Merge pull request #149 from ViMo018/patch-1

Add solution for counting divisible pairs

Author: zonedoutdg

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

@@ -0,0 +1,107 @@
+#include<bits/stdc++.h>
+#include<queue>
+using namespace std;
+typedef unordered_map<int, int> umii;
+typedef unordered_map<long long, long long> umll;
+typedef unordered_map<char, long long> umci;
+typedef vector<pair<int, int>> vpi;
+typedef vector<int> vi;
+typedef long long ll;
+typedef vector<long long> vll;
+typedef unordered_map<int , bool> umib;
+#define sum(v) accumulate(v.begin(), v.end(), 0)
+#define endl '\n'
+#define f0(i, n) for(long long i = 0; i < n; i++)
+#define f1(i, n) for(long long i = 1; i < n; i++)
+#define as(v) sort(v.begin(), v.end())
+#define all(x) (x).begin(), (x).end()
+#define pb push_back
+template<class T> umll frequency(vector<T> &v) {umll freq;for(auto &x:v) freq[x]++; return freq;}
+template<class T> umci S_frequency(vector<T> &v) {umci freq;for(auto &x:v) freq[x]++; return freq;}
+template <class T> void input(vector<T> &v){for(auto &x:v)cin>>x;}
+ll power(ll x, ll y){ ll res = 1; while (y > 0){ if (y & 1) res = (ll)(res*x); y = y>>1; x = (ll)(x*x); } return res; }
+void pvll(const vector<long long> &arr){for(auto it : arr){cout << it << " ";}cout << endl;}
+void pvi(const vector<int> &arr){for(auto it : arr){cout << it << " ";}cout << endl;}
+static const int MAXN = 100000;
+
+
+void solve(){
+
+//-------------INPUT-------------
+    ll n; 
+    cin >> n;
+    ll k;
+    cin >> k;
+    vll v(n);
+    input(v);
+    //We need to count the no of pairs such that their product makes some number k- power means the product of the two numbers should have
+    //factors frequency that is divisible by k
+    vll spf(MAXN+1);
+    for(ll i=1;i<=MAXN;i++)
+    {
+        spf[i]=i;
+    }
+    for(ll i=2;i*i<=MAXN;i++)
+    {
+        if(spf[i] == i)
+        {
+            for(ll j=i*i;j<=MAXN;j+=i){
+                if(spf[j]==j)
+                {
+                    spf[j] = i;
+                }
+            }
+        }
+    }
+    map<vector<pair<ll,ll>>,ll>freq;
+
+    ll ans=0;
+
+    for(auto &x : v)
+    {
+        map<ll,ll>factor;
+        while(x >1)
+        {
+            ll p = spf[x];
+            ll cnt = 0;
+            while(x%p ==0)
+            {
+                x = x/p;
+                cnt++;
+            }
+            cnt=cnt%k;
+            if(cnt>0)
+            {
+                factor[p] = cnt; 
+            }
+        }
+        vector<pair<ll,ll>>key,need;
+    
+        for(auto &it : factor)
+        {
+            key.push_back(it);
+            ll req = (k-it.second)%k;
+            if(req>0)
+            {
+                need.push_back({it.first,req});
+            }
+        }
+        ans+=freq[need];
+        freq[key]++;
+    }
+    cout << ans << endl;
+
+
+    //link submission :- https://codeforces.com/contest/1225/submission/355215715
+
+//-------------CODE--------------
+    
+
+
+}
+
+
+int main(){
+    //int tt; cin >> tt; while(tt--)
+{solve();};
+}