Implement solution for "Resistor Time Machine" problem

Problems/Mathematics/Day-03/sol/amansharma264/solution1.cpp

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+/*
+Problem Name: Resistor Time Machine
+
+Short Problem Statement:
+Mike has an unlimited number of unit resistors (R0 = 1). Using series and parallel
+connections, he wants to construct an equivalent resistance equal to a/b.
+The task is to find the minimum number of unit resistors required to obtain
+the resistance a/b.
+
+Approach (Using Prefix Sums / Euclidean Algorithm Insight):
+The process of building resistance values using series (+1) and parallel
+(R / (R + 1)) operations mirrors the steps of the Euclidean Algorithm.
+At each step:
+- We can add as many series resistors as possible (a / b)
+- Reduce the fraction using modulo
+- Swap roles of numerator and denominator
+
+The total number of operations corresponds to the total number of unit
+resistors required.
+
+This is equivalent to summing all quotients in the Euclidean Algorithm.
+
+Time Complexity:
+O(log(min(a, b)))
+
+Space Complexity:
+O(1)
+
+Example:
+Input: 3 2
+Process:
+3/2 -> 1 + 1/2
+1/2 -> parallel construction
+Total resistors used = 3
+
+Submission Link:
+https://codeforces.com/submissions/amansharma264
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+#define fastio() ios::sync_with_stdio(false); cin.tie(nullptr);
+
+int main() {
+    fastio();
+
+    long long a, b;
+    cin >> a >> b;
+
+    long long ans = 0;
+
+    while (b != 0) {
+        ans += a / b;
+        a %= b;
+        swap(a, b);
+    }
+
+    cout << ans << '\n';
+    return 0;
+}