HDU 5651 xiaoxin juju needs help 逆元

题目链接：

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

bc:http://bestcoder.hdu.edu.cn/contests/contest_showproblem.php?cid=682&pid=1002

xiaoxin juju needs help

 

 Accepts: 150

 

 Submissions: 966

 Time Limit: 2000/1000 MS (Java/Others)

 

 Memory Limit: 65536/65536 K (Java/Others)

Problem Description

As we all known, xiaoxin is a brilliant coder. He knew palindromic strings when he was only a six grade student at elementry school.

This summer he was working at Tencent as an intern. One day his leader came to ask xiaoxin for help. His leader gave him a string and he wanted xiaoxin to generate palindromic strings for him. Once xiaoxin generates a different palindromic string, his leader will give him a watermelon candy. The problem is how many candies xiaoxin's leader needs to buy?

Input

This problem has multi test cases. First line contains a single integer $T(T\le 20)$ which represents the number of test cases. For each test case, there is a single line containing a string $S(1\le length(S)\le 1,000)$.

Output

For each test case, print an integer which is the number of watermelon candies xiaoxin's leader needs to buy after mod $1,000,000,007$.

Sample Input

3
aa
aabb
a

Sample Output

1
2
1

题解：

1、可行性：

统计每个字母出去的次数，如果有两种即以上字母出现的次数为奇数，则一定不可能排出回文串。

2、统计：

由于回文串左右两边必须相同，所以我们考虑一边就可以了（如果为奇数则正中间一个不管就和偶数情况是一样的了）。

则可以转化为排列组合问题，等价于求解: (cnt[i]代表某个字母出现的次数(cnt[i]>0) )

(　)%(1e9+7)

由于涉及到除法，要求逆元：

例子：

　　求a/b(mod n) 如果b与n互质，则求满足bx=1(%n)的一个解x，原式可转化为a/b*bx(%n)，即a*x(%n)；这样就把除法消除了。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int maxn = 1000 + 10;
const int mod = 1e9 + 7;
typedef long long LL;

char str[maxn];
int n;

int cnt[33];
int tmp[33], tot;

LL b[maxn];
//预处理出阶乘
void pre() {
    b[0] = b[1] = 1;
    for (int i = 2; i < maxn; i++) b[i] = (b[i - 1] * i) % mod;
}
//扩展的欧几里得算法求逆元
void gcd(LL a, LL b, LL &d, LL &x, LL &y) {
    if (!b) {
        d = a;
        x = 1; y = 0;
    }
    else {
        gcd(b, a%b, d, y, x);//y=x',x=y';
        //x=y'; y=x'-(a/b)*y';
        y -= (a / b)*x;
    }
}

void init() {
    tot = 0;
    memset(cnt, 0, sizeof(cnt));
}

int main() {
    pre();
    int tc;
    scanf("%d", &tc);
    while (tc--) {
        init();
        scanf("%s", str);
        n = strlen(str);
        for (int i = 0; i < strlen(str); i++) {
            cnt[str[i] - 'a']++;
        }
        int flag = 0;
        for (int i = 0; i < 33; i++) {
            if (cnt[i] % 2) flag++;
            //统计一半的字母出现的次数
            if (cnt[i] > 0) {
                tmp[tot++] = cnt[i] / 2;
            }
        }
        if (flag > 1) { printf("0
"); continue; }
        LL sum = 1;
        for (int i = 0; i < tot; i++) {
            int t = tmp[i];
            sum = (sum*b[t]) % mod;
        }
        LL d, x, y;
        gcd(sum, mod, d, x, y);
        //x有可能是负数，需要处理成正的。
        x = (x%mod + mod) % mod;
        LL ans = (b[n / 2] * x) % mod;
        printf("%lld
", ans);
    }
    return 0;
}