| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 482719 | Fysty | A Difficult(y) Choice (BOI21_books) | 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.
#include <bits/stdc++.h>
//#include "books.h"
#include <random>
#include <chrono>
using namespace std;
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize ("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<double,double> pdd;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template<typename T> void _do(T x){cerr<<x<<"\n";}
template<typename T,typename ...U> void _do(T x,U ...y){cerr<<x<<", ";_do(y...);}
#define dbg(...) cerr<<#__VA_ARGS__<<" = ";_do(__VA_ARGS__);
const int MOD1=1e9+7;
const int MOD2=998244353;
const ll INF=3e18;
const ld PI=3.14159265358979323846;
ll gcd(ll a,ll b){if(b==0) return a; return gcd(b,a%b);}
ll fpow(ll a,ll b,ll m)
{
if(!b) return 1;
ll ans=fpow(a*a%m,b/2,m);
return (b%2?ans*a%m:ans);
}
ll inv(ll a,ll m) {return fpow(a,m-2,m);}
#define MottoHayaku ios::sync_with_stdio(false);cin.tie(0);
#define int ll
#define rep(i,n) for(ll i=0;i<n;i++)
#define rep1(i,n) for(ll i=1;i<=n;i++)
#define repk(i,m,n) for(int i=m;i<n;i++)
#define F first
#define S second
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define uni(c) c.resize(distance(c.begin(),unique(c.begin(),c.end())))
#define unisort(c) sort(c.begin(),c.end()),uni(c)
ll a[100005];
//ll skim(ll n){}
//void answer(vector<ll> ans){}
//void impossible(){}
void solve(int n,int k,ll A,int S)
{
if(n==k)
{
ll tot=0;
vector<ll> ans;
rep1(i,n)
{
a[i]=skim(i);
ans.pb(i);
tot+=a[i];
}
if(tot>=A&&tot<=2*A) answer(ans);
else impossible();
return;
}
ll tot=0;
rep1(i,k)
{
a[i]=skim(i);
tot+=a[i];
}
if(a[k]>=A)
{
if(tot>=A&&tot<=2*A)
{
vector<ll> ans;
rep1(i,k) ans.pb(i);
answer(ans);
}
else impossible();
return;
}
ll l=k+1,r=n+1;
while(l<r)
{
ll mid=l+r>>1;
a[mid]=skim(mid);
if(a[mid]<A) l=mid+1;
else r=mid;
}
if(l!=n+1)
{
if(a[l]==0) a[l]=skim(l);
vector<ll> ans={l};
tot=a[l];
rep1(i,k-1)
{
tot+=a[i];
ans.pb(i);
}
if(tot>=A&&tot<=2*A)
{
answer(ans);
return;
}
}
l--;
ll cnt=0;
for(int i=l;i>=1;i--)
{
if(cnt==k) break;
if(a[i]==0)
{
cnt++;
a[i]=skim(i);
}
}
if(l==k)
{
ll tot=0;
vector<ll> ans;
rep1(i,k)
{
ans.pb(i);
tot+=a[i];
}
if(tot>=A&&tot<=2*A) answer(ans);
else impossible();
return;
}
rep(i,k+1)
{
vector<ll> ans;
ll tot=0;
rep1(j,k-i)
{
tot+=a[j];
ans.pb(j);
}
rep(j,i)
{
tot+a[l-j];
ans.pb(l-j);
}
if(tot>=A&&tot<=2*A)
{
answer(ans);
return;
}
}
impossible();
}