Merge pull request #238 from suzzzal/patch-3

day01sol02

Author: Dheeraj-06

Problems/Mathematics/Day-01/sol/Sujal_Kshatri/day01sol02.cpp

@@ -0,0 +1,80 @@
+/*
+submission link - https://codeforces.com/contest/1225/submission/355299661
+n = number of elements (≤ 100000)
+A = maximum value of ai (≤ 100000)
+
+Time Complexity:
+- Sieve creation: O(A log log A)
+- Factorization of all elements: O(n log A)
+Overall: O(n log A)
+
+Space Complexity:
+- SPF array + maps: O(A + n)
+
+Approach:
+We want to count pairs of numbers whose product is a perfect k-th power.
+
+For each number:
+1. Factorize it using the Smallest Prime Factor (SPF) sieve.
+2. Keep prime exponents modulo k to form a signature.
+3. Compute the required signature needed to complete k.
+4. Count how many times this required signature appeared before.
+5. Add it to the answer and store the current signature.
+*/
+
+#include<bits/stdc++.h>
+using namespace std;
+
+int main(){
+    int size,power;
+    cin>>size>>power;
+
+    vector<int>arr(size);
+    for(int i=0;i<size;i++)cin>>arr[i];
+
+    const int limit=100000;
+    vector<int>spf(limit+1);
+    for(int i=1;i<=limit;i++)spf[i]=i;
+    for(int i=2;i*i<=limit;i++){
+        if(spf[i]==i){
+            for(int j=i*i;j<=limit;j+=i){
+                if(spf[j]==j)spf[j]=i;
+            }
+        }
+    }
+
+    map<vector<pair<int,int>>,long long>store;
+    long long result=0;
+
+    for(int val:arr){
+        map<int,int>count;
+        while(val>1){
+            int prime=spf[val];
+            int exp=0;
+            while(val%prime==0){
+                val/=prime;
+                exp++;
+            }
+            count[prime]+=exp;
+        }
+
+        vector<pair<int,int>>sign,need;
+        for(auto &it:count){
+            int prime=it.first;
+            int rem=it.second%power;
+            if(rem!=0){
+                sign.push_back({prime,rem});
+                need.push_back({prime,(power-rem)%power});
+            }
+        }
+
+        sort(sign.begin(),sign.end());
+        sort(need.begin(),need.end());
+
+        result+=store[need];
+        store[sign]++;
+    }
+
+    cout<<result<<"\n";
+    return 0;
+}