D. Fox And Jumping
time limit per test
memory limit per test
input
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.
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
n (1 ≤ n ≤ 300), number of cards.
n numbers li (1 ≤ li ≤ 109), the jump lengths of cards.
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.
很明显，只要取的数gcd=1就行。
因此dp，f[i]表示gcd=i的最小费用
质数不多随便做。
PS：注意map中没赋值的情况
PS2：l[i]可能重复，小心覆盖
PS3：事先给h[1]赋值INF时，要考虑l[i]中本来就有1的情况
wa到残。。。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<queue>
#include<map>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (300+300+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
priority_queue<int> q;
map<int,int> h;
int gcd(int a,int b){if (b==0) return a;return gcd(b,a%b);}
int n,l[MAXN],c[MAXN];
int main()
{
// freopen("Jumping.in","r",stdin);
// freopen(".out","w",stdout);
cin>>n;
For(i,n) scanf("%d",&l[i]);
For(i,n) scanf("%d",&c[i]);
For(i,n)
{
if (h.find(l[i])!=h.end())
{
h[l[i]]=min(h[l[i]],c[i]);
continue;
}
h[l[i]]=c[i],q.push(l[i]);
}
while (!q.empty())
{
int t=q.top();
q.pop();
For(i,n)
{
int t2=gcd(t,l[i]),c2=h[t]+h[l[i]];
// cout<<t2<<' '<<c2<<endl;
if (h.find(t2)==h.end()||h[t2]>c2)
{
h[t2]=c2;
q.push(t2);
}
}
}
if (h.find(1)==h.end()) cout<<"-1\n";
else cout<<h[1]<<endl;
return 0;
}
