2 seconds
256 megabytes
standard input
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.
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.
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.
3 100 99 9900 1 1 1
2
5 10 20 30 40 50 1 1 1 1 1
-1
7 15015 10010 6006 4290 2730 2310 1 1 1 1 1 1 1 10
6
8 4264 4921 6321 6984 2316 8432 6120 1026 4264 4921 6321 6984 2316 8432 6120 1026
7237
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.
观察后可以发现,如果有a和b两张卡,那么我们可以每次都跳gcd(a, b)
如果要任意格子都能到,说明gcd == 1
于是问题转换成了得到k个数,它们的gcd==1 的最小代价
dp[i] 表示得到i的最小代价
数据有点大,改成用map来转移即可
/*************************************************************************
> File Name: cf289b.cpp
> Author: ALex
> Mail: zchao1995@gmail.com
> Created Time: 2015年02月03日 星期二 02时05分37秒
************************************************************************/
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
map <int, int> dp;
const int N = 330;
int l[N], c[N];
int gcd (int a, int b)
{
return b == 0 ? a : gcd (b, a % b);
}
int main ()
{
int n;
while (~scanf("%d", &n))
{
map <int, int> :: iterator it;
for (int i = 1; i <= n; ++i)
{
scanf("%d", &l[i]);
}
for (int i = 1; i <= n; ++i)
{
scanf("%d", &c[i]);
}
dp.clear();
dp[0] = 0;
for (int i = 1; i <= n; ++i)
{
for (it = dp.begin(); it != dp.end(); ++it)
{
int t = gcd (l[i], it -> first);
if (dp.count (t))
{
dp[t] = min (dp[t], it -> second + c[i]);
}
else
{
dp[t] = it -> second + c[i];
}
}
}
if (!dp.count (1))
{
printf("-1\n");
}
else
{
printf("%d\n", dp[1]);
}
}
return 0;
}