이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ld,ld> pdd;
#define ff(i,a,b) for(int i = a; i <= b; i++)
#define fb(i,b,a) for(int i = b; i >= a; i--)
#define trav(a,x) for (auto& a : x)
#define sz(a) (int)(a).size()
#define pb push_back
#define fi first
#define se second
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k) print the k-th smallest number in os(0-based)
const int mod = 1000000007;
const int inf = 1e9 + 5;
const int mxN = 200005;
int n, q, S, T;
int niz[mxN];
int P[mxN];
int main() {
cin.tie(0)->sync_with_stdio(0);
cin >> n >> q >> S >> T;
ff(i,0,n)cin >> niz[i];
ff(i,1,n)P[i] = niz[i] - niz[i - 1];
ll uk = 0;
ff(i,1,n){
if(P[i] >= 0)uk -= 1ll * P[i] * S;
else uk -= 1ll * P[i] * T;
}
while(q--){
int l, r, val;
cin >> l >> r >> val;
if(P[l] >= 0)uk += 1ll * P[l] * S;
else uk += 1ll * P[l] * T;
if(r != n){
if(P[r + 1] >= 0)uk += 1ll * P[r + 1] * S;
else uk += 1ll * P[r + 1] * T;
}
P[l] += val; if(r != n)P[r + 1] -= val;
if(P[l] >= 0)uk -= 1ll * P[l] * S;
else uk -= 1ll * P[l] * T;
if(r != n){
if(P[r + 1] >= 0)uk -= 1ll * P[r + 1] * S;
else uk -= 1ll * P[r + 1] * T;
}
cout << uk << '\n';
}
return 0;
}
/*
3 5 1 2
0 4 1 8
1 2 2
1 1 -2
2 3 5
1 2 -1
1 3 5
// probati bojenje sahovski
*/
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