| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 1163517 | pan | popa (BOI18_popa) | C++20 | 0 ms | 0 KiB |
#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i)
#define RFOR(i, x, n) for (ll i =x; i>=n; --i)
#pragma GCC optimize("O3","unroll-loops")
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<pi, ll> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
/*
int v[10005];
int le[10005], ri[10005];
int gcd(int a, int b)
{
return b == 0 ? a : gcd(b, a % b);
}
int query(ll l1, ll r1, ll l2, ll r2)
{
ll v1 = 0;
for (ll i=l1; i<=r1; ++i) v1 = gcd(v1, v[i]);
ll v2 = 0;
for (ll i=l2; i<=r2; ++i) v2 = gcd(v2, v[i]);
return v1==v2;
}
*/
int solve(int N, int* Left, int* Right)
{
ll n = N;
for (ll i=0; i<n; ++i) Left[i] = Right[i] = -1;
deque<ll> dq;
for (ll i=0; i<n; ++i)
{
ll eaten = -1;
while (dq.size() && query(dq.back(), i, i, i))
{
eaten = dq.back();
dq.pop_back();
}
if (eaten!=-1) Left[i] = eaten;
if (dq.size()) Right[dq.back()] = i;
dq.pb(i);
}
return dq[0];
}
/*
int main()
{
ll n;
cin >> n;
for (ll i=0; i<n; ++i) cin >> v[i];
ll root = solve(n, le, ri);
cout << root << endl;
for (ll i=0; i<n; ++i) cout << le[i] << ' ' ;
cout << endl;
for (ll i=0; i<n; ++i) cout << ri[i] << ' ' ;
cout << endl;
return 0;
}
*/