Submission #1027968

#	Time	Username	Problem	Language	Result	Execution time	Memory
1027968		GrindMachine	Seesaw (JOI22_seesaw)	C++17	100 / 100	361 ms	62136 KiB

seesaw

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#endif

/*

refs:
https://www.luogu.com.cn/blog/gyh20/joi-open-2022-ti-xie

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

struct frac{
    __int128 a,b;
    frac(ll a_, ll b_){
        a = a_, b = b_;
    }
    bool operator<(const frac &f) const{
        return a*f.b < b*f.a;
    }
    bool operator>(const frac &f) const{
        return a*f.b > b*f.a;
    }
    frac operator-(const frac &f){
        auto [a2,b2] = f;
        return frac(a*b2-a2*b,b*b2);
    }
};

void solve(int test_case)
{
    ll n; cin >> n;
    vector<ll> a(n+5);
    rep1(i,n) cin >> a[i];

    vector<ll> p(n+5);
    rep1(i,n) p[i] = p[i-1]+a[i];

    frac g(p[n],n);
    vector<pair<frac,ll>> b;

    rep1(len,n-1){
        ll lo = 1, hi = n-len+1;         
        ll pos = 0;
        while(lo <= hi){
            ll mid = (lo+hi) >> 1;
            ll s = p[mid+len-1]-p[mid-1];
            frac f(s,len);
            if(f < g){
                pos = mid;
                lo = mid+1;
            }
            else{
                hi = mid-1;
            }
        }

        for(int i = pos; i <= pos+1; ++i){
            if(i < 1 or i+len-1 > n) conts;
            ll s = p[i+len-1]-p[i-1];
            frac f(s,len);
            b.pb({f,len});
        }
    }

    b.pb({g,n});
    sort(all(b));
    vector<frac> here[n+5];
    reverse(all(b));
    for(auto [f,len] : b){
        here[len].pb(f);
    }
    reverse(all(b));

    set<frac> st;
    rep1(len,n){
        st.insert(here[len].back());
    }

    frac best(inf2,1);

    for(auto [f,len] : b){
        auto mx = *st.rbegin();
        auto diff = mx-f;
        if(diff < best) best = diff;

        here[len].pop_back();
        if(here[len].empty()) break;
        st.erase(f);
        st.insert(here[len].back());
    }

    ld ans = (ld)best.a/best.b;
    cout << fixed << setprecision(11);
    cout << ans << endl;

    /*

    frac best(inf2,1);

    rep(mask,1<<(n-1)){
        frac mn = g, mx = g;
        bool ok = true;

        rep1(len,n-1){
            ll i = 0;
            if(mask&(1<<(len-1))){
                i = b[len]+1;
            }
            else{
                i = b[len];
            }

            if(i < 1 or i+len-1 > n){
                ok = false;
                break;
            }

            frac f(p[i+len-1]-p[i-1],len);
            if(f < mn) mn = f;
            if(f > mx) mx = f;
        }

        if(!ok) conts;
        frac res = mx-mn;
        if(res < best) best = res;
    }

    ld ans = (ld)best.a/best.b;
    cout << fixed << setprecision(11);
    cout << ans << endl;

    */
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

• Subtask 1: 1 / 1
• Subtask 2: 33 / 33
• Subtask 3: 33 / 33
• Subtask 4: 33 / 33