solved the second problem

Author: ishantripathi64

Problems/Mathematics/Day-01/sol/Ishan Tripathi/solution2.cpp

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+#include <bits/stdc++.h>
+
+#include <ext/pb_ds/assoc_container.hpp>
+#include <ext/pb_ds/tree_policy.hpp>
+
+using namespace std;
+using namespace __gnu_pbds;
+
+typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds; // find_by_order, order_of_key
+// less makes it sorted greater makes it reverse sorted less equal makes it like multiset
+// we can also change the data type of the pbds
+
+// Fast I/O
+#define FAST_IO                  \
+    ios::sync_with_stdio(false); \
+    cin.tie(NULL);
+#define pb push_back
+#define in insert
+#define int long long
+#define all(x) (x).begin(), (x).end()
+
+// Typedefs for convenience
+typedef vector<int> vi;
+typedef pair<int, int> pii;
+
+// Constants
+const int INF = 1e18;
+const int MOD = 1e9 + 7;
+
+// Debug (can be disabled in contests)
+#ifdef DEBUG
+#define dbg(x) cerr << #x << " = " << x << '\n';
+#else
+#define dbg(x)
+#endif
+
+void solve()
+{
+    int n, k;
+    cin >> n >> k;
+    vi v(n);
+    for (int i = 0; i < n; i++)
+    {
+        cin >> v[i];
+    }
+    vi prime(1e5 + 1);
+    map<int, vector<pair<int, int>>> facs;
+
+    for (int i = 2; i <= 1e5; i++)
+    {
+        if (prime[i] == 0)
+        {
+            for (int j = i; j <= 1e5; j += i)
+            {
+                if (j != i)
+                    prime[j] = 1;
+
+                int temp = j;
+                int cnt = 0;
+                while (temp % i == 0)
+                {
+                    temp /= i;
+                    cnt++;
+                }
+
+                cnt %= k;
+                if (cnt > 0)
+                    facs[j].pb({i, cnt});
+            }
+        }
+    }
+    int u = 0;
+    int ans = 0;
+    map<vector<pair<int, int>>, int> mp;
+    for (int i = 0; i < n; i++)
+    {
+        vector<pair<int, int>> s = facs[v[i]];
+        mp[s]++;
+        // if (s.size() == 0)
+        // {
+        //     cout<<v[i]<<" "<<endl;
+        //     u++;
+        // }
+    }
+    for (auto &p : mp)
+    {
+        vector<pair<int, int>> s = p.first;
+        for (int i = 0; i < s.size(); i++)
+        {
+            s[i].second = (k - s[i].second);
+        }
+        if (s != p.first)
+            ans += mp[s] * p.second;
+        else
+            ans += (p.second) * (p.second - 1) / 2;
+        p.second = 0;
+    }
+
+    cout << ans << endl;
+}
+
+signed main()
+{
+    FAST_IO
+    int t = 1;
+    // cin >> t;
+    while (t--)
+        solve();
+    return 0;
+}
+/*
+    time complexity : 1e5*log1e5
+    space complexity : 1e5 ish
+    so basically what I did was I calculated prime factorization of each and every number till 1e5
+    then I took modulo of each exponent with k since while multiplying the sum of exponent of each prime factor must be a multiple of k 
+
+    so then I used a map of vector pair to store prime fac of each number
+    then I just took into count those numeber which are present in the array using another map
+    now in the end I just itereated the map and counted good pairs
+
+
+    subbmission:
+    https://codeforces.com/contest/1225/submission/355181685
+
+*/