Codeforces Round #290 (Div. 2)-D. Fox And Jumping
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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.
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.
3100 99 99001 1 1
2
510 20 30 40 501 1 1 1 1
-1
715015 10010 6006 4290 2730 2310 11 1 1 1 1 1 10
6
84264 4921 6321 6984 2316 8432 6120 10264264 4921 6321 6984 2316 8432 6120 1026
7237
#include <bits/stdc++.h>#define maxn 100005#define MOD 1000000007using 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;}
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