From 83db78e6301b14c87d55869c56d1d49b11d71c5d Mon Sep 17 00:00:00 2001
From: Ankur-Dwivedi22 <ankur.2201005cs@iiitbh.ac.in>
Date: Sat, 6 Apr 2024 13:17:23 +0530
Subject: [PATCH] chore: completed day 4 hard task author: Ankur-Dwivedi22

---
 Hard/Day4/Solution.cpp | 66 +++++++++++++++++++++++++++++++++++++++++-
 1 file changed, 65 insertions(+), 1 deletion(-)

diff --git a/Hard/Day4/Solution.cpp b/Hard/Day4/Solution.cpp
index 5223d7e..031864d 100644
--- a/Hard/Day4/Solution.cpp
+++ b/Hard/Day4/Solution.cpp
@@ -1 +1,65 @@
-// Write Your Code Here
+#include <iostream>
+#include <vector>
+
+using namespace std;
+using ll = long long;
+
+// Function to calculate the number of pairs
+ll calculatePairs(ll n, ll k) {
+    if (n == 0 || k == 0) {
+        return 0;
+    }
+    ll x = n / k;
+    ll l = n % k;
+    ll r = k - l;
+    ll L = (x + 1) * l, R = x * r;
+    return R * L + (R - x) * R / 2 + L * (L - x - 1) / 2;
+}
+
+void solve() {
+    // Input the number of machines, cost per brick, and additional cost
+    int num_machines;
+    long long brick_cost, additional_cost;
+    cin >> num_machines >> brick_cost >> additional_cost;
+
+    int n = 0;
+    vector<int> machine_count(num_machines);
+    // Input machine counts and find the maximum count
+    for (int i = 0; i < num_machines; ++i) {
+        cin >> machine_count[i];
+        n = max(n, machine_count[i]);
+    }
+
+    // Initialize vectors to store pairs and additional costs
+    vector<long long> pair(n + 1);
+    vector<long long> add(n + 1);
+
+    // Calculate pairs and additional costs for each machine
+    for (int i = 0; i < num_machines; ++i) {
+        for (int j = 1; j <= machine_count[i]; ++j) {
+            pair[j] += calculatePairs(machine_count[i], j);
+        }
+        add[machine_count[i]] += calculatePairs(machine_count[i], machine_count[i]);
+    }
+
+    // Calculate the maximum profit
+    long long max_profit = 0;
+    long long other = 0;
+    for (int i = 1; i <= n; ++i) {
+        max_profit = max(max_profit, brick_cost * (pair[i] + other) - additional_cost * (i - 1));
+        other += add[i];
+    }
+    // Output the maximum profit
+    cout << max_profit << endl;
+}
+
+int main() {
+    // Input the number of test cases
+    int num_test_cases;
+    cin >> num_test_cases;
+    // Solve each test case
+    while (num_test_cases--) {
+        solve();
+    }
+    return 0;
+}