HDU 5651 逆元

Posted 2020-06-29 半根毛线code

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？

 

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<map>
 5 #define ll __int64
 6 using namespace std;
 7 int n;
 8 char a[1005];
 9 #define mod 1000000007
10 map<char,int> mp;
11 ll quickmod(ll a,ll b)
12 {
13     ll sum=1;
14     while(b)
15     {
16         if(b&1)
17             sum=(sum*a)%mod;
18         b>>=1;
19         a=(a*a)%mod;
20     }
21     return sum;
22 }
23 
24 int main()
25 {
26     scanf("%d",&n);
27     for(int i=1;i<=n;i++)
28     {
29         mp.clear();
30         int jishu=0;
31         int sum=0;
32         scanf("%s",a);
33         int len=strlen(a);
34         for(int j=0;j<len;j++)
35             mp[a[j]]++;
36         for(int j=‘a‘;j<=‘z‘;j++)
37         {
38             if(mp[j]%2==1)
39              {
40              jishu++;
41              }
42         }
43         if(len%2==0)
44         {
45             if(jishu!=0)
46             {
47             cout<<"0"<<endl;
48             continue;}
49         }
50         else
51         {
52             if(jishu>1)
53             {cout<<"0"<<endl;
54             continue;}
55         }
56       for(int j=‘a‘;j<=‘z‘;j++)
57       {
58            sum=sum+mp[j]/2;
59       }
60       ll ans=1;
61       ll gg=1;
62       for(int j=‘a‘;j<=‘z‘;j++)
63       {
64       for(int k=1;k<=mp[j]/2;k++)
65          gg=gg*k%mod;
66 
67       }
68       for(int j=1;j<=sum;j++)
69       {
70           ans=ans*j%mod;
71       }
72       printf("%I64d\n",(ans*(quickmod(gg,mod-2)%mod))%mod);
73     }
74     return 0;
75 }

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