Merge pull request #320 from ayush-mg/main

Create Solution2.cpp

Author: zonedoutdg

Problems/Mathematics/Day-03/Ayush Mishra/Solution2.cpp

@@ -0,0 +1,50 @@
+
+
+// Submission Link: - 
+// https://codeforces.com/contest/343/submission/355549255
+// Explanation of Logic:
+
+// Time Complexity - O(log(min(a,b)))
+// Space Complexity - O(1)
+
+// 1. The If Condition (Perfect Division):
+//    - checks if 'a' is perfectly divisible by 'b' (remainder is 0).
+//    - If true, this represents a whole number resistance. It means the 
+//      target can be built purely with series resistors, with no parallel 
+//      branches needed.
+//    - We simply output 'a / b' as the total count.
+
+// 2. The Else Block (Euclidean Division Lemma):
+//    - If there is a remainder, we enter the loop to handle the fractional part
+//      (representing the parallel branches).
+//    - 'count' starts with the integer part of the first division.
+//    - Inside the while loop:
+//      'd / c' finds the integer part of the inverted remainder.
+//       We add this to 'count'.
+//       We swap variables to perform the mathematical inversion (1/x) 
+//       for the next iteration.
+
+      
+
+#include<bits/stdc++.h>
+using namespace std;
+int main(){
+    long long a,b;
+    cin>>a>>b;
+    if(a%b == 0){
+        cout<<a/b;
+    }
+    else {
+        long long c = a%b;
+        long long d = b;
+        long long count = a/b;
+        while(c != 0){
+            count += d/c;
+            long long temp = d%c;
+            d = c;
+            c = temp;
+        }
+        cout<<count;
+    }
+    return 0;
+}