Merge pull request #1 from Aiyaan-Mahajan/Aiyaan-Mahajan-patch-1

Q2 day 1 solved

Author: Aiyaan-Mahajan

Problems/Mathematics/Day-01/sol/Aiyaan_Mahajan/Q2sol.cpp

@@ -0,0 +1,68 @@
+/*
+Q2 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=𝑥𝑘
+.*/
+
+#include <bits/stdc++.h>
+using namespace std;
+ 
+// Function to factorize a number and return prime exponents modulo k
+map<int, int> factorizeMod(long long x, int k) {
+    map<int, int> factors;
+    for (long long p = 2; p * p <= x; ++p) {
+        int count = 0;
+        while (x % p == 0) {
+            x /= p;
+            count++;
+        }
+        if (count % k != 0) {
+            factors[p] = count % k; // only store modulo k
+        }
+    }
+    if (x > 1) factors[x] = 1 % k; // remaining prime factor
+    return factors;
+}
+ 
+// Function to compute the complement signature modulo k
+map<int, int> complementSignature(const map<int, int>& sig, int k) {
+    map<int, int> comp;
+    for (auto &[prime, exp] : sig) {
+        comp[prime] = (k - exp) % k;
+    }
+    return comp;
+}
+ 
+int main() {
+ 
+    int n, k;
+    cin >> n >> k;
+    vector<long long> a(n);
+    for (int i = 0; i < n; ++i) cin >> a[i];
+ 
+    // Map to count how many times each signature has occurred
+    map<map<int,int>, long long> signatureCount;
+    long long totalPairs = 0;
+ 
+    for (int i = 0; i < n; ++i) {
+        map<int,int> sig = factorizeMod(a[i], k);
+        map<int,int> comp = complementSignature(sig, k);
+ 
+        // Add pairs that match the complement
+        totalPairs += signatureCount[comp];
+ 
+        // Increment count of current signature
+        signatureCount[sig]++;
+    }
+ 
+    cout << totalPairs << "\n";
+    return 0;
+}
+
+/*
+My submission : https://codeforces.com/problemset/submission/1225/355325134
+*/