Submission #573135

#	Time	Username	Problem	Language	Result	Execution time	Memory
573135		piOOE	Fancy Fence (CEOI20_fancyfence)	C++17	100 / 100	31 ms	5456 KiB

fancyfence

#include <bits/stdc++.h>

using namespace std;

#define sz(x) ((int)size(x))
#define all(x) begin(x), end(x)
#define trace(x) cout << #x << ": " << (x) << endl;

typedef long long ll;

mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());

int rand(int l, int r) { return (int) ((ll) rnd() % (r - l + 1)) + l; }

const int N = 100001;
ll mod = 1e9 + 7;

int h[N], R[N], L[N];

ll w[N];
ll pref[N];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> h[i];
    }
    for (int i = 1; i <= n; ++i) {
        cin >> w[i];
    }
    int id = 1;
    for (int i = 1; i <= n;) {
        int j = i;
        ll sum = 0;
        while (j <= n && h[j] == h[i]) {
            sum += w[j];
            j += 1;
        }
        h[id] = h[i];
        w[id] = sum;
        ++id;
        i = j;
    }
    n = id - 1;
    vector<int> st;
    for (int i = 1; i <= n; ++i) {
        while (!st.empty() && h[st.back()] > h[i]) {
            R[st.back()] = i;
            st.pop_back();
        }
        if (st.empty()) {
            L[i] = 0;
        } else {
            L[i] = st.back();
        }
        st.push_back(i);
    }
    for (int x : st) {
        R[x] = n + 1;
    }
    for (int i = 1; i <= n; ++i) {
        w[i] %= mod;
        pref[i] = pref[i - 1] + w[i];
    }
    ll ans = 0;
    const ll inv2 = (mod + 1) / 2;
    for (int i = 1; i <= n; ++i) {
        ll sumR = (pref[R[i] - 1] - pref[i]) % mod;
        ll sumL = (pref[i - 1] - pref[L[i]]) % mod;
        ll x = (h[i] * (ll)(h[i] + 1)) % mod * inv2 % mod;
        ll now = (sumR * (sumL + w[i])) % mod;
        now = (now + (sumL + w[i]) * (sumL + w[i] + 1) % mod * inv2 % mod - sumL * (sumL + 1) % mod * inv2 % mod) % mod;
        now = (now + mod) % mod;
        ans = (ans + now * x) % mod;
    }
    cout << ans;
    return 0;
}

• 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