Submission #295261

#	Time	Username	Problem	Language	Result	Execution time	Memory
295261		NaimSS	Hacker (BOI15_hac)	C++14	40 / 100	16 ms	512 KiB

hac

#include <bits/stdc++.h>
#define ld long double
#define endl "\n"
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define pb(x) push_back(x)
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define all(v) v.begin(),v.end()
#define ff first
#define ss second
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define per(i, a, b) for(int i = b-1; i>=a ; i--)
#define trav(a, x) for(auto& a : x)
#define allin(a , x) for(auto a : x)
#define Unique(v) sort(all(v));v.erase(unique(all(v)),v.end());
#define sz(v) (int)v.size()
//#define int long long
using namespace std;
typedef vector<int> vi;
#define y1 abacaba
#define left sefude
#define db(x) cerr << #x <<" == "<<x << endl;
#define db2(x,y) cerr<<#x <<" == "<<x<<", "<<#y<<" == "<<y<<endl;
#define db3(x,y,z) cerr << #x<<" == "<<x<<", "<<#y<<" == "<<y<<", "<<#z<<" == "<<z<<endl; 
#define prl cerr<<"called: "<< __LINE__<<endl;
typedef long long ll;
typedef pair<int,int> pii;
inline ll mod(ll n, ll m ){ ll ret = n%m; if(ret < 0) ret += m; return ret; }
ll gcd(ll a, ll b){return (b == 0LL ? a : gcd(b, a%b));}
ll exp(ll a,ll b,ll m){
    if(b==0LL) return 1LL;
    if(b==1LL) return mod(a,m);
    ll k = mod(exp(a,b/2,m),m);
    if(b&1LL){
        return mod(a*mod(k*k,m),m);
    }
    else return mod(k*k,m);
}

const int N = 5050 * 3;
int a[N];
int pre[N];
int32_t main(){
  fastio;
  int n;
  cin>>n;
  for(int i=1;i<=n;i++){
    cin >> a[i];

    a[i+n] = a[i+n*2] = a[i];
  }
  for(int i=1;i<=3*n;i++){
    pre[i] = pre[i-1] + a[i];
  }
  int res=0;
  int tam = (n+1)/2;
  for(int i=n+1;i<=2*n;i++){
    int best = 1e9 + 7;
    for(int k=0;k<tam;k++){
      best = min(best,pre[i + k] - pre[i - (tam - k)]);
    }
    res = max(res,best);
  }
  cout << res << endl;
  // menor subarray de tamanho (n+1)/2 que contém POS
  // math -> gcd it all
  // Did u check N=1? Did you switch N,M?
}

• Subtask 1: 20 / 20
• Subtask 2: 20 / 20
• Subtask 3: 0 / 20
• Subtask 4: 0 / 40