150 lines
3.3 KiB
C++
150 lines
3.3 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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auto lb = [](auto... args) {
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return std::lower_bound(args...);
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};
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auto ub = [](auto... args) {
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return std::upper_bound(args...);
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};
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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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ll n, c;
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cin >> n >> c;
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ll sum = 0, sumsq = 0;
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ll s;
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for (int i = 0; i < n; ++i) {
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cin >> s;
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sum += s;
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sumsq += s * s;
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}
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ld C = sumsq - c;
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ld A = 4 * n;
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// NOTE: tbh, look back at solutions,m not sure what went exactly wrong here.
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// most definitely just not taking my time
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// NOTE: didn't check WHY round down/why is this right, just that it *looked* right in the examples
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// this logic is a surefire way to skip certainty and absolutely confuse yourself
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ld B = 4 * sum;
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ll root = (-B + sqrtl(B * B - 4 * A * C)) / (2 * A);
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prln("{}", root);
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// NOTE: tbh, look back at solutions,m not sure what went exactly wrong here.
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// most definitely just not taking my time
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// NOTE: didn't check WHY round down/why is this right, just that it *looked* right in the examples
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// this logic is a surefire way to skip certainty and absolutely confuse yourself
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/*
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p ez, if anything should've just been locked and written the equation
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division and long double - fucked
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say paintings are a, b, c
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ax^2+bx+c
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area c= (a + 2w) ^ 2 + ...
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=a^2+4w^2+4aw + ...
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c = a^2+b^2+...+4(a+b+...) * w + 4nw^2
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0 = sumsq - c
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a^2+b^2+c^2=sum of painting area
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(a+2w)^2+(b+2w)^2+...= c; find w quickly
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a^2+2aw+4w^2+...
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c = = a^2 + 4w^2+2aw + ...
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=a^2+b^2+c^2 + ... 2aw + 2bw + 2cw + ... + 4nw^2
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0 = sumsq + 4nw^2 + 2w(a+b+c+d+....) - c
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quadratic formula
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= 4nw^2+2w(a+b+c+d)
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= 4nw^2+2w(a+b+...) - c +a^2+b^2+c^2+...
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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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