Codefroces-510D

Fox Ciel is playing a game. In this game there is an infinite long tape with cells indexed by integers (positive, negative and zero). At the beginning she is standing at the cell 0.

There are also n cards, each card has 2 attributes: length li and cost ci. If she pays ci dollars then she can apply i-th card. After applying i-th card she becomes able to make jumps of length li, i. e. from cell x to cell (x - li) or cell (x + li).

She wants to be able to jump to any cell on the tape (possibly, visiting some intermediate cells). For achieving this goal, she wants to buy some cards, paying as little money as possible.

If this is possible, calculate the minimal cost.

Input
The first line contains an integer n (1 ≤ n ≤ 300), number of cards.

The second line contains n numbers li (1 ≤ li ≤ 109), the jump lengths of cards.

The third line contains n numbers ci (1 ≤ ci ≤ 105), the costs of cards.

Output
If it is impossible to buy some cards and become able to jump to any cell, output -1. Otherwise output the minimal cost of buying such set of cards.

Examples
input

3
100 99 9900
1 1 1

output

2

input

5
10 20 30 40 50
1 1 1 1 1

output

-1

input

7
15015 10010 6006 4290 2730 2310 1
1 1 1 1 1 1 10

output

6

input

8
4264 4921 6321 6984 2316 8432 6120 1026
4264 4921 6321 6984 2316 8432 6120 1026

output

7237

Note
In first sample test, buying one card is not enough: for example, if you buy a card with length 100, you can’t jump to any cell whose index is not a multiple of 100. The best way is to buy first and second card, that will make you be able to jump to any cell.

In the second sample test, even if you buy all cards, you can’t jump to any cell whose index is not a multiple of 10, so you should output -1.

题目大意：有N张卡片，每张卡片有一个l表示可以通过这张卡片向前走l步或向后走l步，和一个c表示如果选这张卡片需要花费c金币，然后问我们可以最少花费多少钱购买卡片使得我们可以到达一个无穷长的磁带的任意一个地方。
解题思路：如果我们可以到达任意地方，那么就需要使得我们选择的卡片的GCD等于1，所以我们可以枚举加入卡片，在加入一张卡片之后，就将之前加入卡片之后得到的gcd与这张卡片的l求gcd，然后取该gcd的最小值，最后答案保存在gcd为1的空间里。
代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <bitset>
using namespace std;
#define int long long
const int maxn = 3e5 + 7;
const int inf = ((1ll*1)<<50);
int n, m;
int buf[110];
map<int, int> dp;
void add(int x,int v)
{
    if(dp.count(x)==0){
        dp[x] = v;
    }
    dp[x] = min(dp[x], v);
}
int a[400],c[400];
signed main()
{
    int n;
    cin>>n;
    for (int i = 1;i<=n;i++){
        cin >> a[i];
    }
    for (int j = 1;j<=n;j++){
        cin >> c[j];
        add(a[j], c[j]);
    }
    for (int i = 1; i <= n;i++){
        for(auto j:dp){
            int x = __gcd(a[i], j.first);
            add(x, j.second + c[i]);
        }
    }
    if(dp.count(1)==0){
        puts("-1");
    }
    else{
        printf("%lld\n", dp[1]);
    }
}