BestCoder Round #77 (div.2)

xiaoxin juju needs help

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

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?

 

[align=left]Input[/align]
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).

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

3
aa
aabb
a

 

[align=left]Sample Output[/align]

1
2
1

 

[align=left]Source[/align]
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;
}