제출 #57750

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
57750		Benq	Poklon (COCI17_poklon)	C++14	140 / 140	732 ms	43408 KiB

poklon

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 500001;

template<class T, int SZ> struct BIT {
    T bit[SZ+1];
    
    BIT() { memset(bit,0,sizeof bit); }
    
    void upd(int k, T val) { // add val to index k
        for( ;k <= SZ; k += (k&-k)) bit[k] += val;
    }
    
    T query(int k) {
        T temp = 0;
        for (;k > 0;k -= (k&-k)) temp += bit[k];
        return temp;
    }
    T query(int l, int r) { return query(r)-query(l-1); } // range query [l,r]
};

BIT<int,MX> B;
map<int,int> m;
int N,Q, A[MX], ans[MX];
vi oc[MX];
vpi todo[MX];

void ad(int L, int R, int x) {
    B.upd(L,x);
    B.upd(R+1,-x);
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> Q;
    int co = 0;
    F0R(i,N) {
        cin >> A[i];
        if (!m.count(A[i])) m[A[i]] = co++;
        A[i] = m[A[i]];
    }
    F0R(i,Q) {
        int a,b; cin >> a >> b;
        todo[a-1].pb({b-1,i});
    }
    F0Rd(i,N) {
        int t = A[i];
        if (sz(oc[t]) >= 2) ad(oc[t][sz(oc[t])-2],(sz(oc[t]) >= 3 ? oc[t][sz(oc[t])-3] : N)-1,-1);
        oc[t].pb(i);
        if (sz(oc[t]) >= 2) ad(oc[t][sz(oc[t])-2],(sz(oc[t]) >= 3 ? oc[t][sz(oc[t])-3] : N)-1,1);
        for (auto a: todo[i]) ans[a.s] = B.query(a.f);
    }
    
    F0R(i,Q) cout << ans[i] << "\n";
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

• Subtask 1: 140 / 140