제출 #1153632

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1153632		MPG	Foehn Phenomena (JOI17_foehn_phenomena)	C++20	100 / 100	94 ms	7240 KiB

foehn_phenomena

#pragma GCC optimization ("unroll-loops")
#pragma GCC optimization ("O1, O2, O3, Ofast")
#pragma GCC optimization ("trapv")
#pragma GCC optimization ("sse, sse2, sse3, sse4, sse4.1, sse4.2, avx")



#include <bits/stdc++.h>
using namespace std;
typedef long long                           ll;
#define                                     max_heap priority_queue<ll>
#define                                     min_heap priority_queue<pair <ll, ll>, vector<pair <ll, ll>>, greater<pair <ll, ll>>> // size(), push(), top(), pop();
//#define                                     min_heap priority_queue<ll, vector<ll>, greater<ll>>
#define                                     sariE cin.tie(NULL); cout.tie(NULL); ios_base::sync_with_stdio(false);
#define                                     filE freopen("in.txt", "r", stdin); freopen("out1.txt", "w", stdout);
#define                                     endl '\n'
#define                                     md(a) (a % mod + mod) % mod
//cout << setprecision(5) << f;
ll const maxn = 2e5 + 110;
ll const inf = 2e18;
ll const loG = 23;
ll const mod = 998244353; //1e9 + 9, // 1e9 + 7;
ll const sq = 750;
ll power(ll a, ll b, ll mod){if (b == 0) return 1; if (b == 1) return a; ll x = power(a, b / 2, mod); return (((x * x) % mod) * (b % 2 ? a : 1)) % mod;}







ll n, q, s, t, asl[maxn], dp[maxn], ans;

ll trans(ll d){
    if (d > 0){
        return -d * s; 
    }
    else
        return abs(d * t);
}


int main(){
sariE;// filE;

cin >> n >> q >> s >> t;
for (int i = 0; i < n + 1; i++){
    cin >> asl[i];
}
for (int i = 0; i < n; i++){
    dp[i] = asl[i + 1] - asl[i];
    ans += trans(dp[i]);
}
//cout << ans << endl;
while (q--){
    ll l, r, x; cin >> l >> r >> x;
    ans = ans - trans(dp[l - 1]) + trans(dp[l - 1] + x);
    dp[l - 1] += x;
    if (r < n){
        ans =  ans - trans(dp[r]) + trans(dp[r] - x);
        dp[r] -= x;
    }
    cout << ans << endl;
}




}

• Subtask 1: 30 / 30
• Subtask 2: 10 / 10
• Subtask 3: 60 / 60