add day3 problem-2

Author: codeboy-pro

Problems/Mathematics/Day-03/sol/PradipMaity/Solution2.cpp

@@ -0,0 +1,92 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+/*
+ We want minimum number of 1-ohm resistors needed
+ to build resistance = a / b
+
+ Allowed operations (each consumes exactly 1 resistor):
+
+ 1. Series:
+      R = Re + 1
+      (fraction increases, used when R > 1)
+
+ 2.Parallel:
+      R = Re / (Re + 1)
+      (fraction becomes smaller, used when R < 1)
+
+-----------------------------------------------
+ KEY OBSERVATION
+
+ Reverse-thinking makes problem easy:
+
+ Instead of building a/b from 1,
+ go BACKWARDS from a/b to 1/1.
+
+ When:
+      a >= b   → fraction > 1
+ means fraction was formed by SERIES operations
+ so we repeatedly do:
+      a = a - b
+ Each subtraction corresponds to ONE resistor
+
+ Number of such subtractions = a / b
+ (integer division)
+
+ Then we reduce:
+      a %= b
+
+-----------------------------------------------
+
+ When:
+      b > a   → fraction < 1
+ means fraction was formed by PARALLEL operations
+ so we repeatedly do:
+      b = b - a
+
+ Number of such subtractions = b / a
+
+ Then we reduce:
+      b %= a
+
+-----------------------------------------------
+
+This is EXACTLY what Euclidean Algorithm does.
+But instead of counting steps,
+we SUM UP all quotients.
+
+The sum = minimum number of resistors.
+
+And problem guarantees there is always a solution,
+so it always eventually reaches (1,1).
+*/
+
+long long minResistors(long long a, long long b) {
+
+    long long ans = 0;
+
+    while (a && b) {
+
+        // If numerator >= denominator
+        if (a >= b) {
+            ans += a / b;   // number of series operations needed
+            a %= b;         // continue reducing
+        }
+        else {
+            ans += b / a;   // number of parallel operations needed
+            b %= a;         // continue reducing
+        }
+    }
+
+    return ans;
+}
+
+int main() {
+    long long a, b;
+    cin >> a >> b;
+
+    cout << minResistors(a, b);
+}
+
+
+// Submission Link:- (https://codeforces.com/contest/343/my)