Submission #226995

#	Time	Username	Problem	Language	Result	Execution time	Memory
226995		qkxwsm	Climbers (RMI18_climbers)	C++14	0 / 100	7 ms	512 KiB

climbers

#include <bits/stdc++.h>

using namespace std;

template<class T, class U>
void ckmin(T &a, U b)
{
    if (a > b) a = b;
}

template<class T, class U>
void ckmax(T &a, U b)
{
    if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define SZ(x) ((int) (x).size())
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;

const int MAXN = 5013;

int N;
int arr[MAXN];
vi compress;
int lpref[MAXN], rpref[MAXN];
ll ans;

int indexof(int x)
{
    return LB(ALL(compress), x) - compress.begin();
}
int32_t main()
{
    cout << fixed << setprecision(12);
    cerr << fixed << setprecision(4);
    ios_base::sync_with_stdio(false); cin.tie(0);
    cin >> N;
    FOR(i, 0, N)
    {
        cin >> arr[i];
        compress.PB(arr[i]);
    }
    int idx = max_element(arr, arr + N) - arr;
    sort(ALL(compress));
    compress.erase(unique(ALL(compress)), compress.end());
    FOR(i, 0, idx)
    {
        int lo = indexof(arr[i]), hi = indexof(arr[i + 1]);
        if (lo > hi) swap(lo, hi);
        lpref[lo]++; lpref[hi]--;
    }
    FOR(i, idx, N - 1)
    {
        int lo = indexof(arr[i]), hi = indexof(arr[i + 1]);
        if (lo > hi) swap(lo, hi);
        rpref[lo]++; rpref[hi]--;
    }
    FOR(i, 1, SZ(compress))
    {
        lpref[i] += lpref[i - 1];
        rpref[i] += rpref[i - 1];
    }
    FOR(i, 0, SZ(compress) - 1)
    {
        int h = compress[i + 1] - compress[i];
        ans += h * max(lpref[i], rpref[i]);
        // cerr << h << ' ' << max(lpref[i], rpref[i]) << endl;
    }
    cout << ans << '\n';
    return 0;
}

• Subtask 1: 0 / 25
• Subtask 2: 0 / 25
• Subtask 3: 0 / 50