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[];
 #define mod 1000000007
 map<char,int> mp;
 ll quickmod(ll a,ll b)
 {
     ll sum=;
     while(b)
     {
         if(b&)
             sum=(sum*a)%mod;
         b>>=;
         a=(a*a)%mod;
     }
     return sum;
 }

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

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