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Hasirama don't feel bad. see the success rate. here is a hint if you are still struggling: Find a primitive root first. Then the powers modulo p of that primitive root generates all the numbers from 0 to 1. But when does it generate another primitive root? See if you can detect the pattern
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subhajit104 #include <bits/stdc++.h> using namespace std; #define ll long long int ll n , s_primitive,number; ll power(ll a, ll b, ll mod){ if(b == 0) return 1; ll temp = power(a,b>>1,mod)%mod; temp = (temp*temp)%mod; if(b&1) temp = temp*a%mod; return temp; }
sgrebenkin You made my day. My solution was not passing a couple of tests because of TLE.
My implementation of this function was a problem.
morganalice725 You have given very best and more relevant code which I have created for this question's reply but after see your answer I can't share it because you have already give it here. But I want to say, you are doing good work, I am student of an academic course and sometime teacher gives me an uk cipd assignment topic related to coding if I get stuck in any point so discuss that with you and I am sure you will solve that issue quickly.
brucemcgreggor1 A method of solving a problem which depends on the smaller occurrences of the same problem known as Recursion. Most of the College Life Network programming languages must allow this method. A function called itself within its own code.
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subhajit104 algo: 1. find all the unique prime factor of the n-1.
ll number = n-1 ; vector<ll> fact; for(ll i = 2 ; i*i <= number ; i++){ // O(n) if( number%i == 0 ) fact.push_back(i); while(number%i == 0) number /= i; } if(number > 2) fact.push_back(number);
- then find the smallest primitive root using the giving link in the editorial.
ll s_primitive; for(ll i = 2 ; i < n ; i++){ bool flag = true; for(auto prime: fact){ if(power(i,n/prime,n) == 1){ flag = false; break; } } if(flag){ s_primitive = i; break; } } cout << s_primitive << " ";
- find the number of the primitove roots. We know that any prime p has ϕ(p−1) primitive roots. We also know that the prime power decomposition of p - 1 can be written as: p−1=p1^(e1),p2^(e2)..., and we then know that ϕ(p−1)=p1^(e1−1)(p1−1)p2^(e2−1)(p2−1)...pk^(ek−1)*(pk−1)
n--; for(auto fa:fact) n = n/fa*(fa-1); cout << n << endl;
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weskerhluffy This video explains very well the primitive root topic and how to calculate the number of primitive roots https://www.youtube.com/watch?v=IviYLdqmIdw&t=6s . P.S.: There are a lot of spam in this discussion
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