子串，容易想到 (SAM)
首先类似模板题一样的记录 (num(v)) 表示有多少子串经过 (v)
(sum(v)) 表示状态 (v) 在字符串中的出现次数
如果 (T = 1) 显然 (sum(v)=1)
否则，发现 (sum(v)) 实际就是其后继节点的 (sum_{}num(y))
然后 (O(n)) 即可扫除答案
#include <map>
#include <set>
#include <ctime>
#include <queue>
#include <stack>
#include <cmath>
#include <vector>
#include <bitset>
#include <cstdio>
#include <cctype>
#include <string>
#include <numeric>
#include <cstring>
#include <cassert>
#include <climits>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std ;
//#define int long long
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (a); i >= (b); i--)
#define loop(s, v, it) for (s::iterator it = v.begin(); it != v.end(); it++)
#define cont(i, x) for (int i = head[x]; i; i = e[i].nxt)
#define clr(a) memset(a, 0, sizeof(a))
#define ass(a, sum) memset(a, sum, sizeof(a))
#define lowbit(x) (x & -x)
#define all(x) x.begin(), x.end()
#define ub upper_bound
#define lb lower_bound
#define pq priority_queue
#define mp make_pair
#define pb push_back
#define pof pop_front
#define pob pop_back
#define fi first
#define se second
#define iv inline void
#define enter cout << endl
#define siz(x) ((int)x.size())
#define file(x) freopen(x".in", "r", stdin),freopen(x".out", "w", stdout)
typedef long long ll ;
typedef unsigned long long ull ;
typedef pair <int, int> pii ;
typedef vector <int> vi ;
typedef vector <pii> vii ;
typedef queue <int> qi ;
typedef queue <pii> qii ;
typedef set <int> si ;
typedef map <int, int> mii ;
typedef map <string, int> msi ;
const int N = 1000010 ;
const int INF = 0x3f3f3f3f ;
const int iinf = 1 << 30 ;
const ll linf = 2e18 ;
const int MOD = 1000000007 ;
const double eps = 1e-7 ;
void douout(double x){ printf("%lf
", x + 0.0000000001) ; }
template <class T> void print(T a) { cout << a << endl ; exit(0) ; }
template <class T> void chmin(T &a, T b) { if (a > b) a = b ; }
template <class T> void chmax(T &a, T b) { if (a < b) a = b ; }
template <class T> void upd(T &a, T b) { (a += b) %= MOD ; }
template <class T> void mul(T &a, T b) { a = (ll) a * b % MOD ; }
int n, t, k, p, np, q, nq, last = 1, cnt = 1 ;
int sum[N], num[N], c[N], a[N] ;
char s[N] ;
struct node {
int to[26], fa, len ;
} po[N] ;
void add(int c) {
p = last ;
last = np = ++cnt ;
po[np].len = po[p].len + 1 ; num[np] = 1 ;
for (; p && !po[p].to[c]; p = po[p].fa) po[p].to[c] = np ;
if (!p) po[np].fa = 1 ;
else {
q = po[p].to[c] ;
if (po[q].len == po[p].len + 1) po[np].fa = q ;
else {
nq = ++cnt ;
po[nq] = po[q] ;
po[nq].len = po[p].len + 1 ;
po[np].fa = po[q].fa = nq ;
for (; p && po[p].to[c] == q; p = po[p].fa) po[p].to[c] = nq ;
}
}
}
void work() {
rep(i, 1, cnt) c[po[i].len]++ ;
rep(i, 1, cnt) c[i] += c[i - 1] ;
rep(i, 1, cnt) a[c[po[i].len]--] = i ;
per(i, cnt, 1) {
if (t) num[po[a[i]].fa] += num[a[i]] ;
else num[a[i]] = 1 ;
}
num[1] = 0 ;
per(i, cnt, 1) {
sum[a[i]] = num[a[i]] ;
rep(j, 0, 25)
if (po[a[i]].to[j])
sum[a[i]] += sum[po[a[i]].to[j]] ;
}
}
void solve() {
if (k > sum[1]) {
puts("-1") ;
return ;
}
int now = 1 ;
k -= num[now] ;
while (k > 0) {
int p = 0 ;
while (k > sum[po[now].to[p]]) {
k -= sum[po[now].to[p]] ;
p++ ;
}
now = po[now].to[p] ;
putchar('a' + p) ;
k -= num[now] ;
}
}
signed main(){
// file("test") ;
scanf("%s%d%d", s + 1, &t, &k) ;
n = strlen(s + 1) ;
rep(i, 1, n) add(s[i] - 'a') ;
work() ;
solve() ;
return 0 ;
}
/*
写代码时请注意：
1.ll？数组大小，边界？数据范围？
2.精度？
3.特判？
4.至少做一些
思考提醒：
1.最大值最小->二分？
2.可以贪心么？不行dp可以么
3.可以优化么
4.维护区间用什么数据结构？
5.统计方案是用dp？模了么？
6.逆向思维？
*/