题意：给出300个数，每个数有权值，求从中挑出k个数使得它们的gcd为1的最小代价。。。。
解法：因为每个数对应的素数并不是很多，因此dp的状态并不是很多。。。
#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 305
#define maxm 200005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
//head
map<pair<int, int>, int> dp;
int a[maxn];
int b[maxn];
int n;
void read()
{
scanf("%d", &n);
for(int i = 0; i < n; i++) scanf("%d", &a[i]);
for(int i = 0; i < n; i++) scanf("%d", &b[i]);
}
int dfs(int g, int now)
{
if(dp.count(mp(g, now))) return dp[mp(g, now)];
if(g == 1) return 0;
if(now == n) return INF;
int res = min(dfs(g, now + 1), dfs(__gcd(g, a[now]), now + 1) + b[now]);
return dp[mp(g, now)] = res;
}
void work()
{
int ans = 0;
for(int i = 0; i < n; i++) ans = __gcd(ans, a[i]);
if(ans != 1) printf("-1\n");
else printf("%d\n", dfs(0, 0));
}
int main()
{
read();
work();
return 0;
}
