Merge pull request #103 from Abhay-S-R/feature/divisor_analysis

Feat: Add solution for Divisor analysis

Author: ks-iitjmu

7_mathematics/divisor_analysis.cpp

@@ -1,14 +1,89 @@
-// Kunal Sharma IIT Jammu
 #include <bits/stdc++.h>
 using namespace std;
 
 typedef long long ll;
+#define mod 1000000007
+
+ll aPowBMod(ll base, ll exp)
+{
+    ll res = 1;
+    base %= mod;
+    while (exp > 0)
+    {
+        if (exp % 2 != 0)
+            res = (res * base) % mod;
+        base = (base * base) % mod;
+        exp /= 2;
+    }
+    return res;
+}
 
 int main()
 {
     ios::sync_with_stdio(0);
     cin.tie(0);
     cout.tie(0);
-    
+
+    ll n;
+    cin >> n;
+    vector<pair<ll, ll>> values(n);
+    for (int i = 0; i < n; i++)
+    {
+        cin >> values[i].first >> values[i].second;
+    }
+
+    ll numOfDivisors = 1;
+    ll numOfDivisors_exp = 1;
+    bool is_tao_even = false;
+    for (auto &pr : values)
+    {
+        ll exp = pr.second;
+        if((exp+1) % 2 == 0) is_tao_even = true;
+        numOfDivisors = (numOfDivisors * (exp + 1)) % mod;
+        numOfDivisors_exp = (numOfDivisors_exp * (exp + 1)) % (mod-1);
+    }
+
+    ll sum = 1;
+    for (auto &[p, e] : values)
+    {
+        ll num = (aPowBMod(p, e + 1) - 1 + mod) % mod; // To prevent negative, if aPowBMod(p, e+1) == 0
+        ll dem_inv = aPowBMod(p - 1, mod - 2);
+        ll complete = (num * dem_inv) % mod;
+        sum = (sum * complete) % mod;
+    }
+
+    ll half_tao = 1;
+    if(is_tao_even){
+        bool flag = false;
+        for(auto &pr : values){
+            ll exp = pr.second;
+            if(!flag && (exp+1) % 2 ==0){
+                exp/=2;
+                flag = true;
+            }
+            half_tao = (half_tao * (exp+1)) % (mod-1);
+        }
+    }
+
+    ll prod = 1;
+    ll final_exp;
+    for (auto &[p, e] : values)
+    {
+        if (e % 2 == 0)
+        {
+            ll e_half =e / 2;
+            final_exp = (e_half * numOfDivisors_exp) % (mod - 1);
+        }
+        else
+        {
+            ll e_mod = e % (mod-1);
+            final_exp = (half_tao * e_mod) % (mod - 1);
+        }
+        ll term = aPowBMod(p, final_exp);
+        prod = (prod * term) % mod;
+    }
+
+    cout<<numOfDivisors<<" "<<sum<<" "<<prod<<'\n';
+
     return 0;
 }