BZOJ 2160: 拉拉队排练

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=2160

构造完回文树后，从后往前扫一遍，令cnt[fail[i]]+=cnt[i]，可得到每个串的出现次数。那么对于回文树中的一个结点，其长度为len[i],出现次数为cnt[i],那我们拿出前K个长度为奇数的串相乘就可以了。快速幂一下。

#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdio>
#define rep(i,l,r) for (int i=l;i<=r;i++)
#define down(i,l,r) for (int i=l;i>=r;i--)
#define clr(x,y) memset(x,y,sizeof(x))
#define ll long long
#define maxn 1000500
#define mm 19930726
using namespace std;
ll ans,k;
int n,m,len;
char ch[maxn];
struct data{ll x,y;
}a[maxn];
ll read(){
    ll x=0,f=1; char ch=getchar();
    while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while (isdigit(ch)) {x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
struct PT{
    ll sum;
    int n,tot,fail[maxn],l[maxn],nxt[maxn][30],s[maxn],last;
    ll cnt[maxn];
    int newnode(int len){
        clr(nxt[tot],0);
        l[tot]=len; cnt[tot]=0;
        return tot++;
    }
    void init(){
        tot=0; sum=0; clr(fail,0);
        newnode(0); newnode(-1);
        fail[0]=1;
        last=1;
        s[0]=-1;
        n=0;
    }
    int getfail(int v){
        while (s[n-l[v]-1]!=s[n]) v=fail[v];
        return v;
    }
    void add(int c){
        s[++n]=c;
        int cur=getfail(last);
        if (!nxt[cur][c]){
            int now=newnode(l[cur]+2);
            fail[now]=nxt[getfail(fail[cur])][c];
            nxt[cur][c]=now;
        } last=nxt[cur][c];
        cnt[last]++;
    }
}T;
void p(ll x,ll y){
    while (y){
        if (y&1) ans=ans*x%mm;
        x=x*x%mm;
        y/=2;
    }
}
bool cmp(data a,data b){
    return a.x>b.x;
}
int main(){
    len=read(); k=read();
    scanf("%s",ch); 
    T.init();
    rep(i,0,len-1) T.add(ch[i]-'a');
    down(i,T.tot-1,0) {
        T.cnt[T.fail[i]]+=T.cnt[i];
        if (i>1&&T.l[i]&1) a[++m]=(data){T.l[i],T.cnt[i]};
    }
    sort(a+1,a+1+m,cmp);
    ans=1;
    rep(i,1,m){
        if (a[i].y>=k) {
            p(a[i].x,k);
            k=0;
            break;
        }
        else if (a[i].y<k){
            p(a[i].x,a[i].y);
            k-=a[i].y;
        }
    }
    if (k) puts("-1");
    else printf("%lld\n",ans);
    return 0;
}