d3q2 done 👍

Author: Aaryan-Degama

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

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+// link : https://codeforces.com/contest/343/submission/355845658
+
+
+#include <bits/stdc++.h>
+using namespace std;
+ 
+int main() {
+    long long a,b;
+    cin>>a>>b;
+    long long ans=0;
+    while(a>0 && b>0) {
+        if(a>b) {
+            ans+=a/b;
+            a=a%b;
+        }
+        else {
+            ans+=b/a;
+            b=b%a;
+        }
+    }
+    cout<<ans<<endl;
+}
+
+
+
+/*
+Explanation:
+
+We need to construct a resistance equal to a / b using unit resistors.
+Series and parallel connections correspond to increasing or reducing the
+fraction exactly like steps in the Euclidean Algorithm.
+
+If a > b:
+- We can add 1 in series floor(a / b) times
+- a becomes a % b
+
+If b > a:
+- We can add 1 in parallel floor(b / a) times
+- b becomes b % a
+
+Each division step adds the quotient to the number of resistors used.
+Thus, the answer is the sum of all quotients during the Euclidean process.
+
+The fraction is irreducible, so the algorithm always terminates.
+
+Time Complexity:
+O(log(min(a, b)))
+
+Space Complexity:
+O(1)
+*/