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CodeForces 512B(区间dp)

D - Fox And Jumping

                      Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

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 applyingi-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 Input

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

区间dp
 1 //2016.8.6
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<map>
 5 #include<algorithm>
 6 
 7 using namespace std;
 8 
 9 int l[305], c[305];
10 map<int, int>dp1, dp2;
11 
12 int gcd(int a, int b)
13 {
14     return b==0?a:gcd(b, a%b);
15 }
16 
17 int main()
18 {
19     int n;
20     while(cin>>n)
21     {
22         dp1.clear();
23         for(int i = 1; i <= n; i++)
24           scanf("%d", &l[i]);
25         for(int i = 1; i <= n; i++)
26           scanf("%d", &c[i]);
27         dp1[0] = 0;
28         for(int i = 1; i <= n; i++)
29         {
30             map<int, int>::iterator it;
31             for(it = dp1.begin(); it != dp1.end(); it++)
32             {
33                 int g = gcd(it->first, l[i]);
34                 if(dp1.count(g))
35                   dp1[g] = min(it->second+c[i], dp1[g]);
36                 else dp1[g] = it->second+c[i];
37             }
38         }
39         if(dp1.count(1))cout<<dp1[1]<<endl;
40         else cout<<-1<<endl;
41     }
42 
43     return 0;
44 }

 

posted @ 2016-08-06 22:15  Penn000  阅读(264)  评论(0编辑  收藏  举报