Implement solution for "D. Perfect K-th Power Pairs" problem

Author: amansharma264

Problems/Mathematics/Day-01/sol/amansharma264/solution2.cpp

@@ -0,0 +1,143 @@
+/*
+====================================================
+PROBLEM STATEMENT:
+====================================================
+You are given an integer n and an integer k (k ≥ 2),
+along with an array of n positive integers.
+
+Your task is to count the number of unordered pairs
+(i, j) such that:
+- 1 ≤ i < j ≤ n
+- ai × aj is a perfect k-th power
+
+A number is a perfect k-th power if it can be written
+as x^k for some integer x.
+
+----------------------------------------------------
+PROBLEM LINK:
+https://codeforces.com/problemset/problem/1225/D
+====================================================
+*/
+
+/*
+====================================================
+APPROACH:
+====================================================
+Key Idea:
+For a product ai × aj to be a perfect k-th power,
+the total exponent of every prime factor in the
+product must be divisible by k.
+
+Steps:
+1. For each number ai, perform prime factorization.
+2. Reduce each prime’s exponent modulo k.
+3. Store the resulting factorization as "cur".
+4. Compute the complementary factorization "need"
+   such that:
+       cur × need → all exponents divisible by k
+5. Use a map to count how many times "need"
+   has appeared before.
+6. Add that count to the answer.
+7. Insert "cur" into the map.
+
+----------------------------------------------------
+WHY THIS WORKS:
+- Pairing numbers whose prime exponents complement
+  each other modulo k ensures the product becomes
+  a perfect k-th power.
+====================================================
+*/
+
+/*
+====================================================
+TIME & SPACE COMPLEXITY:
+====================================================
+Time Complexity:
+O(n * sqrt(ai))   (prime factorization per element)
+
+Space Complexity:
+O(n)              (map storage)
+====================================================
+*/
+
+/*
+====================================================
+EXAMPLE:
+====================================================
+Input:
+6 3
+1 3 9 8 24 1
+
+Output:
+5
+
+Explanation:
+There are 5 unordered pairs whose product is a
+perfect cube.
+====================================================
+*/
+
+/*
+====================================================
+SUBMISSION LINK:
+====================================================
+(Add your Codeforces submission link here after AC)
+====================================================
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+#define fastio() ios::sync_with_stdio(false); cin.tie(nullptr);
+
+int main() {
+    fastio();
+
+    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> mp;
+    long long ans = 0;
+
+    for (int x : a) {
+        vector<pair<int,int>> cur, need;
+
+        int temp = x;
+        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) {
+            int cnt = 1 % k;
+            if (cnt) {
+                cur.push_back({temp, cnt});
+                need.push_back({temp, (k - cnt) % k});
+            }
+        }
+
+        sort(cur.begin(), cur.end());
+        sort(need.begin(), need.end());
+
+        ans += mp[need];
+        mp[cur]++;
+    }
+
+    cout << ans << '\n';
+    return 0;
+}