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 alsocards, each card has 2 attributes: lengthand costci. If she paysdollars then she can apply-th card she becomes able to make jumps of lengthli, i. e. from cellto cellor 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(1≤n≤300), number of cards.
The second line containsnumbers(1≤li≤109), the jump lengths of cards.
The third line containsnumbers(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
3100 99 99001 1 1
output
2
input
510 20 30 40 501 1 1 1 1
output
-1
input
715015 10010 6006 4290 2730 2310 11 1 1 1 1 1 10
output
6
input
84264 4921 6321 6984 2316 8432 6120 10264264 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：给出n个跳的距离，和跳该距离要花的钱，如果已经买过，就不用花钱了，问最少花多少钱，可以跳到任意位置。否则输出-1
能跳到任意位置，也就是说，它要跳的距离会形成一个+1，，或-1，只有这样才能跳到所有的位置。
那么也就是说，要找到花费最少的几个数，他们的最大公倍数为1，这样就能组合出+1或-1。
使用dp找出到达1的最小花费。
#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <map>using namespace std ;int a[400] , c[400] ;map <int,int> Map ;map <int,int>::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] ) ;elseMap[ 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] ) ;elseMap[ m ] = iter->second + c[i] ;}}k = Map[1] ;if( k )printf("%d\n", k) ;elseprintf("-1\n") ;return 0 ;}
但我自信，我能点亮心烛，化解心灵的困惑。
