155 lines
3 KiB
C++
155 lines
3 KiB
C++
#include <bits/stdc++.h> // {{{
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// https://codeforces.com/blog/entry/96344
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#pragma GCC optimize("O2,unroll-loops")
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#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
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using namespace std;
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template <typename T>
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[[nodiscard]] static T MIN() {
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return std::numeric_limits<T>::min();
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}
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template <typename T>
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[[nodiscard]] static T MAX() {
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return std::numeric_limits<T>::max();
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}
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template <typename T>
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[[nodiscard]] static T sc(auto&& x) {
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return static_cast<T>(x);
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}
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template <typename T>
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[[nodiscard]] static T sz(auto&& x) {
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return static_cast<T>(x.size());
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}
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#define prln(...) std::println(__VA_ARGS__)
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#define pr(...) std::print(__VA_ARGS__)
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#ifdef LOCAL
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#define dbgln(...) std::println(__VA_ARGS__)
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#define dbg(...) std::print(__VA_ARGS__)
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#endif
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inline static void NO() {
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prln("NO");
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}
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inline static void YES() {
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prln("YES");
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}
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using ll = long long;
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using ld = long double;
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template <typename T>
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using ve = std::vector<T>;
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template <typename T, size_t N>
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using ar = std::array<T, N>;
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template <typename T1, typename T2>
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using pa = std::pair<T1, T2>;
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template <typename... Ts>
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using tu = std::tuple<Ts...>;
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template <typename... Ts>
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using dq = std::deque<Ts...>;
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template <typename... Ts>
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using qu = std::queue<Ts...>;
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template <typename... Ts>
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using pq = std::priority_queue<Ts...>;
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template <typename... Ts>
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using st = std::stack<Ts...>;
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#define ff first
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#define ss second
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#define eb emplace_back
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#define pb push_back
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#define all(x) (x).begin(), (x).end()
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#define rall(x) (x).rbegin(), (x).rend()
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// }}}
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void solve() {
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int n;
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cin >> n;
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ve<pa<ll, int>> a(n);
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for (int i = 0; i < n; ++i) {
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cin >> a[i].ff;
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a[i].ss = i;
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}
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sort(all(a));
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ve<ll> prefix(n + 1, 0), postfix(n + 2, 0);
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for (int i = 0; i < n; ++i) {
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prefix[i + 1] = prefix[i] + a[i].ff;
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}
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for (int i = n - 1; i >= 0; --i) {
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postfix[i + 1] = postfix[i + 2] + a[i].ff;
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}
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ve<ll> ans(n);
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for (int i = 0; i < n; ++i) {
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ll x = n + a[i].ff * (i + 1) - prefix[i + 1];
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x += postfix[i + 2] - a[i].ff * (n - i - 1);
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ans[a[i].ss] = x;
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}
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for (auto& e : ans)
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pr("{} ", e);
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prln();
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/*
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1 3 4
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3 + 0 +
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to query @ i, want to get info related to
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a[i] - a[0], a[i] - a[1], ....
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prefix differences + postfix differences for each index
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n + a[i] - a[0] + a[i] - a[1] + ... + a[i] - a[i] + a[j] - a[i] + a[j + 1]
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- a[i]
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= n + a[i] * (i + 1) - prefix[i] (inclusive) + postfix[i] (exclusive) -
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a[i] * ``
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n * a[i] - sum of prefix before i + sum of postfix after i
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sum a[i] - a[j] + 1 for j in range(i + 1)
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count how manny points before us? * a[i]
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remove prefix sum up to j
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- and postfix! <- note this
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*-**
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-
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---
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----
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---
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-
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--
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----
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--
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-
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really, we're just adding the inclusive distance between ALL pairs
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*/
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}
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int main() { // {{{
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cin.tie(nullptr)->sync_with_stdio(false);
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cin.exceptions(cin.failbit);
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int t = 1;
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cin >> t;
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while (t--) {
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solve();
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}
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return 0;
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}
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// }}}
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