이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "gap.h"
#include <bits/stdc++.h>
#define bupol __builtin_popcount
//#define int long long
#define ll long long
#define ld long double
#define fi first
#define se second
#define pb push_back
#define lf (id<<1)
#define rg ((id<<1)|1)
#define md ((l+r)>>1)
using namespace std;
const int MAXN = 1e5+5;
const int MAXK = 205;
const int LOG = 20;
const int MOD = 1e9+7;
const int SQRT = 520;
const ll INF = 1e18+10;
typedef pair<ll,ll> pii;
typedef pair<ll,pii> ipii;
ll mn, mx;
ll x, y;
int n;
set<ll> s; ll ans = INF;
void sol(ll l, ll r){ // cari semua di range l-r
//cout << l << ' '<< r << " p\n";
if(l > r || r-l+1 <= ans) return;
MinMax(l, r, &mn, &mx);
if(mn > mx){ // pasti berurutan
ans = max(ans, mn-mx);
s.insert(mn); s.insert(mx);
return;
}
//if(mn==-1 || mx==-1) return; // gk ada lagi di range ini
if(s.find(mn)!=s.end() && s.find(mx)!=s.end()) return;
s.insert(mn); s.insert(mx);
int tmn = mn, tmx = mx;
tmn++; tmx--;
sol(tmn, ((tmn+tmx)>>1) );
//cout << ((tmn+tmx)>>1)+1 << ' '<< tmx << " oo\n";
sol(((tmn+tmx)>>1)+1, tmx);
}
long long findGap(int T, int N)
{
n = N;
sol(0, INF);
//for(auto in : s) cout << in << " xx\n";
auto it = s.begin(), it2 = s.begin(); it2++;
while(it2 != s.end()){
if((*it)!=-1) ans = max(ans, (*it2)-(*it));
it++; it2++;
}
return ans;
}
// for(int i=1; i<=(n+1)/2; i++){
// MinMax(x, y, &mn, &mx);
// a[i] = mn; a[n-i+1] = mx;
// //cout << mn << ' '<< mx << " p\n";
// x = mn+1; y = mx-1;
// }
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |