Submission #931695

#	Time	Username	Problem	Language	Result	Execution time	Memory
931695		jcelin	Fancy Fence (CEOI20_fancyfence)	C++14	100 / 100	22 ms	13784 KiB

fancyfence

#include <bits/stdc++.h>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/tree_policy.hpp>

using namespace std;
//using namespace __gnu_pbds;

typedef long long ll;
typedef long double ld;

#define ii pair<int,int>
#define pll pair<ll,ll>
#define vi vector<int>
#define vii vector<ii>
#define vll vector<ll>
#define vpll vector<pll>
#define msi multiset<int>
#define si set<int>
#define PB push_back
#define PF push_front
#define PPB pop_back
#define PPF pop_front
#define X first
#define Y second
#define MP make_pair
#define FOR(i, a, b) for (int i = int(a); i < int(b); i++)
#define REP(i, n) FOR(i, 0, n)
#define all(x) (x).begin(), (x).end()

const int mod = 1e9 + 7;
const int MOD = 998244353;
const int inf = mod;
const ll INF = 1e18;
const int MAXN = 1e6 + 7;
const int logo = 17;
const int off = 1 << logo;
const int trsz = off << 1;
const int dx[] = {1, -1, 0, 0};
const int dy[] = {0, 0, 1, -1};


vpll st;
int ans, n, in2;

ll pf[MAXN], w[MAXN], h[MAXN], dp[MAXN];

int add(ll a, ll b){
	return (a + b) % mod;
}

int mul(ll a, ll b){
	return (a * b) % mod;
}

int sub(ll a, ll b){
	return (a - b + mod) % mod;
}

int exp(int b, ll e){
	int ret = 1;
	while(e){
		if(e & 1) ret = mul(ret, b);
		b = mul(b, b);
		e /= 2;
	}
	return ret;
}

int ch2(ll a){
	return mul(in2, mul(a, a + 1));
}

void solve(){
	cin >> n;
	in2 = exp(2, mod - 2);
	for(int i=1; i<=n; i++) cin >> h[i];
	for(int i=1; i<=n; i++) cin >> w[i], pf[i] = add(pf[i - 1], w[i]);
	st.PB({0, 0});
	for(int i=1; i<=n; i++){
		while(h[st.back().X] > h[i]) st.PPB();
		ll su = add(st.back().Y, mul(ch2(h[i]), sub(pf[i - 1], pf[st.back().X])));
		dp[i] = add(mul(su, w[i]), mul(ch2(h[i]), ch2(w[i])));
		st.PB({i, add(su, mul(ch2(h[i]), w[i]))});
		ans = add(ans, dp[i]);
	}
	cout << ans << "\n";
}


int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	int t=1;
	//cin >> t;
	while(t--)solve();
	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