| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 497596 | armashka | Hedgehog Daniyar and Algorithms (IZhO19_sortbooks) | C++17 | 0 ms | 0 KiB |
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>
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define fast ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define all(v) v.begin(),v.end()
#define pb push_back
#define sz size()
#define ft first
#define sd second
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef unsigned long long ull;
const int N = 1e6 + 5;
const ll M = 1e8;
const ll inf = 1e18;
const ll mod = 1e9;
const double Pi = acos(-1);
ll binpow(ll x, ll ti) { ll res = 1;while (ti){if(ti & 1)res *= x;x *= x;ti >>= 1; x %= mod; res %= mod;} return res;}
ll binmul(ll x, ll ti) { ll res = 0;while (ti){if(ti & 1)res += x;x += x;ti >>= 1; x %= mod; res %= mod;} return res;}
ll nok(ll a, ll b) { return (a*b)/__gcd(abs(a),abs(b)) * (a*b > 0 ? 1 : -1); }
bool odd(ll n) { return (n % 2 == 1); }
bool even(ll n) { return (n % 2 == 0); }
ll n, m, a[N], q[N], mx[N][30], lg[N], ans[N], dp[5005][5005], maaax, t[N * 4];
vector <pair<pll, ll>> pos[N];
ll get(ll l, ll r) {
ll len = (r - l + 1);
ll k = lg[len];
return max(mx[l][k], mx[r - (1 << k) + 1][k]);
}
void upd(ll pos, ll val, ll v = 1, ll tl = 1, ll tr = n) {
if (tl == tr) {
t[v] = val;
return;
}
ll tm = (tl + tr) / 2;
if (pos <= tm) upd(pos, val, v + v, tl, tm);
else upd(pos, val, v + v + 1, tm + 1, tr);
t[v] = t[v + v] + t[v + v + 1];
}
ll get(ll l, ll r, ll v = 1, ll tl = 1, ll tr = n) {
if (r < tl || tr < l) return 0;
if (l <= tl && tr <= r) return t[v];
ll tm = (tl + tr) / 2;
return get(l, r, v + v, tl, tm) + get(l, r, v + v + 1, tm + 1, tr);
}
const void solve(/*Armashka*/) {
cin >> n >> m;
for (int i = 2; i <= n; ++ i) lg[i] = lg[i / 2] + 1;
for (int i = 1; i <= n; ++ i) {
cin >> a[i];
maaax = max(maaax, a[i]);
mx[i][0] = a[i];
}
if (n <= 5000) {
for (int l = 1; l <= n; ++ l) {
set <ll> st;
st.insert(a[l]);
for (int r = l + 1; r <= n; ++ r) {
dp[l][r] = dp[l][r - 1];
if (*st.rbegin() > a[r]) dp[l][r] = max(dp[l][r], *st.rbegin() + a[r]);
st.insert(a[r]);
}
}
while (m --) {
ll l, r, k;
cin >> l >> r >> k;
if (dp[l][r] <= k) cout << "1\n";
else cout << "0\n";
}
return;
}
if (mx <= 1000) {
for (int i = 1; i <= 25; ++ i) {
for (int j = 1; j + (1 << (i - 1)) - 1 <= n; ++ j) {
mx[j][i] = max(mx[j][i - 1], mx[j + (1 << (i - 1))][i - 1]);
}
}
for (int i = 1; i <= m; ++ i) {
ll l, r, k;
cin >> l >> r >> k;
pos[r].pb({{l, k}, i});
}
for (int i = 1; i <= n; ++ i) {
q[a[i]] = i;
while (pos[i].sz) {
ll l = pos[i].back().ft.ft, k = pos[i].back().ft.sd, id = pos[i].back().sd;
pos[i].pop_back();
ll res = 0;
for (int j = 1; j <= 1000; ++ j) {
if (q[j] <= l) continue;
ll cur = get(l, q[j]);
if (cur > j) res = max(res, cur + j);
}
if (res <= k) ans[id] = 1;
}
}
for (int i = 1; i <= m; ++ i) cout << ans[i] << "\n";
return;
}
for (int i = 1; i < n; ++ i) upd(i, (a[i] > a[i + 1]));
for (int i = 1; i <= m; ++ i) {
ll l, r, k;
cin >> l >> r >> k;
if (get(l + 1, r) > 0) cout << "0\n";
else cout << "1\n";
}
}
signed main() {
fast;
//freopen("divide.in", "r", stdin);
//freopen("divide.out", "w", stdout);
int tt = 1;
//cin >> tt;
while (tt --) {
solve();
}
}
/*
5 4 4
1 2
3 1
3 4
5 3
4 5 2 3
2 1 3 1
1 3 5
2 3 4 5
2 1 3 1
*/