[树形DP][后缀自动机][后缀树] CodeChef January Challenge 2018 KILLKTH

$Solution$

反串的后缀自动机的parent树就是原串的后缀树。
按dfs序记录子树大小前缀和，二分位置。
定位后就可以在节点内二分（可以直接算的啦）

#include <bits/stdc++.h>
#define show(x) cerr << #x << " = " << x << endl
using namespace std;
typedef long long ll;
typedef pair<int, int> Pairs;

const int N = 404040;

inline char get(void) {
    static char buf[100000], *S = buf, *T = buf;
    if (S == T) {
        T = (S = buf) + fread(buf, 1, 100000, stdin);
        if (S == T) return EOF;
    }
    return *S++;
}
template<typename T>
inline void read(T &x) {
    static char c; x = 0; ll sgn = 0;
    for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;
    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
    if (sgn) x = -x;
}
inline void read(char *s) {
    static char c; ll len = 0;
    for (c = get(); c < 'a' || c > 'z'; c = get());
    for (; c >= 'a' && c <= 'z'; c = get()) s[len++] = c;
    s[len] = 0;
}

inline ll S(int l, int r) {
    return (ll)(l + r) * (r - l + 1) / 2;
}

namespace SAM {
    int s[N];
    int rgt[N], r[N];
    ll size[N], sum[N];
    int par[N], mx[N];
    int buc[N], pre[N], id[N];
    int go[N][27], to[N][27];
    int Tcnt, last, root, len, clc;
    inline int extend(int ii) {
        int key = s[ii];
        int p = last, np = ++Tcnt;
        r[np] = ii;
        mx[np] = mx[p] + 1; ++rgt[np];
        for (; p && !go[p][key]; p = par[p]) go[p][key] = np;
        if (p) {
            int q = go[p][key];
            if (mx[q] != mx[p] + 1) {
                int nq = ++Tcnt;
                mx[nq] = mx[p] + 1; r[nq] = r[q];
                memcpy(go[nq], go[q], sizeof go[q]);
                par[nq] = par[q];
                par[q] = par[np] = nq;
                for (; p && go[p][key] == q; p = par[p]) go[p][key] = nq;
            } else {
                par[np] = q;
            }
        } else {
            par[np] = root;
        }
        return last = np;
    }
  inline void sort(void) {
    for (int i = 1; i <= Tcnt; i++) ++buc[mx[i]];
    for (int i = 1; i <= Tcnt; i++) buc[i] += buc[i - 1];
    for (int i = Tcnt; i; i--) id[buc[mx[i]]--] = i;
  }
    inline void build(void) {
        for (int i = Tcnt; i; i--) {
            int x = id[i];
            rgt[par[x]] += rgt[x];
            to[par[x]][s[r[x] - mx[par[x]]]] = x;
            size[x] = S(mx[par[x]] + 1, mx[x]) * rgt[x];
        }
    }
    inline void dfs(int u) {
        pre[u] = ++clc; id[clc] = u;
        for (int i = 0; i < 26; i++)
            if (to[u][i]) dfs(to[u][i]);
    }

    inline void init(char *begin, char *end) {
        Tcnt = last = root = 1; clc = 0;
        len = end - begin;
        for (int i = 0; i < len; i++) s[i] = begin[i] - 'a';
        reverse(s, s + len);
        for (int i = 0; i < len; i++) extend(i);
        sort(); build(); dfs(root);
        for (int i = 1; i <= Tcnt; i++) sum[i] = sum[i - 1] + size[id[i]];
    }
    inline char kth(ll k) {
        int u = lower_bound(sum + 1, sum + Tcnt + 1, k) - sum;
        k -= sum[u - 1]; u = id[u];
        int L = mx[par[u]] + 1, R = mx[u], Mid, Pos;
        while (L <= R) {
            Mid = (L + R) >> 1;
            if (S(mx[par[u]] + 1, Mid) * rgt[u] >= k) R = (Pos = Mid) - 1;
            else L = Mid + 1;
        }
        k -= S(mx[par[u]] + 1, Pos - 1) * rgt[u];
        k = (k - 1) % Pos + 1;
        return s[r[u] - k + 1] + 'a';
    }
}

char ch[N];
char c;
int n, q, lst;
ll x, y;

int main(void) {
    freopen("1.in", "r", stdin);
    freopen("1.out", "w", stdout);
    read(ch); n = strlen(ch);
    SAM::init(ch, ch + n);
    read(q);
    while (q--) {
        read(x); read(y);
        c = SAM::kth((ll)x * lst % y + 1);
        //c = SAM::kth(x);
        lst += c;
        putchar(c);
        putchar('\n');
    }
    return 0;
}