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

最新推荐文章于 2022-12-16 16:19:21 发布

原创最新推荐文章于 2022-12-16 16:19:21 发布 · 325 阅读

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本文介绍了一种通过购买卡片使狐狸能够到达任意位置的游戏策略。关键在于找到一组卡片，其长度值的组合能覆盖所有可能的位置，特别是包含长度为1的情况。文章详细解释了算法流程，并提供了完整的代码实现。

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原题链接

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

只有找到一组数他们加减之后为1即可，那么这组数的最大公约数必须为1

#include <bits/stdc++.h>
#define maxn 100005
#define MOD 1000000007
using namespace std;
typedef long long ll;

int k1[305], k2[305];
map<int, int> m;
int gcd(int a, int b){
	return b ? gcd(b, a % b) : a;
}
int main(){
//	freopen("in.txt", "r", stdin);
	int n;
	scanf("%d", &n);
	for(int i = 0; i < n; i++)
	 scanf("%d", k1+i);
	for(int i = 0; i < n; i++)
	 scanf("%d", k2+i);
	map<int, int> ::iterator iter;
	for(int i = 0; i < n; i++){
		if(m.count(k1[i]))
		  m[k1[i]] = min(m[k1[i]], k2[i]);
		else
		 m[k1[i]] = k2[i];
		iter = m.begin();
		for(; iter != m.end(); ++iter){
			int a = gcd(iter->first, k1[i]);
			if(m.count(a))
			 m[a] = min(m[a], iter->second+k2[i]);
			else
			 m[a] = iter->second + k2[i];
		}
	}
	if(m.count(1) == 0)
	 puts("-1");
	else
	 printf("%d\n", m[1]);
	return 0;
}