hdu5651

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链接:hdu5651

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

    题解

  • 如果奇数大于1肯定无解，然后所有字母出现次数除2，按照不相异的元素的全排列公式进行计算，然后用乘法逆元做一下，
  • 乘法逆元用拓展欧几里得做
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#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#define rep(i,x,y) for(register int i=x;i<=y;++i)
#define repd(i,x,y) for(register int i=x;i>=y;--i)
#define ll long long
using namespace std;
const int N=1e5;
const ll p=1000000007;
char str[N];
int a[30],T,len,cnt,flag;

int exgcd(ll a,ll b,ll &d,ll &x,ll &y){
    if(b==0){x=1; y=0; d=a;}
    else {exgcd(b,a%b,d,x,y); ll t=x;x=y;y=t-a/b*x;}
}

ll inv(ll a,ll p){
    ll d,x,y;
    exgcd(a,p,d,x,y);
    return d==1?(x%p+p)%p:-1;
}
void readin(){
    cin>>str;
    len=strlen(str);cnt=flag=0;
    memset(a,0,sizeof(a));
    rep(i,0,len-1)a[str[i]-'a']++;
    rep(i,0,29){
        if(a[i]&1)flag++;
        if(a[i]){
            a[i]>>=1;
            cnt+=a[i];
        }
    }
}

ll sum(ll X){
    ll tp=1;
    rep(i,2,X)tp=tp*i%p;
    return tp;
}

int main(){
    ios::sync_with_stdio(false);
    cin>>T;
    while(T--){
        readin();
        if(flag>1){
            cout<<0<<endl;
            continue;
        }
        rep(i,0,26)if(a[i]!=0){
            a[i]=sum(a[i]);
            a[i]=inv(a[i],p);
        }
        ll ans=sum(cnt);
        rep(i,0,26)if(a[i])ans=ans*a[i]%p;
        cout<<ans<<endl;
    }
    return 0;
}