21.10.28模拟 String Coloring AGC26 C

将2*n长度的字符串拆成两段，两段分别计算，开个map

在s1==s2前提下，
s1在前半段出现的从左往右出现的字符，s2一定在后半段从右往左出现
即正序s1前半段=逆序s2后半段，正序s2前半段=逆序s1后半段

typedef long long lxl;
map<pair<string,string>, lxl> cnt;
const int maxn = 50;
char s[maxn];

int main() {
	freopen("b.in", "r", stdin);
	freopen("b.out", "w", stdout);

	int n;
	cin>>n>>s;
	for(int i = 0; i < (1 << n); i++) {
		string s1 = "", s2 = "";
		for(int j = 0; j < n; j++) {
			if(i & (1 << j)) s1 += s[j];
			else s2 += s[j];
		}
		cnt[{s1,s2}]++;
	}
	lxl sum = 0;
	for(int i = 0; i < (1 << n); i++) {
		string s1 = "", s2 = "";
		for(int j = n - 1; j >= 0; j--) {
			if(i & (1 << j)) s1 += s[j + n];
			else s2 += s[j + n];
		}
		sum += cnt[{s1,s2}];
	}
	cout<<sum<<'
';
}