Submission #250564

#TimeUsernameProblemLanguageResultExecution timeMemory
250564mode149256Candies (JOI18_candies)C++14
100 / 100
630 ms22392 KiB
/*input 5 3 5 1 7 6 20 623239331 125587558 908010226 866053126 389255266 859393857 596640443 60521559 11284043 930138174 936349374 810093502 521142682 918991183 743833745 739411636 276010057 577098544 551216812 816623724 */ #include <bits/stdc++.h> using namespace std; namespace my_template { typedef long long ll; typedef long double ld; typedef complex<ld> cd; typedef pair<int, int> pi; typedef pair<ll, ll> pl; typedef pair<ld, ld> pd; typedef vector<int> vi; typedef vector<vi> vii; typedef vector<ld> vd; typedef vector<ll> vl; typedef vector<vl> vll; typedef vector<pi> vpi; typedef vector<vpi> vpii; typedef vector<pl> vpl; typedef vector<cd> vcd; typedef vector<pd> vpd; typedef vector<bool> vb; typedef vector<vb> vbb; typedef std::string str; typedef std::vector<str> vs; #define x first #define y second #define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n" const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L; template<typename T> pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); } template<typename T> pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); } template<typename T> T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); } template<typename T> T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); } template<typename T> void print(vector<T> vec, string name = "") { cout << name; for (auto u : vec) cout << u << ' '; cout << '\n'; } } using namespace my_template; const int MOD = 1000000007; const ll INF = 1e15; const int MX = 100101; using lii = list<pl>::iterator; struct comp { bool operator()(const lii &a, const lii &b) { return (*a).x > (*b).x or ((*a).x == (*b).x and (*a).y < (*b).y); } }; int main() { ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); int N; cin >> N; list<pl> sk; set<lii, comp> did; for (int i = 0; i < N; ++i) { ll a; cin >> a; sk.push_back({a, i}); did.insert((--sk.end())); } // printf("%d\n", (*did.begin()).y); ll suma = 0; for (int i = 0; i < (N + 1) / 2; ++i) { auto it = (*did.begin()); ll val = (*it).x; suma += val; did.erase(it); ll kitas = -val; if (it != sk.begin()) { auto prv = prev(it); kitas += (*prv).x; did.erase(prv); sk.erase(prv); } else { kitas += -INF; } if (next(it) != sk.end()) { auto nxt = next(it); kitas += (*nxt).x; did.erase(nxt); sk.erase(nxt); } else { kitas += -INF; } (*it).x = kitas; did.insert(it); printf("%lld\n", suma); } } /* Look for: * special cases (n=1?) * overflow (ll vs int?) * the exact constraints (multiple sets are too slow for n=10^6 :( ) * array bounds */
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