Merge branch 'main' into ahwan

Author: Dheeraj-06

.DS_Store

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+/*We have:
+
+An even number of people n standing in a circle, numbered 1 to n clockwise.
+
+Each person looks directly opposite in the circle.
+
+You are given three numbers: a, b, c.
+
+a is looking at b.
+
+We want to know who c is looking at.
+
+If no circle is possible → return -1.
+
+*/
+
+typedef long long ll;
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+	int t;
+	cin>>t;
+	while(t--){
+	   ll a, b, c;
+        cin >> a >> b >> c;
+
+        ll diff = abs(a - b);
+        ll n = 2 * diff;
+
+        if (diff==0 || max({a,b,c})> n){
+            cout << -1 << endl;
+        } else {
+            ll d = c + diff;
+            if (d > n) d -= n;
+            cout << d << endl;
+	}
+	}
+  return 0;
+}
+
+//O(1) per test case
+//My SUBMISSION : https://codeforces.com/contest/1560/submission/355307599

Problems/.DS_Store

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+// Link to Submission: https://codeforces.com/contest/1225/submission/355410192
+
+/*
+Problem Statement:
+You are given n positive integers a1, …, an, and an integer k ≥ 2.
+Count the number of pairs i, j such that 1 ≤ i < j ≤ n, and there exists an integer x such that ai ⋅ aj = x^k.
+*/
+
+/*
+Brief Explanation:
+To avoid TLE we avoid comparing each combination of elements, rather we calculate the reduced form of each element in the array.
+The reduced form of an element can be calculated as -> A = q^x * p^y... ~ q^(x % k) * p^(y %k)...
+We then get the complement i.e. Ac ~ q^((k - (x % k)) % k) * p^((k - (y % k)) % k)
+While iterating through the array, we maintain a frequency map of reduced forms.
+For each element, we count how many previously seen reduced forms equal its complement and add that to the pair count.
+*/
+
+import java.util.*;
+
+public class Solution2 {
+    static long[] reducedAndComplement(long x, int k) {
+        long rf = 1;
+        long comp = 1;
+
+        for (long p = 2; p * p <= x; p++) {
+            int count = 0;
+
+            while (x % p == 0) {
+                x /= p;
+                count++;
+            }
+
+            count %= k;
+
+            for (int i = 0; i < count; i++)
+                rf *= p;
+
+            int req = (k - count) % k;
+            for (int i = 0; i < req; i++)
+                comp *= p;
+        }
+
+        if (x > 1) {
+            rf *= x;
+            int req = (k - 1) % k;
+            for (int i = 0; i < req; i++)
+                comp *= x;
+        }
+
+        return new long[]{rf, comp};
+    }
+
+    public static void main(String[] args) {
+        Scanner sc = new Scanner(System.in);
+        HashMap<Long, Integer> freq = new HashMap<>();
+        int n = sc.nextInt();
+        int k = sc.nextInt();
+        
+        long fc = 0;
+
+        for (int i = 0; i < n; i++) {
+            long a = sc.nextLong();
+            long[] rc = reducedAndComplement(a, k);
+            long r = rc[0];
+            long c = rc[1];
+
+            fc += freq.getOrDefault(c, 0);
+
+            freq.put(r, freq.getOrDefault(r, 0) + 1);
+        }
+
+        System.out.println(fc);
+
+        sc.close();
+    }
+}
+
+// Time Complexity: O(n * sqr.root(A))
+// Space Complexity: O(n)

Problems/Mathematics/.DS_Store

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+/*
+Problem Statement:
+Given three distinct integers a, b, and c placed on a circle of even size,
+numbered clockwise from 1. Person a is opposite to person b.
+Find the number opposite to c. If no such circle exists, return -1.
+
+Prefix Sums:
+Not applicable since the problem is purely mathematical and does not involve
+range queries or cumulative sums.
+
+Time Complexity: O(1)
+Space Complexity: O(1)
+
+Submission Link:
+https://codeforces.com/contest/1560/submission/355443029
+*/
+#include <bits/stdc++.h>
+using namespace std;
+
+int main(){
+
+int t;
+cin>>t;
+while(t--){
+int a,b,c,n;
+cin>>a>>b>>c;
+if(a>b) swap(a,b);    //naming the bigger number b and smaller one a
+
+int SizeOfCircle= 2*(b-a); // thinking: b and a form the diameter of circle 
+
+if(SizeOfCircle==2) cout<<-1<<endl; //edge case when a and b are adjacent
+else{
+if(c>SizeOfCircle|| b>SizeOfCircle) cout<<-1<<endl; // edge case when either c or the bigger number do not reside in the size
+else{
+int ans=0;
+if(c>b) ans=(c-(b-a));      // because a and b form diameter then c lies in upper or lower half and opposite of c also lies based on that
+else if(c<=b) ans=(c+(b-a));
+
+if(ans<=0) ans+=SizeOfCircle;
+else if(ans>SizeOfCircle) ans-=SizeOfCircle;   // fixing ans if it resides outside elements of circle
+
+cout<< ans<<endl;
+}
+
+
+}
+}
+return 0;
+}