Merge pull request #105 from hardik-rana02/hardik

Author: zonedoutdg

Problems/Mathematics/Day-01/sol/Hardik Singh Rana/Solution2.cpp

@@ -0,0 +1,78 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+#define ll long long
+#define pb push_back
+#define f(i,a,b) for(ll i=a;i<b;i++)
+#define fb(i,a,b) for(ll i=a;i>=b;i--)
+#define vi vector<int>
+#define vl vector<ll>
+#define pii pair<int,int>
+#define pll pair<ll,ll>
+#define vpii vector<pii>
+#define vpll vector<pll>
+#define all(x) (x).begin(), (x).end()
+#define fast_io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
+
+const ll MOD = 1e9+7;
+const ll INF = 1e18;
+/*
+We need to count the no of pairs such that their product makes some number's k-th power-this means the product of the two numbers should have
+factors frequency that is divisible by k. For every number first factorize it using spf(also store its factors mod k) and also store how much
+factor more it needs to make the k-th power add this for all numbers.
+
+*/
+
+/* Submission Link: https://codeforces.com/contest/1225/submission/355183203 */
+void solve() {
+    int n,k;
+    cin>>n>>k;
+    vi a(n);
+    f(i,0,n) cin>>a[i];
+    vi spf(1e5+1);
+    f(i,1,1e5+1) spf[i]=i;
+    for(int i=2;i*i<=1e5;i++){
+        if(spf[i]==i){
+            for(int j=i*i;j<=1e5;j+=i){
+                if(spf[j]==j){
+                    spf[j]=i;
+                }
+            }
+        }
+    }
+    map<vpii,ll> cnt;
+    ll ans=0;
+    for(int x:a){
+        map<int,int> factors;
+        while(x>1){
+            int p=spf[x];
+            int c=0;
+            while(x%p==0){
+                x/=p;
+                c++;
+            }
+            c%=k;
+            if(c>0)
+                factors[p]=c;
+        }
+        vpii key,need;
+        for(auto &it:factors){
+            key.pb(it);
+            int y=(k-it.second)%k;
+            if(y>0){
+                need.pb({it.first,y});
+            }
+        }
+        ans+=cnt[need];
+        cnt[key]++;
+    }
+    cout<<ans<<endl;
+}
+
+int main() {
+    fast_io;
+    int t = 1;
+    //cin >> t;
+    while(t--) solve();
+    return 0;
+}