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
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 }