PAM练习

题：https://www.luogu.com.cn/problem/P1659

题意：问前k大的奇数长度的回文串的长度乘积；

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int M=1e6+6;
const int maxn=26;
const int mod=19930726;
char s[M];
struct node{
    ll val,num;
    bool operator<(const node &b)const{
        return val>b.val;
    }
}b[M];

struct pam{
    int  son[M][maxn],cnt[M],num[M],fail[M],len[M];
    char s[M];
    int tot,last;
    int newnode(int Len){
        for(int i=0;i<maxn;i++)
            son[tot][i]=0;
        cnt[tot]=0;
        num[tot]=0;
        fail[tot]=0;
        len[tot]=Len;
        return tot++;
    }
    void init(){
        s[0]='#';
        tot=0;
        last=0;
        newnode(0);
        newnode(-1);
        fail[0]=1;
    }
    int getfail(int p,int i){
        while(s[i-len[p]-1]!=s[i])
            p=fail[p];
        return p;
    }
    void solve(const char *buf){
        init();
        int n=strlen(buf+1);
        for(int i=1;i<=n;i++){
            s[i]=buf[i]-'a';
            int cur=getfail(last,i);
            if(!son[cur][s[i]]){
                int now=newnode(len[cur]+2);
                fail[now]=son[getfail(fail[cur],i)][s[i]];
                son[cur][s[i]]=now;
                num[now]=num[fail[now]]++;
            }
            cnt[last=son[cur][s[i]]]++;
        }
        for(int i=tot-1;i>=0;i--)
            cnt[fail[i]]+=cnt[i];
    }
}PAM;
ll ksm(ll a,ll b){
    ll t=1ll;
    while(b){
        if(b&1){
            t=(t*a)%mod;
        }
        b>>=1ll;
        a=(a*a)%mod;
    }
    return t;
}
int main(){
    ll n,k;
    scanf("%lld%lld",&n,&k);
    scanf("%s",s+1);
    PAM.solve(s);
    int m=0;
    for(int i=0;i<PAM.tot;i++){
    //    cout<<PAM.len[i]<<"!!"<<PAM.cnt[i]<<endl;
        if(PAM.len[i]&1){
            b[m].val=PAM.len[i];
            b[m].num=PAM.cnt[i];
            m++;
        }
        
    }
    sort(b,b+m);
    ll ans=1ll;
    for(int i=0;i<m;i++){
        ans=(1ll*ans*ksm(b[i].val,min(1ll*b[i].num,k)))%mod;
    //    cout<<ans<<endl;
        k-=b[i].num;
        if(k<=0)
            break;
    }
    printf("%lld\n",ans);
    return 0;
}

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