# |
Submission time |
Handle |
Problem |
Language |
Result |
Execution time |
Memory |
61811 |
2018-07-26T18:21:18 Z |
Benq |
popa (BOI18_popa) |
C++11 |
|
117 ms |
824 KB |
#include "popa.h"
#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 = 100001;
/*vi seq = {12, 4, 16, 2, 2, 20};
int g(int l, int r) {
int ret = 0;
FOR(i,l,r+1) ret = __gcd(ret,seq[i]);
return ret;
}
int query(int a, int b, int c, int d) {
return g(a,b) == g(c,d);
}*/
bool divi(int a, int b) {
return query(b,a,a,a);
}
int join(int a, int b, int* R) { // a is higher
R[a] = b;
return a;
}
int solve(int N, int* Left, int* Right) {
vi st;
F0R(i,N) {
int cur = -1;
while (sz(st) && divi(i,st.back())) {
cur = join(st.back(),cur,Right);
st.pop_back();
}
// if (sz(st)) cout << "OOPS " << i << " " << st.back() << " " << divi(i,st.back()) << "\n";
st.pb(i);
Left[i] = cur, Right[i] = -1;
/*for (int i: st) cout << i << " ";
cout << "\n";*/
}
int cur = -1;
while (sz(st)) {
cur = join(st.back(),cur,Right);
st.pop_back();
}
return cur;
}
/*int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
int* Left = new int[6];
int* Right = new int[6];
cout << solve(6,Left,Right) << "\n";
F0R(i,6) cout << Left[i] << " ";
cout << "\n";
F0R(i,6) cout << Right[i] << " ";
cout << "\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 |
12 ms |
248 KB |
Output is correct |
2 |
Correct |
15 ms |
436 KB |
Output is correct |
3 |
Correct |
13 ms |
436 KB |
Output is correct |
4 |
Correct |
10 ms |
436 KB |
Output is correct |
# |
Verdict |
Execution time |
Memory |
Grader output |
1 |
Correct |
109 ms |
652 KB |
Output is correct |
2 |
Correct |
88 ms |
652 KB |
Output is correct |
3 |
Correct |
87 ms |
692 KB |
Output is correct |
4 |
Correct |
93 ms |
692 KB |
Output is correct |
# |
Verdict |
Execution time |
Memory |
Grader output |
1 |
Correct |
84 ms |
692 KB |
Output is correct |
2 |
Correct |
102 ms |
700 KB |
Output is correct |
3 |
Correct |
117 ms |
824 KB |
Output is correct |
4 |
Correct |
72 ms |
824 KB |
Output is correct |