fix: solve resistor time machine #286

Author: MK-codes365

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;
+}