Merge pull request #290 from MK-codes365/MK-codes365

Fix: Solve Resistor Time Machine (Issue #286)

Author: zonedoutdg

Problems/Mathematics/Day-02/sol/Mukut/Solution2.cpp

@@ -0,0 +1,74 @@
+/*
+Problem: 
+Pick k bloggers from n to maximize total followers, and count how many ways we can do this. 
+The answer should be modulo 10^9 + 7.
+
+Approach:
+To maximize the sum, we must pick bloggers with the highest follower counts. 
+I sort the array in descending order. The "threshold" value will be the followers of the k-th blogger.
+I then count how many bloggers in the entire list have this threshold value and how many 
+we actually need to pick from them to complete our k-blogger selection.
+The result is simply (total_with_threshold) Choose (needed_from_them).
+I used Pascal's triangle to precompute combinations since n is small (up to 1000).
+
+TimeComplexity: O(n^2) precompute + O(n log n) per test case
+Spacecomplexity: O(nCr table size) approx O(n^2)
+
+Submission Link: https://codeforces.com/contest/1475/submission/355409229
+*/
+
+#include <iostream>
+#include <vector>
+#include <algorithm>
+
+using namespace std;
+
+long long nCr[1005][1005];
+const int MOD = 1e9 + 7;
+
+void pre() {
+    for (int i = 0; i <= 1000; i++) {
+        nCr[i][0] = 1;
+        for (int j = 1; j <= i; j++) {
+            nCr[i][j] = (nCr[i - 1][j - 1] + nCr[i - 1][j]) % MOD;
+        }
+    }
+}
+
+void solve() {
+    int n, k;
+    cin >> n >> k;
+    
+    vector<int> a(n);
+    for (int i = 0; i < n; i++) cin >> a[i];
+
+    sort(a.begin(), a.end(), greater<int>());
+
+    int target = a[k - 1]; 
+    int cnt = 0;
+    for (int x : a) {
+        if (x == target) cnt++;
+    }
+
+    int take = 0;
+    for (int i = 0; i < k; i++) {
+        if (a[i] == target) take++;
+    }
+
+    cout << nCr[cnt][take] << endl;
+}
+
+int main() {
+    ios::sync_with_stdio(0);
+    cin.tie(0);
+
+    pre();
+    
+    int t;
+    cin >> t;
+    while (t--) {
+        solve();
+    }
+    
+    return 0;
+}

Problems/Mathematics/Day-03/sol/MK-codes365/Solution2.cpp

@@ -0,0 +1,45 @@
+/*
+SHORT PROBLEM STATEMENT: 
+We need to find the minimum number of 1-ohm resistors to get a resistance of a/b.
+Resistors can be connected in series or parallel.
+
+APPROACH USING PREFIX SUMS: 
+Not applicable here as it's a math/GCD based problem.
+The logic is basiclly like the Euclidean algorithm. If we have to reach a/b, 
+we keep taking the floor part (x/y) as series resistors and then 
+calculate the remainder and swap to simulate the parallel recipricol logic.
+
+TIME & SPACE COMPLEXITY:
+Time: O(log(min(a, b)))
+Space: O(1)
+
+SUBMISSION LINK: https://codeforces.com/contest/343/submission/355457508 
+*/
+
+
+#include <iostream>
+#include <algorithm>
+
+using namespace std;
+
+typedef long long ll;
+
+int main() {
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+
+    ll x, y;
+    if (!(cin >> x >> y)) return 0;
+
+    ll count = 0;
+
+    while (y > 0) {
+        count += (x / y);
+        x %= y;
+        swap(x, y);
+    }
+
+    cout << count << "\n";
+
+    return 0;
+}