codefoces--510D. Fox And Jumping

3680 단어 동적 기획
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.
제목 대의: n개의 점프 거리와 이 거리를 점프하는 데 쓸 돈을 주고 이미 샀으면 돈을 쓰지 않아도 된다. 최소 얼마를 쓰느냐고 물으면 임의의 위치로 점프할 수 있다.그렇지 않으면 출력 -1
임의의 위치로 뛸 수 있다. 즉, 뛸 거리는 +1, 또는 -1을 형성하고 그래야만 모든 위치로 뛸 수 있다.
그러면 가장 적게 드는 몇 개의 수를 찾아야 한다. 그들의 최대 공배수는 1이다. 그러면 +1이나 -1을 조합할 수 있다.
dp를 사용하여 1에 도달하는 최소 비용을 찾아낸다.
 
 
#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
", k) ; else printf("-1
") ; return 0 ; }

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