Merge branch 'main' of https://github.com/aricthecoder/CP-Chronicles into day5sol2

Author: aricthecoder

Problems/Data-structures/Day-04/sol/Himansh_Arora/Solution1.cpp

@@ -0,0 +1,44 @@
+//Problem Link- https://codeforces.com/problemset/problem/1997/C
+//Submission Link- https://codeforces.com/problemset/submission/1997/355848236
+//checked greedy pairing of current index with last one in stack
+//time complexity- O(n)
+
+#include <bits/stdc++.h>
+using namespace std;
+
+void solve() {
+    int n;
+    string s;
+    cin>>n>>s;
+    stack <pair<char,int>> s1;
+    int cost=0;
+    for(int i=0;i<n;i++){
+        if(s1.empty()){
+            s1.push({s[i],i});
+        }else{
+            if(s[i]==')'&&s1.top().first=='_'){
+                cost+=(i-s1.top().second);
+                s1.pop();
+            }else if(s[i]=='('&&s1.top().first=='_'){
+                s1.push({s[i],i});
+            }else if(s[i]=='_'&&s1.top().first=='('){
+                cost+=(i-s1.top().second);
+                s1.pop();            
+            }
+        }
+    }
+
+     cout<<cost<<endl;
+}
+
+signed main() {
+    
+    int t;
+    cin >> t;
+    
+    while (t--) {
+        solve();
+    }
+    
+    return 0;
+}

Problems/Data-structures/Day-04/sol/Naman Pal/day004sol01.cpp

@@ -1,6 +1,5 @@
 // Submission Link: https://codeforces.com/contest/1997/submission/355835686
 
-
 /*
 So first I made the RBS with minimmum cost by putting closing bracket as soon as we get a opening bracket before it, i.e., closing bracket as early as 
 possible(greedy appraoch). After getting corect bracket sequence I put the indexes of '(' in vector o and indexes of ')' in vector c in order they occur. Both 

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

@@ -0,0 +1,51 @@
+// 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)
+*/

Problems/Mathematics/Day-03/sol/Apoorv012/Solution2.cpp

@@ -0,0 +1,32 @@
+/*
+We needed to find out the max no. of resistors needed, to get to a target resistance, with only 1 unit resistors.
+
+I wanted to convert the resistance to 1/1, so I took the larger no. and subtracted it by the maximum multiple of lower number, which is still lower than the larger one, adding 1 each time.
+
+Time: O(log(max(a, b)))
+Space: O(1)
+
+Submission Link: https://codeforces.com/contest/343/submission/355851930
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+using ll = long long;
+
+int main() {
+    ll a, b;
+    cin >> a >> b;
+
+    ll ans = 1;
+    while (a != 1 || b != 1) {
+        if (a > b) swap(a, b);
+        // b is always greater
+
+        ll count = (b-1) / a;
+        ans += count;
+        b -= count * a;
+    }
+
+    cout << ans;
+}

Problems/Mathematics/Day-03/sol/Naman Pal/day003sol02.cpp

@@ -1,4 +1,4 @@
-Submission Link: https://codeforces.com/contest/343/submission/355828800
+// Submission Link: https://codeforces.com/contest/343/submission/355828800
 
 /*
 Problem: Find the mininmum number of 1 ohm resistance needed to make equivalent resistance a/b