ZOJ

题意:给一个字符串,和每个字符代表的val,每个回文串的价值就是前半部分的val26进制%777777777,求价值第k小的回文串
题解:建个pam,然后dfs两边(0,1),统计价值sort一遍就好了
k爆int了,= =白wa了半天

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 777777777
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("c.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100000+10,maxn=100000+10,inf=0x3f3f3f3f;

char s[N];
ll res,f[N],val[30];
pll ans[N];
struct PAM{
    int ch[N][26],fail[N],len[N],s[N];
    ll cnt[N];
    int last,n,p;
    int newnode(int w)
    {
        for(int i=0;i<26;i++)ch[p][i] = 0;
        cnt[p] = 0;
        len[p] = w;
        return p++;
    }
    void init()
    {
        p = last = n = 0;
        newnode(0);
        newnode(-1);
        s[n] = -1;
        fail[0] = 1;
    }
    int getfail(int x)
    {
        while(s[n-len[x]-1] != s[n]) x = fail[x];
        return x;
    }
    void add(int c)
    {
        s[++n] = c;
        int cur = getfail(last);
        if(!ch[cur][c]){
            int now = newnode(len[cur]+2);
            fail[now] = ch[getfail(fail[cur])][c];
            ch[cur][c] = now;
        }
        last = ch[cur][c];
        cnt[last]++;
    }
    void cal()
    {
        for(int i=p-1;i>=0;i--)cnt[fail[i]]+=cnt[i];
    }
    void dfs(int u,int len,ll v)
    {
        for(int i=0;i<26;i++)
            if(ch[u][i])
            {
                ans[++res]=mp((v+f[len]*val[i]%mod)%mod,cnt[ch[u][i]]);
                dfs(ch[u][i],len+1,(v+f[len]*val[i]%mod)%mod);
            }
    }
}pam;
int main()
{
//    fin;fout;
    f[0]=1;
    for(int i=1;i<N;i++)f[i]=f[i-1]*26ll%mod;
    int T;scanf("%d",&T);
    while(T--)
    {
        pam.init();
        int n,m;scanf("%d%d%s",&n,&m,s+1);
        for(int i=1;i<=n;i++)pam.add(s[i]-'a');
        pam.cal();
        while(m--)
        {
            ll k;scanf("%lld",&k);
            for(int i=0;i<26;i++)scanf("%lld",&val[i]);
            res=0;
            pam.dfs(0,0,0);pam.dfs(1,0,0);
//            printf("%d
",res);
            sort(ans+1,ans+1+res);
            ll co=0;
            for(int i=1;i<=res;i++)
            {
                co+=ans[i].se;
                if(co>=k)
                {
                    printf("%lld
",ans[i].fi);
                    break;
                }
            }
        }
        puts("");
    }
    return 0;
}
/********************
10
78 3
abaabbabababcddajkjdlspeoiabaabbabababcddajkjdlspeoiabaabbabababcddajkjdlspeoi
23 25 25 22 23 24 25 16 17 18 19 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
********************/