An anagram is a word or phrase formed by rearranging the letters of another word or phrase, using all the original letters exactly once, such as "post", "stop", and "spot".
You are given a string s consisting of lowercase English letters. Your task is to count how many distinct anagrams of the string s are palindromes.
A palindrome is a word, phrase, number, or other sequence of characters which reads the same backward as forward, such as "madam" or "racecar".
For example, "aabb" has 6 distinct anagrams, which are: "aabb", "abab", "abba", "baab", "baba", "bbaa". Two of the previous anagrams are palindromes, which are: "abba", "baab".
Input
The first line contains an integer T, where T is the number of test cases.
The first line of each test case contains an integer n (1 ≤ n ≤ 20), where n is the length of the string s.
The second line of each test contains a string s of length n consisting of lowercase English letters.
Output
For each test case, print a single line containing how many distinct anagrams of the string s are palindromes.
Example
5
4
aabb
6
ababbc
6
abaaba
12
babacbcbcaca
14
aaaabbcaaaabbc
2
0
3
90
105
代码：
#include <iostream> #include <algorithm> #include <string.h> #include <cstdio> #include <string> #include <cmath> #include <vector> #include <stack> #include <queue> #include <stack> #include <list> #include <map> #include <set> //#include <unordered_map> #define Fbo friend bool operator < (node a, node b) #define mem(a, b) memset(a, b, sizeof(a)) #define FOR(a, b, c) for (int a = b; a <= c; a++) #define RFOR(a, b, c) for (int a = b; a >= c; a--) #define off ios::sync_with_stdio(0) #define sc(a) scanf("%d",&a) #define pr(a) printf("%d\n",a); #define SC(n,m) scanf("%d%d",&n,&m) bool check1(int a) { return (a & (a - 1)) == 0 ? true : false; } using namespace std; typedef pair<int, int> pii; typedef long long ll; const int INF = 0x3f3f3f3f;//1e10 const int mod = 1e9 + 7; const int Maxn = 1e5 + 5; const int N = 2e5 + 5; const double pi = acos(-1.0); const double eps = 1e-8; ll f[20]; ll a[30], b[30]; void factorial(){//阶乘 f[0] = 1; for (int i = 1; i <= 11; i++) { f[i] = f[i - 1] * i; } } int main(){ factorial(); int t; cin >> t; while (t--) { mem(a, 0); mem(b, 0); int n, k = 0; string s; cin >> n >> s; for (int i = 0; i < s.size(); i++) { a[s[i] - ‘a‘]++; } ll odd = 0, cnt = 0; ll ans = 0; FOR(i, 0, 26) { if (a[i]&1) odd++; if (a[i]) b[k++] = a[i] / 2; ans += a[i] / 2; } if (n & 1 && odd == 1 || (n % 2 == 0 && odd == 0)) { cnt = f[ans]; for (int i = 0; i < k; i++) { if (b[i]) { cnt /= f[b[i]]; } } } cout << cnt << endl; } }
Palindromes Building Gym - 101532K (计算回文串数目)
原文：https://www.cnblogs.com/AlexLINS/p/12713957.html