Codeforces Round #290 (Div. 1)B. Fox And Jumping
来源:互联网 发布:如何解析json数据 编辑:程序博客网 时间:2024/05/29 14:25
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: lengthli and costci. If she paysci dollars then she can applyi-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
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;}
- Codeforces Round #290 (Div. 1)B. Fox And Jumping
- Codeforces Round #290 (Div. 1) B. Fox And Jumping
- Codeforces Round #290 (Div. 1) C Fox And Jumping
- Codeforces Round #290 (Div. 2) D Fox And Jumping
- Codeforces Round #290 (Div. 2)D. Fox And Jumping
- Codeforces Round #290 (Div. 2)-D. Fox And Jumping
- Codeforces Round #290 (Div. 2) C. Fox And Names && D. Fox And Jumping
- Codeforces Round #290 (Div. 2)D - Fox And Jumping——数论gcd=1
- Codeforces Round #290 D. Fox And Jumping
- Codeforces Round #290(Div.2) B.Fox And Two Dots
- B. Fox And Two Dots( Codeforces Round #290 (Div. 2))
- B. Fox And Two Dots Codeforces Round #290 (Div. 2)
- Codeforces Round #290 (Div. 2) B. Fox And Two Dots
- Codeforces Round #290 (Div. 2) B. Fox And Two Dots
- Codeforces Round #290 (Div. 2) B. Fox And Two Dots
- Codeforces Round #228 (Div. 1)-B-Fox and Minimal path
- Codeforces Round #228 (Div. 1) B. Fox and Minimal path
- Codeforces Round #290 (Div. 2) D. Fox And Jumping GCD问题
- Cocos2d-X 3.4版-地图无限滚动与边缘检测《赵云要格斗》
- java 中的 '\0'
- block
- HDU 4281 Judges' response 状态压缩 01背包 MTSP
- 笛卡儿积的Java算法实现
- Codeforces Round #290 (Div. 1)B. Fox And Jumping
- linux 进程调度策略
- javascript对url进行encode的两种方式
- POJ 3294 (UVA 11107) Life Forms 后缀数组
- Java线程面试题 Top 50
- hdu 2846 Repository 字典树的一种变形
- Func和Action,委托与lambda表达式,一看就知道
- iOS开发_剪贴板操作_复制粘贴功能
- 猴子爬山的问题