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做過的用到資料結構+演算法最多的一個題……真真是做ACM以來做的最最麻煩的一個題……

說白了其實就是板子大雜燴……但是會吐的那種。。

此外……此題價值75$……不要問我為什麼。。。TOT

現在進入正片——

給你一個長n的字串，僅由小寫字母組成。然後q次詢問。
每次詢問該串的所有迴文子串裡，按字典序從小到大後第k個的迴文串。輸出它的f函式值。注意迴文串不去重。

具體要做的就是，預處理母串的所有迴文串，按字典序排序，然後快速找到第k個，輸出對應f值。

分為如下子問題：
1·預處理母串所有迴文子串——迴文樹，具體百度，然後與迴文樹惡戰一場吧……！￥@#￥。然後可以把本質不同的迴文子串以及對應的出現次數存起來，然後排序。存迴文子串可以存該種迴文子串出現的左右下標——用以排序。

2·把迴文子串按字典序排序——預處理母串的字尾陣列，然後可以搞RMQ，弄出來個快速lcp。這樣對於上一步存起來的本質不同的迴文子串左右下標，通過比較lcp，可以達到快速按字典序排序。

3·快速找到第k個——二分，預處理到某個串所需要的k的下界，這樣可以快速二分出第k個串。

4·f函式——字串雜湊，一套板子生砸上……

然後就出來了……GG。。。

因為是邊學邊造輪子，所以程式碼比較醜陋……
程式碼如下：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
 

#include <vector>
#include <queue>
#include <algorithm>
#define LL long long
#define Pr pair<int,int>

using namespace std;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+7;

const int MAXN = 112345;
int t1[MAXN],t2[MAXN],c[MAXN];

bool cp(int *r,int a,int b,int l)
{
    return
 
 r[a] == r[b] && r[a+l] == r[b+l];
}

void da(char *str,int *sa,int *rk,int *height,int n,int m)
{
    n++;
    int i,j,p,*x = t1,*y = t2;
    for(i = 0; i < m; ++i) c[i] = 0;
    for(i = 0; i < n; ++i) c[x[i] = str[i]]++;
    for(i = 1; i < m; ++i) c[i] += c[i-1];
    for(i = n-1; i >= 0; --i) sa[--c[x[i]]] = i;
    for(j = 1; j <= n; j <<= 1)
    {
        p = 0;

        for(i = n-j; i < n; ++i) y[p++] = i;
        for(i = 0; i < n; ++i) if(sa[i] >= j) y[p++] = sa[i]-j;

        for(i = 0; i < m; ++i) c[i] = 0;
        for(i = 0; i < n; ++i) c[x[y[i]]]++;
        for(i = 1; i < m; ++i) c[i] += c[i-1];
        for(i = n-1; i >= 0; --i) sa[--c[x[y[i]]]] = y[i];

        swap(x,y);
        p = 1;
        x[sa[0]] = 0;

        for(i = 1; i < n; ++i)
            x[sa[i]] = cp(y,sa[i-1],sa[i],j)? p-1: p++;
        if(p >= n) break;
        m = p;
    }
    int k = 0;
    n--;
    for(i = 0; i <= n; ++i) rk[sa[i]] = i;
    for(i = 0; i < n; ++i)
    {
        if(k) k--;
        j = sa[rk[i]-1];
        while(str[i+k] == str[j+k]) k++;
        height[rk[i]] = k;
    }
}

int rk[MAXN],height[MAXN];
int RMQ[MAXN];
int mm[MAXN];
int best[20][MAXN];

void initRMQ(int n)
{
    mm[0] = -1;
    for(int i = 1; i <= n; ++i)
        mm[i] = ((i&(i-1)) == 0)? mm[i-1]+1: mm[i-1];
    for(int i = 1; i <= n; ++i) best[0][i] = i;
    for(int i = 1; i <= mm[n]; i++)
        for(int j = 1; j +(1<<i)-1 <= n; j++)
        {
            int a = best[i-1][j];
            int b = best[i-1][j+(1<<(i-1))];
            if(RMQ[a] < RMQ[b]) best[i][j] = a;
            else best[i][j] = b;
        }
}

int askRMQ(int a,int b)
{
    int t;
    t = mm[b-a+1];
    b -= (1<<t)-1;
    a = best[t][a];
    b = best[t][b];
    return RMQ[a] < RMQ[b]? a: b;
}

int lcp(int a,int b,int len)
{
    if(a == b) return len-a;
    a = rk[a];
    b = rk[b];
    if(a > b) swap(a,b);
    return height[askRMQ(a+1,b)];
}

struct node {
    int next[26];
    int len;
    int sufflink;
    int num,l,r;
};

char s[112345];
node tree[MAXN];
int num;            // node 1 - root with len -1, node 2 - root with len 0
int suff;           // max suffix palindrome

bool addLetter(int pos) {
    int cur = suff, curlen = 0;
    int let = s[pos] - 'a';

