# |
Submission time |
Handle |
Problem |
Language |
Result |
Execution time |
Memory |
57750 |
2018-07-16T01:48:26 Z |
Benq |
Poklon (COCI17_poklon) |
C++14 |
|
732 ms |
43408 KB |
#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
*/
# |
Verdict |
Execution time |
Memory |
Grader output |
1 |
Correct |
22 ms |
25848 KB |
Output is correct |
2 |
Correct |
23 ms |
25848 KB |
Output is correct |
3 |
Correct |
22 ms |
25900 KB |
Output is correct |
4 |
Correct |
25 ms |
26028 KB |
Output is correct |
5 |
Correct |
99 ms |
29396 KB |
Output is correct |
6 |
Correct |
104 ms |
29484 KB |
Output is correct |
7 |
Correct |
224 ms |
33088 KB |
Output is correct |
8 |
Correct |
341 ms |
37232 KB |
Output is correct |
9 |
Correct |
544 ms |
40520 KB |
Output is correct |
10 |
Correct |
732 ms |
43408 KB |
Output is correct |