이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#include <iostream>
#include <queue>
#include <algorithm>
using namespace std;
const int MAXN = 2e6 + 10;
const int MOD = 1e9 + 7;
int stma[(1 << 22)], stupd[(1 << 22)], stidx[(1 << 22)];
void build(int l, int r, int idx) {
if(l == r) {
stidx[idx] = l; return;
}
int mid = (l + r) >> 1;
build(l, mid, (idx<<1)+1);
build(mid+1, r, (idx<<1)+2);
stma[idx] = max(stma[(idx<<1)+1], stma[(idx<<1)+2]);
stidx[idx] = (stma[(idx<<1)+1] == stma[idx] ? stidx[(idx<<1)+1] : stidx[(idx<<1)+2]);
}
void u(int l, int r, int constl, int constr, int idx, int val) {
if(l <= constl && constr <= r) {
stma[idx] += val;
stupd[idx] += val;
return;
}
int mid = (constl + constr) >> 1;
stma[(idx<<1)+1] += stupd[idx];
stma[(idx<<1)+2] += stupd[idx];
stupd[(idx<<1)+1] += stupd[idx];
stupd[(idx<<1)+2] += stupd[idx];
stupd[idx] = 0;
if(mid >= l && r >= constl) u(l, r, constl, mid, (idx<<1)+1, val);
if(constr >= l && r >= mid+1) u(l, r, mid+1, constr, (idx<<1)+2, val);
stma[idx] = max(stma[(idx<<1)+1], stma[(idx<<1)+2]);
stidx[idx] = (stma[(idx<<1)+1] == stma[idx] ? stidx[(idx<<1)+1] : stidx[(idx<<1)+2]);
}
int seg[MAXN << 2];
void u2(int l, int r, int tar, int idx, int val) {
if(l == r) {
seg[idx] = val; return;
}
int mid = (l + r) >> 1;
if(tar <= mid) u2(l, mid, tar, (idx<<1)+1, val);
else u2(mid+1, r, tar, (idx<<1)+2, val);
seg[idx] = min(seg[(idx<<1)+1], seg[(idx<<1)+2]);
}
int qu2(int l, int r, int constl, int constr, int idx, int val) { // first <= v
if(l <= constl && constr <= r) {
if(seg[idx] > val) return -1;
while(constl < constr) {
int mid = (constl + constr) >> 1;
if(seg[(idx<<1)+1] <= val) constr = mid, idx = (idx<<1)+1;
else constl = mid+1, idx = (idx<<1)+2;
}
return constl;
}
int mid = (constl + constr) >> 1;
if(mid < l || r < constl) return qu2(l, r, mid+1, constr, (idx<<1)+2, val);
else if(constr < l || r < mid+1) return qu2(l, r, constl, mid, (idx<<1)+1, val);
else {
int lc = qu2(l, r, constl, mid, (idx<<1)+1, val);
if(lc != -1) return lc;
return qu2(l, r, mid+1, constr, (idx<<1)+2, val);
}
}
inline int read() {
int x = 0, c = getchar(), neg = false;
while(('0' > c || c > '9') && c!='-' && c!=EOF) c = getchar();
if(c == '-') neg = true, c = getchar();
while('0' <= c && c <= '9') x = x*10 + (c^'0'), c = getchar();
if(neg) x = -x;
return x; // returns 0 if EOF
}
int32_t main() {
ios::sync_with_stdio(0); cin.tie(0);
int32_t n, d, t;
n = read();
d = read();
t = read();
pair<int, int> tri[n+1];
for(int i=1; i<=n; i++) {
int32_t x = read();
tri[i] = {x-i, i};
}
sort(tri+1, tri+1+n);
build(1, n-1, 0);
int ans = 0;
int pr = 0;
priority_queue<int> ok;
int sv[n+1];
for(int i=n; i>=1; i--) {
while(pr+1 <= n && tri[pr+1].first <= t-i) {
pr++;
if(tri[pr].second <= i) ok.push(tri[pr].second);
}
int bruh = (ok.size() ? ok.top() : -1);
if(bruh == i) ok.pop();
sv[i]= bruh;
if(bruh != -1) {
ans++;
if(bruh < i) {
u(bruh, i-1, 1, n-1, 0, 1);
}
u2(1, n, i, 0, bruh);
}
else {
u2(1, n, i, 0, MOD);
}
}
while(d--) {
int ss = stidx[0];
int prv=ans;
while(1) {
int idx = qu2(ss+1, n, 1, n, 0, ss);
if(idx == -1) break;
u2(1, n, idx, 0, MOD);
u(sv[idx], idx-1, 1, n-1, 0, -1);
ans--;
}
if(prv == ans) break;
}
cout << ans << "\n";
}
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