Merge branch 'opencodeiiita:main' into main

Author: d2-e2-v2

Problems/Mathematics/Day-01/sol/Abhigyan/Solution1.cpp

@@ -0,0 +1,42 @@
+/* 1 to n people are standing in a circle. n is even.... a is looking at b. 
+If given c is looking at some d person for given a,b then find that d. If such a combination
+of circle does not exist then print -1*/
+
+// Time complexity: O(1)
+// Space complexity: O(1)
+//My submission: https://codeforces.com/contest/1560/submission/355204246
+
+#include <bits/stdc++.h>
+using namespace std;
+
+using ll = long long;
+#define all(x) (x).begin(), (x).end()
+
+const int MOD = 1e9 + 7; // 998244353
+const ll INF = 1e18;
+
+void solve() {
+    int a,b,c;
+    cin>>a>>b>>c;
+    int n=2*abs(a-b);
+    if(a>n || b>n ||c>n ) cout<<-1<<endl;
+    else{
+        int x=c-n/2,y=c+n/2;
+        if(x>=1 && x<=n) cout<<x<<endl;
+        else cout<<y<<endl;
+    }
+}
+
+int main() {
+    // Fast I/O
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int t;
+    cin >> t;
+    while (t--) {
+        solve();
+    }
+    
+    return 0;
+}

Problems/Mathematics/Day-01/sol/Adarsh Yadav/solution.cpp

@@ -0,0 +1,27 @@
+//https://codeforces.com/contest/1560/problem/C
+//https://codeforces.com/contest/1560/submission/355210608
+// TC : O(1) SC: O(1)
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+	int t;
+	cin>>t;
+	while(t--){
+	    int a,b,c;
+	    cin>>a>>b>>c;
+	    int k=abs(a-b);
+	    if(a>2*k || b>2*k || c>2*k){
+	        cout<<-1<<endl;
+	    }
+	    else{
+	        if((c-k)>0){
+	            cout<<c-k<<endl;
+	        }
+	        else{
+	            cout<<c+k<<endl;
+	        }
+	    }
+	}
+
+}

Problems/Mathematics/Day-01/sol/Aditi_Guin/Solution1.cpp

@@ -0,0 +1,23 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    int t;
+    cin >> t;
+
+    while (t--) {
+        int a, b, c;
+        cin >> a >> b >> c;
+
+        int half = abs(a - b);
+        int n = 2 * half;
+
+        if (c > n) {
+            cout << -1 << "\n";
+        } else {
+            int opposite = (c + half <= n) ? (c + half) : (c - half);
+            cout << opposite << "\n";
+        }
+    }
+    return 0;
+}

Problems/Mathematics/Day-01/sol/Aditi_Guin/Solution2.cpp

@@ -0,0 +1,47 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int n, k;
+    cin >> n >> k;
+
+    vector<int> a(n);
+    for (int &x : a) cin >> x;
+
+    map<vector<pair<int,int>>, long long> freq;
+    long long ans = 0;
+
+    for (int x : a) {
+        int temp = x;
+        vector<pair<int,int>> cur, need;
+
+        for (int p = 2; p * p <= temp; p++) {
+            if (temp % p == 0) {
+                int cnt = 0;
+                while (temp % p == 0) {
+                    temp /= p;
+                    cnt++;
+                }
+                cnt %= k;
+                if (cnt) {
+                    cur.push_back({p, cnt});
+                    need.push_back({p, (k - cnt) % k});
+                }
+            }
+        }
+
+        if (temp > 1) {
+            cur.push_back({temp, 1 % k});
+            need.push_back({temp, (k - 1) % k});
+        }
+
+        ans += freq[need];
+        freq[cur]++;
+    }
+
+    cout << ans << "\n";
+    return 0;
+}

