Submission #213074

#TimeUsernameProblemLanguageResultExecution timeMemory
213074qkxwsmPotatoes and fertilizers (LMIO19_bulves)C++14
100 / 100
250 ms15336 KiB
#include <bits/stdc++.h> using namespace std; template<class T, class U> void ckmin(T &a, U b) { if (a > b) a = b; } template<class T, class U> void ckmax(T &a, U b) { if (a < b) a = b; } #define MP make_pair #define PB push_back #define LB lower_bound #define UB upper_bound #define fi first #define se second #define SZ(x) ((int) (x).size()) #define ALL(x) (x).begin(), (x).end() #define FOR(i, a, b) for (auto i = (a); i < (b); i++) #define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--) typedef long long ll; typedef long double ld; typedef pair<int, int> pii; typedef pair<ll, ll> pll; typedef vector<int> vi; typedef vector<ll> vl; typedef vector<pii> vpi; typedef vector<pll> vpl; const int MAXN = 500013; const int INF = 1000000007; int N; ll arr[MAXN]; ll net[MAXN]; ll ans; vl vec; int32_t main() { cout << fixed << setprecision(12); cerr << fixed << setprecision(4); ios_base::sync_with_stdio(false); cin.tie(0); cin >> N; FOR(i, 1, N + 1) { int a, b; cin >> a >> b; arr[i] = a - b; } //you want to make everything >= 0, and you can move smth 1 left or 1 right at a time. //+1 -1 => +1 +0 //-1 +1 => -1 +0 FOR(i, 1, N + 1) { arr[i] += arr[i - 1]; } FOR(i, 1, N) { if (arr[i] < 0) { ans -= arr[i]; arr[i] = 0; } if (arr[i] > arr[N]) { ans += (arr[i] - arr[N]); arr[i] = arr[N]; } ans += arr[i]; } priority_queue<ll> pq; FOR(i, 1, N) { pq.push(arr[i]); pq.push(arr[i]); pq.pop(); } while(!pq.empty()) { ans -= (pq.top()); pq.pop(); } //make the sequence increasing. or equal. in each move you can change smth by 0 or 1. cout << ans << '\n'; return 0; //make all arr's positive, what's the cost? }
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