added solution for 2nd

Author: MANISH-BEESA

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

@@ -0,0 +1,76 @@
+/*
+n = number of elements (≤ 100,000)
+
+A = maximum value of ai (≤ 100,000)
+
+Space Complexity - 𝑂(𝐴+𝑛)
+
+Time Complexity - O(nlogA)​
+
+
+*/
+
+
+#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 i = 0; i < n; i++) cin >> a[i];
+
+    // Smallest Prime Factor sieve
+    const int MAXA = 100000;
+    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<vector<pair<int,int>>, long long> freq;
+    long long ans = 0;
+
+    for (int x : a) {
+        map<int,int> cnt;
+
+        // factorization
+        while (x > 1) {
+            int p = spf[x];
+            int c = 0;
+            while (x % p == 0) {
+                x /= p;
+                c++;
+            }
+            cnt[p] += c;
+        }
+
+        vector<pair<int,int>> sig, need;
+
+        for (auto &it : cnt) {
+            int p = it.first;
+            int e = it.second % k;
+            if (e != 0) {
+                sig.push_back({p, e});
+                need.push_back({p, (k - e) % k});
+            }
+        }
+
+        sort(sig.begin(), sig.end());
+        sort(need.begin(), need.end());
+
+        ans += freq[need];
+        freq[sig]++;
+    }
+
+    cout << ans << "\n";
+    return 0;
+}