hdu 5651(逆元+排列组合）

                             xiaoxin juju needs help

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 984    Accepted Submission(s): 276

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≤20) which represents the number of test cases.
For each test case, there is a single line containing a string S(1≤length(S)≤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

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5651
题意：
给你遗传由若干个小写字母组成的字符串，你可以任意变动互相之间的位子，但不能删除或添加元素，问可以组成多少种不同回文串。
思路：
很简单的排列组合，太简单我就不把他分到排列组合里去了。
先判断是否可以组成回文串，如果有一个以上的奇数个同一的字母，则无论如何都不会组成回文串。
剩下的就枚举回文串的一边，所有的元素都取半。
所有元素的一半的和的阶乘除于每个元素一半的的阶乘即是答案。
这就是问题的所在，除于之后是要取模的。
除法取模问题当然是逆元了，谈到逆元当然是费马小定理喽。
费马小定理：
   费马小定理(Fermat Theory)是
   数论中的一个重要
   定理，其内容为： 假如p是
   质数，且(a,p)=1，
那么 a
   (p-1)≡1（mod p）。即：假如a是
   整数，p是
   质数，且a,p
   互质(即两者只有一个
   公约数1)，那么a的(p-1)次方除以p的
   余数
   恒等于1。
代码：
   
#include<stdio.h>
#include<string.h>
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[0]=1;
    f[1]=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;
}
int main()
{
    int T;
    init();
    while(scanf("%d",&T)!=EOF)
    while(T--)
    {
        scanf("%s",s);
        int n=strlen(s);
        memset(num,0,sizeof(num));
        for(int i=0;i<n;i++)
           num[s[i]-'a']++;
        int len=0;
        for(int i=0;i<26;i++)
        if(num[i]%2)
        len++;
        if(len>1)
        {
            printf("0\n");
            continue;
        }
        int total=0;
        for(int i=0;i<26;i++)
        {
            num[i]/=2;
            total+=num[i];
        }
        ll res=f[total];
        for(int i=0;i<26;i++)
        if(num[i])
        {
            res=res*cal(f[num[i]])%mod;
        }
        printf("%lld\n",res);
    }
    return 0;
}