1203C - Common Divisors - CodeForces Solution


implementation math *1300

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C++ Code:

#include <cstdio>
typedef long long ll;

ll gcd (ll a, ll b){
    return (b == 0) ? a : gcd (b, a%b);
}

int main(){

    ll n; scanf("%lld", &n);
    ll g(0);
    for(ll p = 0; p < n; p++){
        ll x; scanf("%lld", &x);
        g = gcd(g, x);
    }
    ll cnt(0);
    for(ll p = 1; p * p <= g; p++){
        if(g % p){
            continue;
        }
        cnt += 1 + (p * p != g);
    }

    printf("%lld\n", cnt);

    return 0;
}


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