Problems/Mathematics/Day-01/sol/Aditya_Singh/solution002.cpp

@@ -0,0 +1,87 @@
+/*
+Problem:
+You are given an array of n positive integers and an integer k (k >= 2).
+Count the number of unordered pairs (i, j) such that ai * aj is a perfect k-th power.
+
+Approach:
+1. Precompute smallest prime factor (SPF) up to max(ai).
+2. For each number, compute its prime factorization.
+3. Reduce exponents modulo k to form a "signature".
+4. Compute the complementary signature needed to make exponents divisible by k.
+5. Use a map to count matching signatures seen so far.
+
+Time Complexity: O(n log A)
+Space Complexity: O(n)
+
+Problem Link:
+https://codeforces.com/problemset/problem/1225/D
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+static const int MAXA = 100000;
+
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int n, k;
+    cin >> n >> k;
+
+    vector<int> a(n);
+    for (int i = 0; i < n; i++) {
+        cin >> a[i];
+    }
+
+    // Step 1: Compute Smallest Prime Factor (SPF)
+    vector<int> spf(MAXA + 1);
+    for (int i = 1; i <= MAXA; i++) spf[i] = i;
+
+    for (int i = 2; i * i <= MAXA; i++) {
+        if (spf[i] == i) {
+            for (int j = i * i; j <= MAXA; j += i) {
+                if (spf[j] == j)
+                    spf[j] = i;
+            }
+        }
+    }
+
+    // Map to store frequency of signatures
+    map<vector<pair<int,int>>, long long> freq;
+    long long answer = 0;
+
+    // Step 2–5: Process each number
+    for (int x : a) {
+        map<int,int> factorCount;
+
+        // Factorize using SPF
+        while (x > 1) {
+            int p = spf[x];
+            factorCount[p]++;
+            x /= p;
+        }
+
+        vector<pair<int,int>> signature;
+        vector<pair<int,int>> complement;
+
+        for (auto &it : factorCount) {
+            int prime = it.first;
+            int exp = it.second % k;
+
+            if (exp != 0) {
+                signature.push_back({prime, exp});
+                complement.push_back({prime, (k - exp) % k});
+            }
+        }
+
+        // Count valid pairs
+        answer += freq[complement];
+
+        // Store current signature
+        freq[signature]++;
+    }
+
+    cout << answer << "\n";
+    return 0;
+}

Problems/Mathematics/Day-01/sol/Avaneesh Verma/solution002.py

@@ -0,0 +1,46 @@
+#include <bits/stdc++.h>
+using namespace std;
+ 
+int main() {
+    int n, k;
+    cin >> n >> k;
+ 
+    vector<int> a(n);
+    for (int i = 0; i < n; i++)
+        cin >> a[i];
+ 
+    map<vector<pair<int,int>>, long long> freq;
+    long long ans = 0;
+ 
+    for (int i = 0; i < n; i++) {
+        int x = a[i];
+        vector<pair<int,int>> cur, need;
+ 
+        for (int p = 2; p * p <= x; p++) {
+            if (x % p == 0) {
+                int cnt = 0;
+                while (x % p == 0) {
+                    x /= p;
+                    cnt++;
+                }
+                cnt %= k;
+                if (cnt > 0) {
+                    cur.push_back({p, cnt});
+                    need.push_back({p, (k - cnt) % k});
+                }
+            }
+        }
+ 
+        if (x > 1) {
+            cur.push_back({x, 1});
+            need.push_back({x, (k - 1) % k});
+        }
+        sort(cur.begin(), cur.end());
+        sort(need.begin(), need.end());
+ 
+        ans += freq[need];
+        freq[cur]++;
+    }
+    cout << ans << "\n";
+    return 0;
+}

Problems/Mathematics/Day-01/sol/Ayush Mishra/Solution1.cpp

@@ -0,0 +1,43 @@
+// submission link
+// https://codeforces.com/contest/1560/submission/355180160
+
+// Space Complexity = O(1)
+// Time Complexity = O(1) for main solution code O(t) for overall code
+
+
+
+// The if condition checks if the circle is valid and if the inputs exist within that circle.
+// abs(b-a)<=1: Ensures a and b are not adjacent or the same.
+// 2*(abs(b-a)-1)+2 < c || 2*(abs(b-a)-1)+2 < a || 2*(abs(b-a)-1)+2 < b.  checks that the input numbers a, b, and c are not larger than the total size of the circle.
+// If any of these are true, it prints -1 .
+
+// The line int k = 2*(abs(b-a)-1) + 2 calculates the total number of items in the circle.
+
+// If the inputs are valid, it calculates the number opposite to c.
+// The halfway distance is k / 2.
+// If c is in the first half (c<= k/2), the opposite is c + k/2.
+// If c is in the second half (c > k/2), the opposite is c - k/2.
+
+
+
+
+#include<bits/stdc++.h>
+using namespace std;
+int main(){
+    int t;
+    cin>>t;
+    while(t--){
+        int a,b,c;
+        cin>>a>>b>>c;
+        if(abs(b-a)<=1 || 2*(abs(b-a)-1)+2 < c || 2*(abs(b-a)-1)+2 < a || 2*(abs(b-a)-1)+2 < b ){
+            cout<<-1<<"\n";
+        }
+        else {
+            int k =  2*(abs(b-a)-1) + 2;
+            if(c<=k/2) cout<<c+k/2 <<"\n";
+            else cout<<c-k/2<<"\n";
+        }
+
+    }
+    return 0;
+}