**E**uler **T**otient **F**unction (ETF) Φ(n) for an input n is count of numbers in {1, 2, 3, …, n} that are relatively prime to n, i.e., the numbers whose GCD (Greatest Common Divisor) with n is 1.

Φ(5) = 4 gcd(1, 5) is 1, gcd(2, 5) is 1, gcd(3, 5) is 1 and gcd(4, 5) is 1 Φ(6) = 2 gcd(1, 6) is 1 and gcd(5, 6) is 1,

We have discussed different methods to compute Euler Totient function that work well for single input. In problems where we have to call Euler’s Totient Function many times like 10^5 times, simple solution will result in TLE(Time limit Exceeded). The idea is to use Sieve of Eratosthenes.

Find all prime numbers upto maximum limit say 10^5 using Sieve of Eratosthenes.

To compute Φ(n), we do following.

- Initialize result as n.
- Iterate through all primes smaller than or equal to square root of n (This is where it is different from simple methods. Instead of iterating through all numbers less than or equal to square root, we iterate through only primes). Let the current prime number be p. We check if p divides n, if yes, we remove all occurrences of p from n by repeatedly dividing it with n. We also reduce our result by n/p (these many numbers will not have GCD as 1 with n).
- Finally we return result.

// C++ program to efficiently compute values // of euler totient function for multiple inputs. #include <bits/stdc++.h> using namespace std; #define ll long long const int MAX = 100001; // Stores prime numbers upto MAX - 1 values vector<ll> p; // Finds prime numbers upto MAX-1 and // stores them in vector p void sieve() { ll isPrime[MAX+1]; for (ll i = 2; i<= MAX; i++) { // if prime[i] is not marked before if (isPrime[i] == 0) { // fill vector for every newly // encountered prime p.push_back(i); // run this loop till square root of MAX, // mark the index i * j as not prime for (ll j = 2; i * j<= MAX; j++) isPrime[i * j]= 1; } } } // function to find totient of n ll phi(ll n) { ll res = n; // this loop runs sqrt(n / ln(n)) times for (ll i=0; p[i]*p[i] <= n; i++) { if (n % p[i]== 0) { // subtract multiples of p[i] from r res -= (res / p[i]); // Remove all occurrences of p[i] in n while (n % p[i]== 0) n /= p[i]; } } // when n has prime factor greater // than sqrt(n) if (n > 1) res -= (res / n); return res; } // Driver code int main() { // preprocess all prime numbers upto 10 ^ 5 sieve(); cout << phi(11) << "\n"; cout << phi(21) << "\n"; cout << phi(31) << "\n"; cout << phi(41) << "\n"; cout << phi(51) << "\n"; cout << phi(61) << "\n"; cout << phi(91) << "\n"; cout << phi(101) << "\n"; return 0; }

**Output:**

10 12 30 40 32 60 72 100

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