Submission #216731

#TimeUsernameProblemLanguageResultExecution timeMemory
216731eggag32Hedgehog Daniyar and Algorithms (IZhO19_sortbooks)C++17
0 / 100
150 ms18664 KiB
#pragma GCC optimize ("O3") #pragma GCC target ("sse4") #include <bits/stdc++.h> using namespace std; typedef long long ll; typedef long double ld; typedef vector<int> vi; typedef pair<int, int> pi; #define debug(x) cerr << #x << ": " << x << endl; #define debug2(x, y) debug(x) debug(y); #define repn(i, a, b) for(int i = (int)(a); i < (int)(b); i++) #define rep(i, a) for(int i = 0; i < (int)(a); i++) #define all(v) v.begin(), v.end() #define mp make_pair #define pb push_back #define lb lower_bound #define ub upper_bound #define fi first #define se second #define sq(x) ((x) * (x)) template<class T> T gcd(T a, T b){ return ((b == 0) ? a : gcd(b, a % b)); } int main(){ ios_base::sync_with_stdio(false); cin.tie(0); //freopen("input.in", "r", stdin); //freopen("output.out", "w", stdout); int n, m; cin >> n >> m; vi w(n); rep(i, n) cin >> w[i]; //so the answer is (mx[i] + mn[i]) <= k[i] rep(j, m){ int l, r, k; cin >> l >> r >> k; vi cr; repn(i, l - 1, r) cr.pb(w[i]); sort(all(cr)); int mn = cr[0], mx = 0; repn(i, l - 1, r) if(w[i] != cr[i]){ mx = max(mx, w[i]); } if(k >= (mx + mn)) cout << 1 << endl; else cout << 0 << endl; } return 0; } /* Things to look out for: - Integer overflows - Array bounds - Special cases Be careful! */
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