Add solution2

This file implements a solution for a Codeforces problem using various mathematical utilities such as GCD, LCM, modular arithmetic, and prime factorization. It includes functions for binary exponentiation, factorial calculation, and counting specific combinations based on prime factors.

Author: PHOX-9

Problems/Mathematics/Day-01/sol/Ibrahim Raza Beg/Solution2.cpp

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+//https://codeforces.com/contest/1225/submission/355266023
+
+#include <bits/stdc++.h>
+using namespace std;
+
+#define int long long
+#define ll long long
+#define ull unsigned long long
+#define ld long double
+#define pii pair<int,int>
+#define vi vector<int>
+#define vvi vector<vector<int>>
+#define vpi vector<pair<int,int>>
+#define all(x) (x).begin(), (x).end()
+
+
+const long long MAXN = 2e6;
+const long long MOD = 1e9 + 7;
+const int INF = 1e18;
+const ld EPS = 1e-9;
+
+
+#define fastio ios::sync_with_stdio(false); cin.tie(nullptr);
+
+#ifndef ONLINE_JUDGE
+    #define debug(x) cerr << #x << " = " << x << "\n";
+#else
+    #define debug(x)
+#endif
+
+// Utility functions
+int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
+int lcm(int a, int b) { return a / gcd(a, b) * b; }
+int mod_add(int a, int b, int m=MOD) { return ((a % m) + (b % m) + m) % m; }
+int mod_sub(int a, int b, int m=MOD) { return ((a % m) - (b % m) + m) % m; }
+int mod_mul(int a, int b, int m=MOD) { return ((a % m) * (b % m)) % m; }
+
+//Binary Exponentiation
+int binexp(int a, int b, int m=MOD) {
+    int res = 1;
+    a %= m;
+    while(b > 0) {
+        if(b & 1) res = (res * a) % m;
+        a = (a * a) % m;
+        b >>= 1;
+    }
+    return res;
+}
+
+int mod_inv(int a, int m=MOD) {
+    return binexp(a, m - 2, m);
+}
+
+long long fac[MAXN + 1];
+long long inv[MAXN + 1];
+
+long long exp(long long x, long long n, long long m) {
+  x %= m;
+  long long res = 1;
+  while (n > 0) {
+    if (n % 2 == 1) { res = res * x % m; }
+    x = x * x % m;
+    n /= 2;
+  }
+  return res;
+}
+
+void factorial() {
+  fac[0] = 1;
+  for (long long i = 1; i <= MAXN; i++) { fac[i] = fac[i - 1] * i % MOD; }
+}
+
+void inverses() {
+  inv[MAXN] = exp(fac[MAXN], MOD - 2, MOD);
+  for (long long i = MAXN; i >= 1; i--) { inv[i - 1] = inv[i] * i % MOD; }
+}
+
+long long choose(long long n, long long r) {
+  if (r > n)return 0ll;
+  return (fac[n] * inv[r] % MOD * inv[n - r] % MOD) % MOD;
+}
+
+long long catalan(long long n) {
+  return (exp(n + 1, MOD - 2, MOD) % MOD * choose(2 * n, n) % MOD) % MOD;
+}
+
+//Returns all prime numbers <=N
+vector<int> sieve(int n) {
+    vector<bool> prime(n + 1, true);
+    for (int p=2;p*p<=n;p++) {
+        if (prime[p] == true) {
+            
+            for (int i=p*p;i<=n;i+=p)
+                prime[i] = false;
+        }
+    }
+    
+    vector<int> res;
+    for (int p = 2; p <= n; p++){
+        if (prime[p]){ 
+            res.push_back(p);
+        }
+    }
+    return res;
+}
+
+//Prime factors of all numbers upto N
+vector<vector<int>> primefactors(int N)
+{
+    vvi pfac(N + 1);
+        for (int i=2;i<=N;i++){
+        if (!pfac[i].empty())
+            continue;
+
+        for (int j = i; j <= N; j += i)
+            pfac[j].push_back(i);
+    }
+            
+    return pfac;
+}
+
+//smallest prime factor of a number
+int spf(int n) {
+    if (n % 2 == 0) return 2;
+    for (int i = 3; i * i <= n; i += 2) {
+        if (n % i == 0) return i;
+    }
+    return n;
+}
+
+int power2(int p)
+{
+    int v=1ll<<p;
+    return v;
+}
+
+//Sort functions
+
+void sort(vector<int>&a)
+{
+    sort(a.begin(),a.end());
+}
+
+void psort(vector<pair<int,int>>&a)
+{
+    sort(a.begin(),a.end());
+}
+
+void rsort(vector<int>&a)
+{
+    sort(a.rbegin(),a.rend());
+}
+
+void rpsort(vector<pair<int,int>>&a)
+{
+    sort(a.rbegin(),a.rend());
+}
+
+//2-D Vector Declaration: 
+//vvi mat(n,vi(m));     // nxm matrix, all initialized to 0
+
+
+/***************Code***************/
+static const int len=100001;
+vi pf(len,0);
+void spfv()
+{
+    int i,j,l=1e5;
+    for(i=2;i<=l;i++)
+    {
+        if(pf[i]==0)
+        {
+            for(j=i;j<=l;j+=i)
+            {
+                if(pf[j]==0)
+                pf[j]=i;
+            }
+        }
+    }
+}
+
+void solve() {
+    int n,i,k;
+    cin>>n>>k;
+    vi a(n);
+    for(i=0;i<n;i++) 
+    cin>>a[i];
+
+    map<vpi,int> cnt;
+    int ans=0;
+
+    for(i=0;i<n;i++)
+    {
+        int v=a[i];
+        map<int,int> mp;
+        while(v>1)
+        {
+            int p=pf[v];
+            mp[p]++;
+            v=v/p;
+        }
+
+        vpi hv,req;
+        for(auto j:mp)
+        {
+            int p,r;
+            p=j.first;
+            r=j.second%k;
+            if(r!=0)
+        {
+            hv.push_back({p,r});
+            req.push_back({p,(k-r)%k});
+        }
+        }
+        ans+=cnt[req];
+        cnt[hv]++;
+    }
+        cout<<ans<<endl;
+}
+
+
+signed main() {
+    fastio;
+    int t = 1;
+    spfv();
+    while (t--) {
+        solve();
+    }
+    return 0;
+}