Submission #497582

#TimeUsernameProblemLanguageResultExecution timeMemory
497582armashkaHedgehog Daniyar and Algorithms (IZhO19_sortbooks)C++17
0 / 100
384 ms262148 KiB
#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]; 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]); } const void solve(/*Armashka*/) { cin >> n >> m; lg[0] = 1; for (int i = 2; i <= n; ++ i) lg[i] = lg[i / 2] + 1; for (int i = 1; i <= n; ++ i) { cin >> a[i]; mx[i][0] = a[i]; } for (int i = 1; i <= 25; ++ i) { for (int j = 1; j + (1 << (i - 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[a[i]]); if (cur > a[i]) res = max(res, cur); } if (res <= k) ans[id] = 1; } } for (int i = 1; i <= m; ++ i) cout << ans[i] << "\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 */
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