    while (true) {
        curlen = tree[cur].len;
        if (pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos])
            break;
        cur = tree[cur].sufflink;
    }
    if (tree[cur].next[let]) {
        suff = tree[cur].next[let];
        tree[suff].num++;
        return false;
    }

    num++;
    suff = num;
    tree[num].len = tree[cur].len + 2;
    tree[num].r = pos;
    tree[num].l = pos-tree[num].len+1;
    tree[cur].next[let] = num;

    if (tree[num].len == 1) {
        tree[num].sufflink = 2;
        tree[num].num = 1;
        return true;
    }

    tree[num].num++;
    while (true) {
        cur = tree[cur].sufflink;
        curlen = tree[cur].len;
        if (pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos]) {
            tree[num].sufflink = tree[cur].next[let];
            break;
        }
    }
    return true;
}

void initTree() {
    num = 2; suff = 2;
    tree[1].len = -1; tree[1].sufflink = 1;
    tree[2].len = 0; tree[2].sufflink = 1;
}

int len;
int r[MAXN];
int sa[MAXN];
vector <int> adj[MAXN];
vector <pair<Pr,int> > vc;

void dfs(int u)
{
    for(int i = 0; i < adj[u].size(); ++i)
    {
        dfs(adj[u][i]);
        tree[u].num += tree[adj[u][i]].num;
    }
}

bool cmp(pair<Pr,int> a,pair<Pr,int> b)
{
    int l1,l2,r1,r2,len1,len2;
    l1 = a.first.first;
    l2 = b.first.first;
    r1 = a.first.second;
    r2 = b.first.second;
    len1 = r1-l1+1;
    len2 = r2-l2+1;

    int cp = lcp(l1,l2,len);

    if(cp >= len1 || cp >= len2) return len1 <= len2;
    else return s[l1+cp] <= s[l2+cp];
}

int P[MAXN],HashF[MAXN],HashR[MAXN];
class RollingHash {
    public:
        RollingHash() {
            prime = 100001;
            mod1 = 1000000007;
            mod2 = 1897266401;
            P[0] = 1;
            for(int i=1; i<MAXN; i++) {
                P[i] = 1LL * P[i-1] * prime % mod1;
            }
        }

        void Construct() {
            HashF[0] = HashR[ len+1 ] = 0;
            for(int i=1; i<=len; i++) {
                HashF[i] = ( 1LL * HashF[i-1] * prime + s[i-1] ) % mod1;
                HashR[len-i+1] = ( 1LL * HashR[len-i+2] * prime + s[ len - i ] ) % mod1;
            }
        }

        int GetForwardHash( int l, int r ) {
            if( l == 1 ) return HashF[r];
            int hash = HashF[r] - 1LL * HashF[l-1] * P[ r - l + 1 ] % mod1;
            if( hash < 0 ) hash += mod1;
            return hash;
        }
        int GetBackwardHash( int l, int r ) {
            if( r == len ) return HashR[l];
            int hash = HashR[l] - 1LL * HashR[r+1] * P[ r - l + 1 ] % mod1;
            if( hash < 0 ) hash += mod1;
            return hash;
        }
        bool IsPalin( int l, int r ) {
            if( r < l ) return true;
            return (GetForwardHash(l, r) == GetBackwardHash(l, r));
        }

    private:
        int prime, mod1, mod2;
};

vector <pair<LL,int> > by;
LL ans[MAXN];
LL cnt;

void init()
{
    da(s,sa,rk,height,len,128);
    for(int i = 1; i <= len; ++i) RMQ[i] = height[i];
    initRMQ(len);
    initTree();
    for (int i = 0; i < len; i++)
        addLetter(i);

    for(int i = 2; i <= num; ++i)
        adj[tree[i].sufflink].push_back(i);

    dfs(1);

    for(int i = 3; i <= num; ++i)
        vc.push_back(pair<Pr,int> (Pr(tree[i].l,tree[i].r),tree[i].num));

    sort(vc.begin(),vc.end(),cmp);

    RollingHash obj;
    obj.Construct();

    cnt = 1;
    for(int i = 0; i < vc.size(); ++i)
    {
        by.push_back(pair<LL,int> (cnt,i));
        ans[i] = obj.GetForwardHash(vc[i].first.first+1,vc[i].first.second+1);
        //printf("%d %d %lld\n",vc[i].first.first+1,vc[i].first.second+1,ans[i]);
        cnt += vc[i].second;
    }

}



int main()
{
    int q;
    LL k;

    scanf("%d%d",&len,&q);
    scanf("%s",s);

    init();

    int pos;

    while(q--)
    {
        scanf("%lld",&k);
        if(k >= cnt) puts("-1");
        else
        {
            pos = -1;

            int l = 0,r = by.size()-1;

            while(l <= r)
            {
                int mid = (l+r)>>1;
                if(k >= by[mid].first)
                {
                    pos = mid;
                    l = mid+1;
                }
                else r = mid-1;
            }

            printf("%lld\n",ans[by[pos].second]);
        }
    }


    return 0;
}