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codefoces--510D. Fox And Jumping

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codefoces--510D. Fox And Jumping

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 l_i and cost c_i. If she pays c_i dollars then she can apply i-th card. After applying i-th card she becomes able to make jumps of length l_i, i. e. from cell x to cell (x?-?l_i) or cell (x?+?l_i).

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 l_i (1?≤?l_i?≤?10⁹), the jump lengths of cards.

The third line contains n numbers c_i (1?≤?c_i?≤?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

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

input

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

output

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.

題目大意：給出n個跳的距離，和跳該距離要花的錢，如果已經買過，就不用花錢了，問最少花多少錢，可以跳到任意位置。否則輸出-1

能跳到任意位置，也就是說，它要跳的距離會形成一個+1，或-1，只有這樣才能跳到所有的位置。

那麼也就是說，要找到花費最少的幾個數，他們的最大公倍數為1，這樣就能組合出+1或-1。

使用dp找出到達1的最小花費。

#include 
#include 
#include 
#include 
#include 
using namespace std ;
int a[400] , c[400] ;
map  Map ;
map ::iterator iter ;
int gcd(int a,int b)
{
    return b == 0 ? a : gcd(b,a%b) ;
}
int main()
{
    int i , n , m , k ;
    Map.clear() ;
    scanf("%d", &n) ;
    for(i = 0 ; i < n ; i++)
        scanf("%d", &a[i]) ;
    for(i = 0 ; i < n ; i++)
        scanf("%d", &c[i]) ;
    for(i = 0 ; i < n ; i++)
    {
        k = Map[ a[i] ] ;
        if( k )
            Map[ a[i] ] = min( k , c[i] ) ;
        else
            Map[ a[i] ] = c[i] ;
        for( iter = Map.begin() ; iter != Map.end() ; iter++)
        {
            m = gcd(a[i],iter->first) ;
            k = Map[ m ] ;
            if( k )
                Map[ m ] = min( k , iter->second + c[i] ) ;
            else
                Map[ m ] = iter->second + c[i] ;
        }
    }
    k = Map[1] ;
    if( k )
        printf("%d\n", k) ;
    else
        printf("-1\n") ;
    return 0 ;
}