Submission #831332

#	Time	Username	Problem	Language	Result	Execution time	Memory
831332		maomao90	Seesaw (JOI22_seesaw)	C++17	34 / 100	2065 ms	33280 KiB

seesaw

// Hallelujah, praise the one who set me free
// Hallelujah, death has lost its grip on me
// You have broken every chain, There's salvation in your name
// Jesus Christ, my living hope
#include <bits/stdc++.h> 
using namespace std;

#define REP(i, s, e) for (int i = (s); i < (e); i++)
#define RREP(i, s, e) for (int i = (s); i >= (e); i--)
template <class T>
inline bool mnto(T& a, T b) {return a > b ? a = b, 1 : 0;}
template <class T>
inline bool mxto(T& a, T b) {return a < b ? a = b, 1: 0;}

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
#define FI first
#define SE second
typedef pair<int, int> ii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> iii;
#define ALL(_a) _a.begin(), _a.end()
#define SZ(_a) (int) _a.size()
#define pb push_back
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<ii> vii;
typedef vector<iii> viii;

#ifndef DEBUG
#define cerr if (0) cerr
#endif

const int INF = 1000000005;
const ll LINF = 1000000000000000005ll;
const int MAXN = 200005;

int n;
int a[MAXN];
vector<ld> cand;
ld ans;

int main() {
#ifndef DEBUG
    ios::sync_with_stdio(0), cin.tie(0);
#endif
    cin >> n;
    REP (i, 0, n) {
        cin >> a[i];
    }
    REP (i, 0, n) {
        ll sm = 0;
        REP (j, i, n) {
            sm += a[j];
            cand.pb((ld) sm / (j - i + 1));
        }
    }
    ans = LINF;
    for (ld mx : cand) {
        ll sm = 0;
        int l = 0, r = n - 1;
        REP (i, 0, n) {
            sm += a[i];
        }
        ld mn = (ld) sm / n;
        if (mn > mx) {
            continue;
        }
        REP (z, 0, n - 1) {
            if (sm - a[l] <= mx * (r - l)) {
                sm -= a[l++];
            } else {
                sm -= a[r--];
            }
            mnto(mn, (ld) sm / (r - l + 1));
        }
        mnto(ans, mx - mn);
    }
    cout << setprecision(15) << ans << '\n';
    return 0;
}

• Subtask 1: 1 / 1
• Subtask 2: 33 / 33
• Subtask 3: 0 / 33
• Subtask 4: 0 / 33