이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
/* Speech to the young */
//#include <bits/stdc++.h>
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <cassert>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <fstream>
#include <unordered_map>
using namespace std;
#define pb push_back
#define all(x) x.begin(),x.end()
#define F first
#define S second
#define YOSIK() ios_base::sync_with_stdio(0),cin.tie(0)
#define int long long
#define pans cout << "\n------ans-------\n"
const int N = 1e6 + 10;
const int INF = 1e18 + 7;
const int MOD = 1e9 + 7;
const int P = 31;
int n, m, t[N * 4];
int a[N];
void build (int v, int tl, int tr) {
if (tl == tr) {
if (tl == 1) return;
if (a[tl] < a[tl - 1]) t[v] = 1;
return;
}
int mid = (tl + tr) >> 1;
build (v + v, tl, mid), build (v + v + 1, mid + 1, tr);
t[v] = t[v + v] + t[v + v + 1];
return;
}
int get (int v, int tl, int tr, int l, int r) {
if (l <= tl && tr <= r) return t[v];
if (r < tl || tr < l) return 0;
int mid = (tl + tr) >> 1;
return get (v + v, tl, mid, l, r) + get (v + v + 1, mid + 1, tr, l, r);
}
void Solution () {
cin >> n >> m;
int mn = INF;
for (int i = 1; i <= n; ++ i) {
cin >> a[i];
mn = min (mn, a[i]);
}
set <int> s;
int ok = 0;
build (1, 1, n);
for (int i = 1; i <= m; ++ i) {
int l, r, k;
cin >> l >> r >> k;
if (k < mn) {
if (get (1, 1, n, l + 1, r) > 0) cout << 0 << '\n';
else cout << 1 << '\n';
} else {
ok = 0;
s.clear();
for (int j = r; j >= l; -- j) {
auto it = s.lower_bound (a[j]);
if (it == s.begin()) {
s.insert(a[j]);
continue;
}
it --;
if (*(it) + a[j] > k) {
ok = 1;
break;
}
s.insert (a[j]);
}
cout << (ok ? "0\n" : "1\n");
}
}
return;
}
signed main () {
YOSIK();
// precalc();
int T = 1;
//cin >> T;
while (T --) Solution ();
exit (0);
}
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