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build a suffix automata.
T = 0, each substring operator once the operator or each substring substring \ (endpos \) set size times.
With \ (f [i] \) represents the junction \ (I \) represented by \ (endpos \) set size, the \ (f [i] \) for the parent tree subtree \ (F \) of and (T time = 0, f [i] are 1).
With \ (g [i] \) represents the junction \ (I \) number substring starting, the \ (g [i] \) of \ (f [i] \) plus nodes \ (I \) all the edges of the \ (g [v] \) sum.
The first run is similar to balanced trees \ (k \) small.
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#define O(x) cout << #x << "=" << x << endl;
using namespace std;
const int MAXN = 1000010;
struct SAM{
int ch[26];
int len, fa;
}sam[MAXN << 1];
int las = 1, cnt = 1, f[MAXN << 1], g[MAXN << 1];
struct Edge{
int next, to;
}e[MAXN << 1];
int head[MAXN << 1], num;
inline void Add(int from, int to){
e[++num].to = to; e[num].next = head[from]; head[from] = num;
}
inline void add(int c){
int p = las; int np = las = ++cnt;
sam[np].len = sam[p].len + 1; f[cnt] = 1;
for(; p && !sam[p].ch[c]; p = sam[p].fa) sam[p].ch[c] = np;
if(!p) sam[np].fa = 1;
else{
int q = sam[p].ch[c];
if(sam[q].len == sam[p].len + 1) sam[np].fa = q;
else{
int nq = ++cnt; sam[nq] = sam[q];
sam[nq].len = sam[p].len + 1;
sam[q].fa = sam[np].fa = nq;
for(; p && sam[p].ch[c] == q; p = sam[p].fa) sam[p].ch[c] = nq;
}
}
}
char a[MAXN];
int t, k;
void dfs(int u){
for(int i = head[u]; i; i = e[i].next){
dfs(e[i].to);
f[u] += f[e[i].to];
}
}
void Dfs(int u){
g[u] = f[u];
for(int i = 0; i < 26; ++i)
if(sam[u].ch[i]){
if(!g[sam[u].ch[i]]) Dfs(sam[u].ch[i]);
g[u] += g[sam[u].ch[i]];
}
}
void DFS(int u, int res){
if(res <= f[u]) return;
res -= f[u];
for(int i = 0; i < 26; ++i)
if(sam[u].ch[i])
if(g[sam[u].ch[i]] >= res){
putchar(i + 'a');
DFS(sam[u].ch[i], res);
return;
}
else res -= g[sam[u].ch[i]];
printf("-1\n");
}
int main(){
scanf("%s", a + 1);
scanf("%d%d", &t, &k);
int len = strlen(a + 1);
for(int i = 1; i <= len; ++i)
add(a[i] - 'a');
for(int i = 2; i <= cnt; ++i)
Add(sam[i].fa, i);
if(t) dfs(1);
else for(int i = 2; i <= cnt; ++i) f[i] = 1;
f[1] = 0;
Dfs(1); DFS(1, k);
return 0;
}