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| 1 | +/* |
| 2 | +Problem Statement: |
| 3 | +You are given an integer n and an integer k (k ≥ 2), along with an array of n |
| 4 | +positive integers. |
| 5 | +
|
| 6 | +Your task is to count the number of unordered index pairs (i, j) such that: |
| 7 | +1 ≤ i < j ≤ n |
| 8 | +The product ai × aj is a perfect k-th power. |
| 9 | +
|
| 10 | +--------------------------------------- |
| 11 | +Approach: |
| 12 | +A number is a perfect k-th power if, in its prime factorization, |
| 13 | +the exponent of every prime is divisible by k. |
| 14 | +
|
| 15 | +For each number: |
| 16 | +1. Factorize it using smallest prime factor (SPF). |
| 17 | +2. Reduce each prime’s exponent modulo k. |
| 18 | +3. Store this reduced form as a "signature". |
| 19 | +
|
| 20 | +For a pair (ai, aj) to form a perfect k-th power: |
| 21 | +- Their signatures must complement each other such that |
| 22 | + the sum of exponents (mod k) becomes 0 for every prime. |
| 23 | +
|
| 24 | +--------------------------------------- |
| 25 | +Time Complexity: |
| 26 | +O(n log a) |
| 27 | +
|
| 28 | +Space Complexity: |
| 29 | +O(n) |
| 30 | +--------------------------------------- |
| 31 | +*/ |
| 32 | + |
| 33 | +import java.util.*; |
| 34 | + |
| 35 | +public class Solution2 { |
| 36 | + |
| 37 | + static int MAX = 100000; |
| 38 | + static int[] spf = new int[MAX + 1]; |
| 39 | + |
| 40 | + // Precompute smallest prime factor for every number |
| 41 | + static void computeSPF() { |
| 42 | + for (int i = 1; i <= MAX; i++) |
| 43 | + spf[i] = i; |
| 44 | + |
| 45 | + for (int i = 2; i * i <= MAX; i++) { |
| 46 | + if (spf[i] == i) { |
| 47 | + for (int j = i * i; j <= MAX; j += i) { |
| 48 | + if (spf[j] == j) |
| 49 | + spf[j] = i; |
| 50 | + } |
| 51 | + } |
| 52 | + } |
| 53 | + } |
| 54 | + |
| 55 | + // Build reduced prime-exponent signature |
| 56 | + static Map<Integer, Integer> buildSignature(int x, int k) { |
| 57 | + Map<Integer, Integer> map = new HashMap<>(); |
| 58 | + while (x > 1) { |
| 59 | + int p = spf[x]; |
| 60 | + int cnt = 0; |
| 61 | + while (x % p == 0) { |
| 62 | + x /= p; |
| 63 | + cnt++; |
| 64 | + } |
| 65 | + cnt %= k; |
| 66 | + if (cnt > 0) |
| 67 | + map.put(p, cnt); |
| 68 | + } |
| 69 | + return map; |
| 70 | + } |
| 71 | + |
| 72 | + // Build complement signature |
| 73 | + static Map<Integer, Integer> buildComplement(Map<Integer, Integer> sig, int k) { |
| 74 | + Map<Integer, Integer> comp = new HashMap<>(); |
| 75 | + for (Map.Entry<Integer, Integer> e : sig.entrySet()) { |
| 76 | + comp.put(e.getKey(), (k - e.getValue()) % k); |
| 77 | + } |
| 78 | + return comp; |
| 79 | + } |
| 80 | + |
| 81 | + public static void main(String[] args) { |
| 82 | + Scanner sc = new Scanner(System.in); |
| 83 | + computeSPF(); |
| 84 | + |
| 85 | + int n = sc.nextInt(); |
| 86 | + int k = sc.nextInt(); |
| 87 | + |
| 88 | + Map<Map<Integer, Integer>, Long> freq = new HashMap<>(); |
| 89 | + long ans = 0; |
| 90 | + |
| 91 | + for (int i = 0; i < n; i++) { |
| 92 | + int x = sc.nextInt(); |
| 93 | + |
| 94 | + Map<Integer, Integer> sig = buildSignature(x, k); |
| 95 | + Map<Integer, Integer> comp = buildComplement(sig, k); |
| 96 | + |
| 97 | + ans += freq.getOrDefault(comp, 0L); |
| 98 | + freq.put(sig, freq.getOrDefault(sig, 0L) + 1); |
| 99 | + } |
| 100 | + |
| 101 | + System.out.println(ans); |
| 102 | + sc.close(); |
| 103 | + } |
| 104 | +} |
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