Create solution002.cpp

Author: adityasingh-0803

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

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+/*
+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;
+}