이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define lc (i << 1)
#define rc (i << 1 | 1)
using namespace std;
using ll = long long;
using pii = pair<int, int>;
const int MN = 1e6 + 5, LN = 17, MOD = 1e9 + 7, INF = 0x3f3f3f3f, BSZ = 320;
char s[MN];
int spt[MN][21], dep[MN];
int cur;
void Init() {
}
// simiulate it as a tree, append on to previous node
void TypeLetter(char L) {
cur++;
s[cur] = L;
spt[cur][0] = cur - 1;
dep[cur] = dep[cur - 1] + 1;
for (int j = 1; j < LN; j++)
spt[cur][j] = spt[spt[cur][j - 1]][j - 1];
}
// undo will also act as an append, we go back U additions (therefore including all the undos as well)
// we can delete everything in the subtree by readding it to its parent
void UndoCommands(int U) {
int last = cur - U;
cur++;
s[cur] = s[last];
int par = spt[last][0];
spt[cur][0] = par;
dep[cur] = dep[par] + 1;
for (int j = 1; j < LN; j++)
spt[cur][j] = spt[spt[cur][j - 1]][j - 1];
}
char GetLetter(int P) {
P++;
int dis = dep[cur] - P;
int now = cur;
for (int j = LN - 1; j >= 0; j--)
if (dis - (1 << j) >= 0) {
now = spt[now][j];
dis -= (1 << j);
}
return s[now];
}
/*
int main() {
Init();
TypeLetter('a');
TypeLetter('b');
cout << GetLetter(1) << '\n';
TypeLetter('d');
UndoCommands(2);
UndoCommands(1);
cout << GetLetter(2) << '\n';
TypeLetter('e');
UndoCommands(1);
UndoCommands(5);
TypeLetter('c');
cout << GetLetter(2) << '\n';
UndoCommands(2) ;
cout << GetLetter(2) << '\n';
}
*/
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