Add solution for resistor fraction problem

Implemented a solution to calculate the minimum number of resistors needed to form a fraction a/b using a method similar to the Euclidean algorithm.

Author: suzzzal

Problems/Mathematics/Day-03/sol/Sujal_Kshatri/day03sol2.cpp

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+// submission link - https://codeforces.com/contest/343/submission/355607613
+
+#include <bits/stdc++.h>
+using namespace std;
+
+/*
+1. We want to make fraction a/b using smallest number of 1 resistors.
+2. We can add resistor in line (series) or side (parallel).
+3. Instead of making from 1, we go backward from a/b.
+4. If a is bigger than b, then we minus b from a.
+5. If b is bigger than a, then we minus a from b.
+6. Every minus means we used one resistor.
+7. We count how many times we can minus using divide.
+8. This work same like Euclid algo (gcd thing).
+9. When one number become zero, we stop.
+10. Total count is our answer.
+*/
+
+int main()
+{
+    unsigned long long a, b;
+    cin >> a >> b;
+
+    long long totalResistors = 0;
+
+    while (a && b)
+    {
+        totalResistors += a / b;
+        a %= b;
+        swap(a, b);
+    }
+
+    cout << totalResistors;
+    return 0;
+}