111 lines
2.1 KiB
C++
111 lines
2.1 KiB
C++
#include <bits/stdc++.h>
|
|
|
|
// https://codeforces.com/blog/entry/96344
|
|
|
|
#pragma GCC optimize("O2,unroll-loops")
|
|
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
|
|
|
|
using namespace std;
|
|
|
|
template <typename... Args>
|
|
void dbg(std::string const &str, Args &&...args) {
|
|
std::cout << std::vformat(str, std::make_format_args(args...));
|
|
}
|
|
|
|
template <typename T>
|
|
void dbg(T const &t) {
|
|
std::cout << t;
|
|
}
|
|
|
|
template <std::ranges::range T>
|
|
void dbgln(T const &t) {
|
|
if constexpr (std::is_convertible_v<T, char const *>) {
|
|
std::cout << t << '\n';
|
|
} else {
|
|
for (auto const &e : t) {
|
|
std::cout << e << ' ';
|
|
}
|
|
std::cout << '\n';
|
|
}
|
|
}
|
|
|
|
void dbgln() {
|
|
std::cout << '\n';
|
|
}
|
|
|
|
template <typename... Args>
|
|
void dbgln(std::string const &str, Args &&...args) {
|
|
dbg(str, std::forward<Args>(args)...);
|
|
cout << '\n';
|
|
}
|
|
|
|
template <typename T>
|
|
void dbgln(T const &t) {
|
|
dbg(t);
|
|
cout << '\n';
|
|
}
|
|
|
|
template <typename T>
|
|
constexpr T MIN = std::numeric_limits<T>::min();
|
|
|
|
template <typename T>
|
|
constexpr T MAX = std::numeric_limits<T>::min();
|
|
|
|
template <typename T>
|
|
static T sc(auto &&x) {
|
|
return static_cast<T>(x);
|
|
}
|
|
|
|
#define ff first
|
|
#define ss second
|
|
#define eb emplace_back
|
|
#define ll long long
|
|
#define ld long double
|
|
#define vec vector
|
|
#define endl '\n'
|
|
|
|
#define all(x) (x).begin(), (x).end()
|
|
#define rall(x) (r).rbegin(), (x).rend()
|
|
#define sz(x) static_cast<int>((x).size())
|
|
#define FOR(a, b, c) for (ll(a) = (b); (a) < (c); ++(a))
|
|
#define ROF(a, b, c) for (int(a) = (b); (a) > (c); --(a))
|
|
|
|
void solve() {
|
|
ll x, y, z, k;
|
|
cin >> x >> y >> z >> k;
|
|
/*
|
|
vol = k <= x * y * z
|
|
|
|
explore all valid box volumes at 0, 0, 0
|
|
|
|
a * b * c = k
|
|
a * b = k / c <-> c = k / (a * b)
|
|
a=1 -> b * c <= k - bsearch?
|
|
*/
|
|
|
|
ll ans = 0;
|
|
|
|
FOR(a, 1, x + 1) {
|
|
FOR(b, 1, y + 1) {
|
|
ll total = a * b;
|
|
if (k % total == 0 && k / total <= z) {
|
|
ans = max(ans, (x - a + 1) * (y - b + 1) * (z - k / total + 1));
|
|
}
|
|
}
|
|
}
|
|
|
|
dbgln(ans);
|
|
}
|
|
|
|
int main() {
|
|
cin.tie(nullptr)->sync_with_stdio(false);
|
|
|
|
int t = 1;
|
|
cin >> t;
|
|
|
|
while (t--) {
|
|
solve();
|
|
}
|
|
|
|
return 0;
|
|
}
|