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113
Codeforces Round 786 (Div. 3)/E. Breaking the Wall.cpp
Normal file
113
Codeforces Round 786 (Div. 3)/E. Breaking the Wall.cpp
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/* Problem URL: https://codeforces.com/contest/1674/problem/E */
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#include <bits/stdc++.h>
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using namespace std;
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#define V vector
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#define rmin(a, b) a = min(a, b)
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#define rmax(a, b) a = max(a, b)
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#define rep(i, lim) for (size_t i = 0; i < (lim); i++)
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#define nrep(i, s, lim) for (size_t i = s; i < (lim); i++)
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#define repv(i, v) for (auto &i : (v))
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#define fillv(v) for (auto &itr_ : (v)) { cin >> itr_; }
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#define sortv(v) sort(v.begin(), v.end())
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#define all(v) (v).begin(), (v).end()
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using vi = vector<int>;
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using vvi = vector<vi>;
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using vvvi = vector<vvi>;
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using vvvvi = vector<vvvi>;
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using ll = long long;
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using vl = vector<ll>;
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using vvl = vector<vl>;
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using vvvl = vector<vvl>;
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using vvvvl = vector<vvvl>;
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template<class v>
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auto operator<<(ostream &os, const vector<v> &vec)->ostream& {
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os << vec[0];
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for (size_t i = 1; i < vec.size(); i++) {
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os << ' ' << vec[i];
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}
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os << '\n';
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return os;
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}
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template<class v>
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auto operator>>(istream &is, vector<v> &vec)->istream& {
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for (auto &i : vec) {
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is >> i;
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}
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return is;
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}
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template<class v>
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auto operator<<(ostream &os, const vector<vector<v>> &vec)->ostream& {
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for (auto &i : vec) {
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os << i[0];
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for (size_t j = 1; j < i.size(); j++) {
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os << ' ' << i[j];
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}
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os << '\n';
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}
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return os;
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}
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template<class v>
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auto operator>>(istream &is, vector<vector<v>> &vec)->istream& {
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for (auto &i : vec) {
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for (auto &j : i) {
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is >> j;
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}
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}
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return is;
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}
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ll inf = INT64_MAX >> 3;
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int main()
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{
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ios::sync_with_stdio(false);
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cin.tie(nullptr);
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int n;
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cin >> n;
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vi a(n);
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cin >> a;
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ll ans = inf;
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ll minimal = inf;
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rep(i, n) {
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ll m = a[i] / 2 + a[i] % 2;
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rmin(ans, minimal + m);
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rmin(minimal, m);
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}
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rep(i, n) {
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ll ans1 = a[i] / 2 + a[i] % 2;
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ll ans2 = inf;
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if (i > 0) {
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ans2 = a[i - 1];
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rmin(ans, ans1 + max((ans2 - ans1) / 2 + (ans2 - ans1) % 2, 0LL));
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}
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ll ans3 = inf;
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if (i < n - 1) {
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ans3 = a[i + 1];
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rmin(ans, ans1 + max((ans3 - ans1) / 2 + (ans3 - ans1) % 2, 0LL));
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}
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rmin(ans, max((ans2 - ans1) / 2 + (ans2 - ans1) % 2, 0LL) + max((ans3 - ans1) / 2 + (ans3 - ans1) % 2, 0LL) + min(ans1, max(ans2, ans3)));
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}
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cout << ans << '\n';
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}
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