HDU

http://acm.hdu.edu.cn/showproblem.php?pid=5651

题意：生成回文串。输出所有回文串的可能数。

题解：mod除法会损失高位，用逆元来代替除法，模板如下

ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<string>

using namespace std;
const int maxn = 1111;
int num[maxn];
//char s[maxn];
typedef long long ll;
const int mod = 1e9 + 7;
ll f[maxn];
void init() {
    f[1] = 1; f[0] = 1;
    for (int i = 2; i < maxn; i++)
        f[i] = f[i - 1] * i%mod;
}
ll cal(ll x) {
    ll res = 1;
    int k = mod - 2;
    while (k) {
        if (k & 1) {
            res *= x;
            res %= mod;
        }
        x*=x;
        x %= mod;
        k >>= 1;
    }
    return res;//cal(x^ mod)
}
int main() {
    int t;
    init();
    cin >> t;
    while (t--) {
        memset(num, 0, sizeof(num));
        string s;
        cin >> s;
        int n = s.length();

        for (int i = 0; i < n; i++) {
            num[s[i]]++;

        }
        int cnt = 0;
        for (int i = 'a'; i <= 'z'; i++) {
            if (num[i] & 1) cnt++;
        }
        if (cnt > 1) {
            cout << 0 << endl; continue;
        }
        int sum = 0;
        for (int i = 'a'; i <= 'z'; i++) {
            num[i] /= 2;
            sum += num[i];
        }
        ll res = f[sum];
        for (int i ='a'; i <='z'; i++)if(num[i]) {
            res = res*cal(f[num[i]]) % mod;
        }
        cout << res<<endl;
        
    }
}