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 integerT(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 mod1,000,000,007.

 

Sample Input

3aaaabba

 

Sample Output

121

 

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 longusing 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;}