HDU 5651 逆元

xiaoxin juju needs help

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 809    Accepted Submission(s): 231

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

 

Source

BestCoder Round #77 (div.2)

 

题意： 给一段只有小写字母的字符串 判断能够组合成多少种回文串

 

题解：首先，如果不止一个字符出现的次数为奇数，则结果为0。 否则，我们把每个字符出现次数除2，也就是考虑一半的情况。 那么结果就是这个可重复集合的排列数了。

A!/(n1！*n2！....)

这里直接除 会出错 需要用到逆元的知识

http://blog.csdn.net/acdreamers/article/details/8220787

另外如何求逆元？

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#define ll __int64
using namespace std;
int n;
char a[1005];
#define mod 1000000007
map<char,int> mp;
ll quickmod(ll a,ll b)
{
    ll sum=1;
    while(b)
    {
        if(b&1)
            sum=(sum*a)%mod;
        b>>=1;
        a=(a*a)%mod;
    }
    return sum;
}

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        mp.clear();
        int jishu=0;
        int sum=0;
        scanf("%s",a);
        int len=strlen(a);
        for(int j=0;j<len;j++)
            mp[a[j]]++;
        for(int j='a';j<='z';j++)
        {
            if(mp[j]%2==1)
             {
             jishu++;
             }
        }
        if(len%2==0)
        {
            if(jishu!=0)
            {
            cout<<"0"<<endl;
            continue;}
        }
        else
        {
            if(jishu>1)
            {cout<<"0"<<endl;
            continue;}
        }
      for(int j='a';j<='z';j++)
      {
           sum=sum+mp[j]/2;
      }
      ll ans=1;
      ll gg=1;
      for(int j='a';j<='z';j++)
      {
      for(int k=1;k<=mp[j]/2;k++)
         gg=gg*k%mod;

      }
      for(int j=1;j<=sum;j++)
      {
          ans=ans*j%mod;
      }
      printf("%I64d
",(ans*(quickmod(gg,mod-2)%mod))%mod);
    }
    return 0;
}