通过质因数分解可以把很大的数唯一分解为有限个质数的乘积,能够很好的比较两个数。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<queue>
#include<vector>
#include<string>
#include<fstream>
using namespace std;
#define rep(i, a, n) for(int i = a; i <= n; ++ i)
#define per(i, a, n) for(int i = n; i >= a; -- i)
typedef long long ll;
const int N = 2e5+105;
const int mod = 998244353;
const double Pi = acos(- 1.0);
const ll INF = 1e12;
const int G = 3, Gi = 332748118;
ll qpow(ll a, ll b) { ll res = 1; while(b){ if(b & 1) res = (res * a) % mod; a = (a * a) % mod; b >>= 1;} return res; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
//
ll n, m1, m2;
ll p[N], c[N], m, res;
map<ll, ll> mp;
vector<ll> base;
void div(ll x){
m = 0;
for(ll i = 2; i * i <= x; ++ i){
if(x % i == 0){
p[++ m] = i;
c[m] = 0;
while(x % i == 0) c[m] ++, x /= i;
}
}
if(x > 1) p[++ m] = x, c[m] = 1;
}
ll solve(ll x){
div(x);
if(m < (ll)base.size()) return INF;
ll l = 1, r = 0, num = 0, temp = 0;
while(r < base.size()){
num = 0;
ll pp = base[r], cc = mp[base[r]];
if(p[l] == pp){
if(cc <= c[l]) num = 0;
else num = (cc + c[l] - 1) / c[l];
temp = max(temp, num);
l ++;
r ++;
}
else if(p[l] < pp){
l ++;
if(l > m){
return INF;
}
}
else{
l ++;
if(l > m){
return INF;
}
}
}
return temp;
}
int main()
{
scanf("%lld",&n);
scanf("%lld%lld",&m1,&m2);
if(m1 == 1){
printf("0
");
return 0;
}
div(m1);
rep(i,1,m){
mp[p[i]] = m2 * c[i];
base.push_back(p[i]);
}
sort(base.begin(),base.end());
res = INF;
rep(i,1,n){
ll s;
scanf("%lld",&s);
res = min(res, solve(s));
}
if(res == INF) printf("-1
");
else printf("%lld
",res);
return 0;
}