题面
Sol
(sam)
求一个串的不重复的第(k)小子串很好办
如果可以相同
那么要算上每个点(前缀)的后缀的个数
那么就是这个(endpos(right))集合的子串的出现次数
# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
# define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
using namespace std;
typedef long long ll;
IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
const int maxn(1e6 + 5);
int trans[26][maxn], fa[maxn], len[maxn], tot = 1, last = 1;
int n, f[maxn], t[maxn], id[maxn], g[maxn];
char s[maxn];
IL void Extend(RG int c){
RG int p = last, np = ++tot; last = tot;
len[np] = len[p] + 1, g[np] = 1;
while(p && !trans[c][p]) trans[c][p] = np, p = fa[p];
if(!p) fa[np] = 1;
else{
RG int q = trans[c][p];
if(len[q] == len[p] + 1) fa[np] = q;
else{
RG int nq = ++tot;
len[nq] = len[p] + 1, fa[nq] = fa[q];
for(RG int i = 0; i < 26; ++i) trans[i][nq] = trans[i][q];
fa[q] = fa[np] = nq;
while(p && trans[c][p] == q) trans[c][p] = nq, p = fa[p];
}
}
}
IL void Calc(RG int k){
RG int nw = 1;
if(k <= f[nw]){
while(k > 0){
for(RG int i = 0; i < 26; ++i){
RG int p = trans[i][nw];
if(p){
if(k > 0 && k <= f[p]){
k -= g[p], putchar(i + 'a'), nw = p;
break;
}
else k -= f[p];
}
}
}
}
else printf("-1");
puts("");
}
int main(RG int argc, RG char* argv[]){
scanf(" %s", s), n = strlen(s);
for(RG int i = 0; i < n; ++i) Extend(s[i] - 'a');
for(RG int i = 1; i <= tot; ++i) ++t[len[i]];
for(RG int i = 1; i <= tot; ++i) t[i] += t[i - 1];
for(RG int i = 1; i <= tot; ++i) id[t[len[i]]--] = i;
if(Input()){
for(RG int i = tot; i; --i)
if(fa[id[i]]) g[fa[id[i]]] += g[id[i]];
f[1] = 0;
for(RG int i = tot; i; --i){
f[id[i]] = g[id[i]];
for(RG int j = 0; j < 26; ++j) f[id[i]] += f[trans[j][id[i]]];
}
}
else{
for(RG int i = 1; i <= tot; ++i) f[i] = 1, g[i] = 1;
for(RG int i = tot; i; --i)
for(RG int j = 0; j < 26; ++j) f[id[i]] += f[trans[j][id[i]]];
}
Calc(Input());
return 0;
}