이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
// Knapsack DP is harder than FFT.
#include<bits/stdc++.h>
using namespace std;
typedef int64_t ll;
typedef pair<int,int> pii; typedef pair<ll,ll> pll;
#define ff first
#define ss second
#define pb emplace_back
#define FOR(i,n) for(int i=0;i<(n);++i)
#define FOO(i,a,b) for(int i=(a);i<=int(b);++i)
#define OOF(i,a,b) for(int i=(a);i>=int(b);--i)
#define AI(x) (x).begin(),(x).end()
template<class I>bool chmax(I&a,I b){return a<b?(a=b,true):false;}
template<class I>bool chmin(I&a,I b){return b<a?(a=b,true):false;}
template<class V>void lisan(V&v){sort(AI(v));v.erase(unique(AI(v)),v.end());}
template<class V,class T>int lspos(const V&v,T x){return lower_bound(AI(v),x)-v.begin();}
template<class...T>void RI(T&...t){(...,(cin>>t));}
template<class...T>void PL(T...t){int c=sizeof...(T);if(!c){cout<<'\n';return;}(...,(cout<<t<<(--c?' ':'\n')));}
constexpr inline ll cdiv(ll x,ll m){return x/m+(x%m?(x<0)^(m>0):0);}
constexpr inline ll mpow(ll x,ll e,ll m){ll r=1;for(x%=m;e;e/=2,x=x*x%m)if(e&1)r=r*x%m;return r;}
#ifdef OWO
#define safe cerr<<"\033[1;32m"<<__PRETTY_FUNCTION__<<" - "<<__LINE__<<" JIZZ\033[0m\n"
#define debug(args...) SDF(#args, args)
#define OIU(args...) ostream& operator<<(ostream&O,args)
#define LKJ(S,B,E,F) template<class...T>OIU(S<T...>s){O<<B;int c=0;for(auto i:s)O<<(c++?", ":"")<<F;return O<<E;}
LKJ(vector,'[',']',i)LKJ(deque,'[',']',i)LKJ(set,'{','}',i)LKJ(multiset,'{','}',i)LKJ(unordered_set,'{','}',i)LKJ(map,'{','}',i.ff<<':'<<i.ss)LKJ(unordered_map,'{','}',i.ff<<':'<<i.ss)
template<class...T>OIU(pair<T...>p){return O<<'('<<p.ff<<','<<p.ss<<')';}
template<class T,size_t N>OIU(array<T,N>a){return O<<vector<T>(AI(a));}
template<class...T>OIU(tuple<T...>t){return O<<'(',apply([&O](T...s){int c=0;(...,(O<<(c++?", ":"")<<s));},t),O<<')';}
template<class...T>void SDF(const char* s,T...a){int c=sizeof...(T);if(!c){cerr<<"\033[1;32mvoid\033[0m\n";return;}(cerr<<"\033[1;32m("<<s<<") = (",...,(cerr<<a<<(--c?", ":")\033[0m\n")));}
#else
#pragma GCC optimize("Ofast")
#define safe ((void)0)
#define debug(...) ((void)0)
#endif
constexpr ll inf = 1e9, INF = 4e18;
const int N = 1e6 + 225, M = 1e9 + 7;
int n, m, a[N];
struct sufMax{
int v[N];
void mod(int p, int val){
for(; p > 0; p -= p & -p) chmax(v[p], val);
}
int query(int p){
int r = 0;
for(; p < N; p += p & -p) chmax(r, v[p]);
return r;
}
} bit;
struct query{
int l, r, k, id;
query(){}
query(int l, int r, int k, int i): l(l), r(r), k(k), id(i){}
bool operator<(const query &q){ return r < q.r;}
};
int32_t main(){
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
cin >> n >> m; FOO(i,1,n) cin >> a[i];
vector<query> qs;
vector<bool> ans(m);
for(int l, r, k, i = 0; i < m; ++i){
cin >> l >> r >> k;
qs.pb(l, r, k, i);
}
sort(AI(qs));
stack<int> stk;
int i = 0;
for(auto q: qs){
auto [l, r, k, id] = q;
debug(l, r, k, id);
while(i < r){
++i;
while(!stk.empty() and a[stk.top()] <= a[i]) stk.pop();
if(!stk.empty()) bit.mod(stk.top(), a[stk.top()] + a[i]);
stk.push(i);
}
ans[id] = (bit.query(l) <= k);
}
for(auto i: ans) cout << i << '\n';
return 0;
}
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