Add a bunch of other problems
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109
Codeforces Round 1053 (Div. 2)/C. Incremental Stay.cpp
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109
Codeforces Round 1053 (Div. 2)/C. Incremental Stay.cpp
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/* Problem URL: https://codeforces.com/contest/2151/problem/C */
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#include <bits/stdc++.h>
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#include <ext/pb_ds/assoc_container.hpp>
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#include <ext/pb_ds/tree_policy.hpp>
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using namespace std;
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using namespace __gnu_pbds;
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template <class T, class comp = less<>>
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using ordered_set = tree<T, null_type , comp , rb_tree_tag , tree_order_statistics_node_update>;
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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 (int i = 0; i < (lim); i++)
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#define nrep(i, s, lim) for (int 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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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 t;
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cin >> t;
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while (t--) {
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int n;
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cin >> n;
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n <<= 1;
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vl fds(n);
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cin >> fds;
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vl count(2);
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for (int i = 0; i < n; i += 2) {
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count[0] += fds[i];
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count[1] += fds[i + 1];
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}
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ll ans = 0;
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rep(k, n >> 1) {
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cout << count[(k & 1) ^ 1] - count[k & 1] + ans << ' ';
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count[k & 1] -= fds[k];
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count[(k & 1) ^ 1] -= fds[n - k - 1];
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ans += fds[n - k - 1] - fds[k];
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}
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cout << '\n';
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}
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}
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