| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 499855 | robell | Gap (APIO16_gap) | C++14 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
/*
Problem:
Problem Link:
Notes:
*/
#pragma GCC optimize("O2")
#include <bits/stdc++.h>
#include "gap.h"
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag, tree_order_statistics_node_update> indexed_set;
typedef long long ll;
#define pb push_back
#define eb emplace_back
#define countbits __builtin_popcount
#define beg0 __builtin_clz
#define terminal0 __builtin_ctz
#define f first
#define s second
int mod=1e9+7;
inline void rv(int &n){
n=0;int m=1;char c=getchar();if (c=='-'){m=-1; c=getchar();}
for (;c>47 && c<58;c=getchar()){n=n*(1<<1)+n*(1<<3)+c-48;}
n*=m;
}
inline void rv(ll &n){
n=0;int m=1;char c=getchar();if (c=='-'){m=-1; c=getchar();}
for (;c>47 && c<58;c=getchar()){n=n*(1<<1)+n*(1<<3)+c-48;}
n*=m;
}
inline void rv(double &n){
n=0;int m=1;char c=getchar();
if (c=='-'){m=-1; c=getchar();}for (;c>47 && c<58;c=getchar()){n=n*(1<<1)+n*(1<<3)+c-48;}
if (c=='.'){double p = 0.1;c=getchar();for (;c>47 && c<58;c=getchar()){n+=((c-48)*p);p/=10;}}
n*=m;
}
inline void rv(float &n){
n=0;int m=1;char c=getchar();
if (c=='-'){m=-1; c=getchar();}for (;c>47 && c<58;c=getchar()){n=n*(1<<1)+n*(1<<3)+c-48;}
if (c=='.'){double p = 0.1;c=getchar();for (;c>47 && c<58;c=getchar()){n+=((c-48)*p);p/=10;}}
n*=m;
}
inline void rv(string &w){w="";char c=getchar();while (c!=' '&&c!='\n'&&c!=EOF){w+=c;c=getchar();}}
inline void rv(char &c){c=getchar();}
template<typename T, typename ...Types>
void rv(T &n, Types&&... args){rv(n);rv(args...);}
void setIO(){
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
}
void setIO(string f){
freopen((f+".in").c_str(),"r",stdin);
freopen((f+".out").c_str(),"w",stdout);
setIO();
}
ll a[100000],j=0;
ll findGap(int T, ll N){
if (T==1) {
ll l=1,r=1e18;
ll mn,mx;
vector<ll> v;
for (ll i=0;i<(N+1)/2;i++) {
MinMax(l,r,&mn,&mx);
a[j++] = mn;
a[j++] = mx;
l=mn+1,r=mx-1;
}
sort(a,a+N);
ll ans=0;
for (ll i=0;i<N-1;i++)ans=max(ans,a[i+1]-a[i]);
return ans;
} else {
ll mn, mx;
MinMax(1,1e18,&mn,&mx);
ll step=(mx-mn+N-2)/(N-1);
ll ans=step,x,y,l=mn,i;
for (i=mn;i+step<mx;i+=step+1){
MinMax(i,i+step,&x,&y);
if(x!=-1){
ans=max(ans,x-l);
l=y;
}
}
MinMax(i, mx, &x, &y);
if (x!=-1)ans=max(ans,x-l);
return ans;
}
}