59     cin>>s;
60     solve();
61     cout << cnt << endl;
62     //cout << "Hello world!" << endl;
63     return 0;
64 }
65 /*
66 4
67 cabaacba
68 */

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