hdu 5651 xiaoxin juju needs help

xiaoxin juju needs help

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1505    Accepted Submission(s): 437

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)

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:  5659 5658 5657 5654 5653 

 

在网上找了模板，虽然我没有看懂，但是直接用了就过了。。。

 

题意：给你一个字符串，可以打乱顺序，问可以组合成几种回文串。

 

附上代码：

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int mod=1e9+7;
typedef long long ll;
//返回d=gcd(a,b);和对应于等式ax+by=d中的x,y
ll extend_gcd(ll a,ll b,ll &x,ll &y)
{
    if(a==0&&b==0) return -1;//无最大公约数
    if(b==0)
    {
        x=1;
        y=0;
        return a;
    }
    ll d=extend_gcd(b,a%b,y,x);
    y-=a/b*x;
    return d;
}
//*********求逆元素*******************
//ax = 1(mod n)
ll mod_reverse(ll a,ll n)
{
    ll x,y;
    ll d=extend_gcd(a,n,x,y);
    if(d==1) return (x%n+n)%n;
    else return -1;
}

ll c(ll m,ll n)
{
    ll i,j,t1,t2,ans;
    t1=t2=1;
    for(i=n; i>=n-m+1; i--) t1=t1*i%mod;
    for(i=1; i<=m; i++) t2=t2*i%mod;
    return  t1*mod_reverse(t2,mod)%mod;
}

int main()
{
    int T,i,j,n,m;
    char ch[1005];
    int a[30];
    scanf("%d",&n);
    while(n--)
    {
        memset(a,0,sizeof(a));
        scanf("%s",ch);
        for(i=0; ch[i]!=''; i++)
            a[ch[i]-'a']++;
        int t=0;
        for(i=0; i<26; i++)
        {
            if(a[i]%2!=0)
                t++;
            if(t>1)
                break;
        }
        if(i!=26)
        {
            printf("0
");
            continue;
        }
        int len=strlen(ch);
        ll sum=1;
        len/=2;
        for(i=0; i<26; i++)
        {
            sum=(sum*c(a[i]/2,len))%mod;
            len-=a[i]/2;
        }
        printf("%d
",sum);
    }
    return 0;
}