제출 #1257640

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1257640		dosts	Fancy Fence (CEOI20_fancyfence)	C++20	100 / 100	17 ms	5920 KiB

fancyfence

#include <bits/stdc++.h>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2")
using namespace std;
#define int long long
#define pii pair<int,int>
#define ff first
#define ss second
#define sp << " " <<    
#define all(cont) cont.begin(),cont.end()
#define vi vector<int>

const int inf = 2e18,N = 5e5+1,MOD =  1e9+7;


int add(int x,int y) {
    x%=MOD;
    y%=MOD;
    return ((x+y >= MOD) ? (x+y-MOD) : x+y);
}

int mult(int x,int y) {
    return ((x%MOD)*(y%MOD))%MOD;
}

int sub(int x,int y) {
    y%=MOD;
    x%=MOD;
    if (x < y) return (MOD+x)-y;
    return x-y;
}

void solve() {  
    int n;
    cin >> n;
    int h[n+1],w[n+1];
    for (int i=1;i<=n;i++) cin >> h[i];
    for (int i=1;i<=n;i++) cin >> w[i];
    int L[n+1],R[n+1];
    vi stk;
    for (int i=1;i<=n;i++) {
        while (!stk.empty() && h[stk.back()] >= h[i]) stk.pop_back();
        if (stk.empty()) L[i] = 1;
        else L[i] = stk.back()+1;
        stk.push_back(i);
    } 
    stk.clear();
    for (int i = n;i>=1;i--) {
        while (!stk.empty() && h[stk.back()] > h[i]) stk.pop_back();
        if (stk.empty()) R[i] = n;
        else R[i] = stk.back()-1;
        stk.push_back(i);
    }

    vi pref(n+1,0);
    for (int i=1;i<=n;i++) pref[i] = pref[i-1]+w[i];
    int ans = 0,anss = 0;
    for (int i=1;i<=n;i++) {
        ans = 0;
        int tot = sub(pref[R[i]],pref[L[i]-1]);
        ans = add(ans,(tot*(tot+1)/2));
        int left = sub(pref[i-1],pref[L[i]-1]);
        ans = sub(ans,(left*(left+1)/2));
        int right = sub(pref[R[i]],pref[i]);
        ans = sub(ans,(right*(right+1)/2));
        ans = mult(ans,h[i]*(h[i]+1)/2);
        anss = add(anss,ans);
    }
    cout << anss << '\n';
}


int32_t main() { 
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    #ifdef Dodi 
        freopen("in.txt","r",stdin);
        freopen("out.txt","w",stdout);
    #endif
    int t = 1;
    //cin >> t;
    while (t --> 0) 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