BestCoder Round #77 (div.2)

xiaoxin juju needs help

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1124    Accepted Submission(s): 323

提示：可重集的全排列+逆元；（板子题，）
弱渣要学逆元啦！
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)

 

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#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <vector>
#define  LL long long
using namespace std;
LL A[1010]= {1,1};
int D[26];
char s[1010];
LL MOD=1e9+7;
LL solve( LL x)
{
    LL n=MOD-2;
    LL ans=1;
    while(n)
    {
        if(n&1)
        {
            (ans*=x)%=MOD;
        }
        (x*=x)%=MOD;
        n>>=1;
    }
    return ans;
}
int main()
{
    for(LL i=2; i<1001; i++)
        (A[i]=A[i-1]*i)%=MOD;
    int t;
    cin>>t;
    while(t--)
    {
        scanf("%s",s);
        memset(D,0,sizeof(D));
        for(int i=0; s[i]; i++)
            D[s[i]-'a']++;
        int odd=0,cnt=0;
        for(int i=0;i<26; i++)
        {
            if(D[i]%2)
                odd++;
            D[i]/=2;
            cnt+=D[i];
        }
        if(odd>1)
        {
            printf("0\n");
            continue;
        }
        LL ans=A[cnt];
        for(int i=0; i<26; i++)
        {
            (ans*=(solve(A[D[i]])))%=MOD;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}