제출 #836619

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
836619		themm1	Fancy Fence (CEOI20_fancyfence)	C++17	100 / 100	68 ms	6900 KiB

fancyfence

#include <bits/stdc++.h>
using namespace std;
// ordered set whith s.order_of_key(x) method, which returns rank of element x in set s
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
*/

// pair printing
template <class T, class U>
ostream& operator<<(ostream& out, const pair<T, U> &par) {out << "(" << par.first << "; " << par.second << ")"; return out;}
// set printing
template <class T>
ostream& operator<<(ostream& out, const set<T> &cont) { out << "{"; for(const auto &x:cont) out << x << ", "; out << "}"; return out; }
// map printing
template <class T, class U>
ostream& operator<<(ostream& out, const map<T, U> &cont) {out << "{"; for(const auto &x:cont) out << x << ", "; out << "}"; return out; }
// unordered_set printing
template <class T>
ostream& operator<<(ostream& out, const unordered_set<T> &cont) {out << "{";for(const auto &x:cont) out << x << ", ";out << "}";return out;}
// unordered_map printing
template <class T, class U>
ostream& operator<<(ostream& out, const unordered_map<T, U> &cont) {out << "{";for(const auto &x:cont) out << x << ", ";out << "}";return out;}
// vector printing
template<class T>
ostream& operator<<(ostream& out, const vector<T> &cont){ out << "[";  for (const auto &x:cont) out << x << ", ";  out << "]"; return out;}

#define print(x) cout << (x) << endl;
#define dmp(x) cerr << #x << " = " << x << endl
#define dmpn(x) cerr << #x << " = " << x << "; "
#define dmpl(x) cerr << "Line " << __LINE__ << ": " << #x << " = " << x << endl

#define int long long
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
#define pb push_back
#define ff first
#define ss second
#define all(x) begin(x), end(x)
#define sz(x) (int)x.size()
#define contains(s,x) ((s).find(x) != (s).end())
const int MOD = 1e9 + 7;

int fast_pow(int a, int b) {
        int ans = 1;
        while (b > 0) {
                if (b % 2 == 1) ans = ((ll)a * (ll)ans) % MOD;
                a = ((ll)a * (ll)a) % MOD;
                b /= 2;
        }
        return ans;
}
 
int divide(int a, int b) {
        return ((ll)a * (ll)fast_pow(b, MOD - 2)) % MOD;
}

int rect_cnt(int h, int w) {
        int a = (h * w) % MOD;
        int b = divide((a * (((w - 1) + (h - 1)) % MOD)) % MOD, 2);
        int c = divide((a * (((h - 1) * (w - 1)) % MOD)) % MOD, 4);
        return (a + b + c) % MOD;
}

void solve() {
        int N;
        cin >> N;
        vector<int> hs(N), ws(N), p(N + 1, 0);
        for (int i = 0; i < N; i++) cin >> hs[i];
        for (int i = 0; i < N; i++) cin >> ws[i];
        for (int i = 0; i < N; i++) p[i + 1] = (p[i] + ws[i]) % MOD;
        // dmp(rect_cnt(3, 1));

        int ans = 0;
        vector<pii> stck;
        for (int i = 0; i < N; i++) {
                // dmp(ans);
                while (!stck.empty() && hs[stck.back().ff] > hs[i]) stck.pop_back();
                // dmpn(i); dmp(stck);

                int h = hs[i], w = ws[i];
                if (stck.empty()) {
                        int new_ending = (rect_cnt(h, 1) * p[i + 1]) % MOD;
                        int added = (rect_cnt(h, w) + (rect_cnt(h, 1) * ((p[i] * w) % MOD)) % MOD) % MOD;
                        // dmpn(new_ending); dmp(added);
                        ans = (ans + added) % MOD;
                        stck.pb({ i, new_ending });
                        continue;
                }
                int j = stck.back().ff;

                // all rectangles with one end at the end of current rectangle
                int new_ending = (rect_cnt(h, 1) * ((p[i + 1] - p[j + 1] + MOD) % MOD)) % MOD; 
                // all rectangles with one end in current rectangle
                int added = (rect_cnt(h, w) + (((stck.back().ss + (rect_cnt(h, 1) *
                        ((p[i] - p[j + 1] + MOD) % MOD)) % MOD) * w) % MOD)) % MOD;
                ans = (ans + added) % MOD;
                // dmpn(new_ending); dmp(added);

                stck.pb({ i, (stck.back().ss + new_ending) % MOD });
        }
        cout << ans << endl;
        // dmp(rect_cnt(3, 2));
        /*
        for (int i = 0; i < N; i++) {
                cout << hs[i] << " x " << ws[i] << ": " << rect_cnt(hs[i], ws[i]) << endl;
        }
        */
}

int32_t main() {
        ios::sync_with_stdio(false);
        cin.tie(0);
        cout.tie(0);

        int t = 1;
        // cin >> t;
        while (t--) {
                solve();
        }
}

• Subtask 1: 0 / 0
• Subtask 2: 12 / 12
• Subtask 3: 13 / 13
• Subtask 4: 15 / 15
• Subtask 5: 15 / 15
• Subtask 6: 18 / 18
• Subtask 7: 27 / 27