Codeforces Round #290 (Div. 2) D. Fox And Jumping GCD问题

D. Fox And Jumping

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

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.

题意是给出n个数，要求找出其中几个数，可以生成数轴上所有的数，其实，发现只要能生成1,就能生成所有的数，要生成1，则必然这n个数gcd为1

这样就转化成了gcd的问题！要求某几个数gcd是1，可以用dp思想，从头到尾枚举所有的数，因为数太大，所以生成的数用map set存，这样查询logn复杂度，总的复杂度为n*m*logn

#define INF			9000000000000000000
#define EPS			(double)1e-9
#define mod			1000000007
#define PI			3.14159265358979
//*******************************************************************************/
#endif
#define N 100005
#define MOD 1000000000000000007
int n,l[N],pri[N];
map<int,int> mymap;
map<int,int>::iterator it;
int main()
{
    while(S(n)!=EOF)
    {
        mymap.clear();
        FI(n){
            S(l[i]);
        }
        FI(n){
            S(pri[i]);
             if(mymap.count(l[i])){
                mymap[l[i]] = min(mymap[l[i]],pri[i]);
             }
             else
                mymap[l[i]]=pri[i];
        }
        FI(n){
            for(it=mymap.begin();it != mymap.end();it++){
                int t = it->first,val = it->second;
                int temp = gcd(t,l[i]);
                if(mymap.count(temp)){
                    mymap[temp] = min(mymap[temp],val + pri[i]);
                }
                else
                    mymap[temp] = val + pri[i];
            }
        }
        printf("%d\n",mymap.count(1)?mymap[1]:-1);
    }
    return 0;
}