题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5651

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

Hint

题意

给你一个串，你可以改变字符位置

问你能够形成多少种回文串。

题解：

首先把答案为0的情况判断掉

然后就很简单了，因为回文嘛，所以左右肯定相同

然后就可以排列组合怼一波了

就相当于选位置，把所有字母安上去。

C(x1,x2)*C(y1,y2)….这种

代码

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<cstring>
using namespace std;
const int mod = 1e9+7;
const int maxn = 1e5+7;
int num[30];
typedef long long ll;
ll fac[maxn];
ll qpow(ll a,ll b)
{
    ll ans=1;a%=mod;
    for(ll i=b;i;i>>=1,a=a*a%mod)
        if(i&1)ans=ans*a%mod;
    return ans;
}
ll C(ll n,ll m)
{
    if(m>n||m<0)return 0;
    ll s1=fac[n],s2=fac[n-m]*fac[m]%mod;
    return s1*qpow(s2,mod-2)%mod;
}
void solve()
{
    memset(num,0,sizeof(num));
    string s;cin>>s;
    for(int i=0;i<s.size();i++)
        num[s[i]-'a']++;
    if(s.size()%2==0)
    {
        for(int i=0;i<26;i++)
            if(num[i]%2==1)
            {
                printf("0\n");
                return;
            }
    }
    else
    {
        int cnt = 0;
        for(int i=0;i<26;i++)
            if(num[i]%2==1)cnt++;
        if(cnt!=1)
        {
            printf("0\n");
            return;
        }
    }
    long long ans = 1;
    long long las = s.size();
    for(int i=0;i<26;i++)
    {
        ans = (ans * C(las/2,num[i]/2))%mod;
        las-=num[i];
    }
    printf("%lld\n",ans);
}
int main()
{
    fac[0]=1;
    for(int i=1;i<maxn;i++)
        fac[i]=fac[i-1]*i%mod;
    int t;scanf("%d",&t);
    while(t--)solve();
    return 0;
}