Submission #795448

#	Time	Username	Problem	Language	Result	Execution time	Memory
795448		lohacho	Real Mountains (CCO23_day1problem2)	C++14	12 / 25	5031 ms	31648 KiB

Main

#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#define int long long
using namespace std;
 
const int mod = (int)1e6 + 3;
 
signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
 
    int n;
    cin >> n;
    vector<int> a(n), srt(n);
    for(int i = 0; i < n; ++i){
        cin >> a[i];
        srt[i] = a[i];
    }
    sort(srt.begin(), srt.end());
    srt.erase(unique(srt.begin(), srt.end()), srt.end());
 
    int ans = 0;
    for(int i = 0; i + 1 < (int)srt.size(); ++i){
        vector<int> lmn(n, -1), rmn(n, -1);
        set<int> st;
 
        for(int j = 0; j < n; ++j){
            auto p = st.lower_bound(a[j] + 1);
            if(p != st.end()){
                lmn[j] = *p;
            }
            st.insert(a[j]);
        }
        st.clear();
        for(int j = n - 1; j >= 0; --j){
            auto p = st.lower_bound(a[j] + 1);
            if(p != st.end()){
                rmn[j] = *p;
            }
            st.insert(a[j]);
        }
 
        int cnt = 0;
        int mn = -1, mx = n;
        for(int j = 0; j < n; ++j){
            if(lmn[j] != -1 && rmn[j] != -1 && a[j] == srt[i]){
                if(mn == -1) mn = j;
                mx = j;
                ++cnt;
                a[j] = srt[i + 1];
            }
        }
        if(cnt){
            if(cnt == 1){
                (ans += (lmn[mn] + rmn[mn]) * (srt[i + 1] - srt[i]) % mod) %= mod;
            
        assert(ans >= 0 && ans < mod);}
            else{
                (ans += (lmn[mn] + rmn[mx] + min(rmn[mn], lmn[mx])) * (srt[i + 1] - srt[i]) % mod + (srt[i] + srt[i + 1] + 1) * (srt[i + 1] - srt[i]) / 2 % mod * (cnt * 2 - 3) % mod) %= mod;
           
        assert(ans >= 0 && ans < mod); }
        assert(ans >= 0 && ans < mod);
            (ans += cnt * ((srt[i] + srt[i + 1] - 1) * (srt[i + 1] - srt[i]) / 2 % mod) % mod) %= mod;
 
            // cout << fir << ' ' << lst << ' ' << cnt << ' ' << ans << endl;
        }
 
        assert(ans >= 0 && ans < mod);
    }
 
    cout << ans << '\n';
    
    return 0;
}

• Subtask 1: 3 / 3
• Subtask 2: 3 / 3
• Subtask 3: 3 / 3
• Subtask 4: 3 / 3
• Subtask 5: 0 / 4
• Subtask 6: 0 / 5
• Subtask 7: 0 / 4