2019 杭电多校 9 1006

题目链接:HDU 6685

比赛链接:2019 Multi-University Training Contest 9

Problem Description

Rikka hates coins, and she used to never carry any coins with her. These days, Rikka is doing her summer internship abroad. Without mobile payment, Rikka has to face strange prices of commodities, and as a result of always using paper currency, she has to face mountainous coins on here table.

In the local currency system, there are \(4\) kinds of coins: \(10\) cents, \(20\) cents, \(50\) cents and \(1\) dollar. Up to now, Rikka has gained at least \(10^{100}\) coins for each kind.

Now, Rikka is going to have dinner in the canteen, and she decides to pay the bill only with coins. There are \(n\) different combos in the canteen and the price of the \(i\)th is \(w_i\) cents. Rikka would like to choose one combo as dinner but she has not decided to choose which one yet. Therefore, she wants to take some coins so that whichever she chooses, she can always pay the bill without receiving any change.

Since Rikka hates coins, she wants to carry as few coins as possible with her. As it is generally known that Rikka is not good at math, she wants you to help her make the decision.

Input

The first line of the input contains a single integer \(T(1\le T\le 500)\), the number of test cases.

For each test case, the first line contains a single integer \(n(1\le n\le 100)\), the number of combos sold in the canteen.

The second line contains \(n\) positive integers \(w_1,…,w_n(1\le w_i\le 10^9)\), which represents the prices.

Output

For each test case, output a single line with a single integer which represents the minimum number of coins. If there is no valid solution, output \(−1\).

Hint

In the first test case, one optimal solution is to bring one coin of \(10\) cents and two coins of \(20\) cents.

In the second test case, one optimal solution is to bring \(5\) coins of one dollar.

Sample Input

3
5
10 20 30 40 50
5
100 200 300 400 500
1
1

Sample Output

3
5
-1

Solution

题意

给出 \(n\) 种物品的价格,现在要从无限枚 \(10\)元,\(20\)元,\(50\)元,\(100\)元的硬币中选出最少的硬币,满足能购买任何一种物品都不用找零。

题解

显然如果个位不为零时没有可行方案。

接下来考虑可行方案的求解。

\(10\) 分的硬币多只会用一个,如果用了两个,直接替换成一个 \(10\) 分一个 \(20\) 分一定不亏。

\(20\) 分的硬币多只会用三个,如果用了四个,直接替换成一个 \(10\) 分两个 \(20\) 分一个 \(50\) 分一定不亏。

\(50\) 分的硬币多只会用一个,如果用了两个,直接替换成一个 \(50\) 分和一个一元一定不亏。

因此,直接暴力枚举 \(10\), \(20\), \(50\) 的硬币的数量即可,整百的部分用一元硬币填充。

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const int maxn = 100 + 5; int w[maxn]; bool judge(int n, int a, int b, int c) {
for(int i = 0; i <= a; ++i) {
for(int j = 0; j <= b; ++j) {
for(int k = 0; k <= c; ++k) {
if(i * 50 + j * 20 + k * 10 == n) {
return true;
}
}
}
}
return false;
} int main() {
std::ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int T;
cin >> T;
while (T--) {
int n;
cin >> n;
int flag = 0;
for(int i = 0; i < n; ++i) {
cin >> w[i];
if(w[i] % 10) {
flag = 1;
}
}
if(flag) {
cout << -1 << endl;
continue;
}
int ans = inf;
for(int j = 0; j <= 1; ++j) {
for(int k = 0; k <= 3; ++k) {
for(int l = 0; l <= 1; ++l) {
int flag = 1;
int cnt = 0;
for(int i = 0; i < n; ++i) {
if(w[i] < 100) {
if(judge(w[i], j, k, l)) {
continue;
} else {
flag = 0;
break;
}
} else {
if(judge(w[i] % 100 + 100, j, k, l)) {
cnt = max(cnt, (w[i] - 100) / 100);
} else if(judge(w[i] % 100, j, k, l)) {
cnt = max(cnt, w[i] / 100);
} else {
flag = 0;
break;
}
}
}
if(flag) {
ans = min(ans, cnt + j + k + l);
}
}
}
}
cout << ans << endl;
}
return 0;
}

HDU 6685 Rikka with Coin (枚举 思维)的更多相关文章

  1. HDU 6088 - Rikka with Rock-paper-scissors | 2017 Multi-University Training Contest 5

    思路和任意模数FFT模板都来自 这里 看了一晚上那篇<再探快速傅里叶变换>还是懵得不行,可能水平还没到- - 只能先存个模板了,这题单模数NTT跑了5.9s,没敢写三模数NTT,可能姿势太 ...

  2. HDU 6085 - Rikka with Candies | 2017 Multi-University Training Contest 5

    看了标程的压位,才知道压位也能很容易写- - /* HDU 6085 - Rikka with Candies [ 压位 ] | 2017 Multi-University Training Cont ...

  3. HDU 5831 Rikka with Parenthesis II(六花与括号II)

    31 Rikka with Parenthesis II (六花与括号II) Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536 ...

  4. 判断相同区间(lazy) 多校8 HDU 5828 Rikka with Sequence

    // 判断相同区间(lazy) 多校8 HDU 5828 Rikka with Sequence // 题意:三种操作,1增加值,2开根,3求和 // 思路:这题与HDU 4027 和HDU 5634 ...

  5. HDU 6091 - Rikka with Match | 2017 Multi-University Training Contest 5

    思路来自 某FXXL 不过复杂度咋算的.. /* HDU 6091 - Rikka with Match [ 树形DP ] | 2017 Multi-University Training Conte ...

  6. HDU 6093 - Rikka with Number | 2017 Multi-University Training Contest 5

    JAVA+大数搞了一遍- - 不是很麻烦- - /* HDU 6093 - Rikka with Number [ 进制转换,康托展开,大数 ] | 2017 Multi-University Tra ...

  7. B - Rikka with Graph HDU - 5631 (并查集+思维)

    As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some mat ...

  8. HDU 6090 17多校5 Rikka with Graph(思维简单题)

    Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he ...

  9. HDU 6095 17多校5 Rikka with Competition(思维简单题)

    Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he ...

随机推荐

  1. Windows-WAMP搭建与配置

    使用 WampServer 整合软件包进行 WAMP 环境搭建 WampServer 是一款由法国人开发的 Apache Web 服务器.PHP 解释器以及 MySQL 数据库的整合软件包.免去了开发 ...

  2. 83、Tensorflow中的变量管理

    ''' Created on Apr 21, 2017 @author: P0079482 ''' #如何通过tf.variable_scope函数来控制tf.ger_variable函数获取已经创建 ...

  3. Ngrinder简单使用

    文章目录 安装 试玩 性能测试 安装 https://github.com/naver/ngrinder/releases 下载对应版本,是一个war包,3.4以上支持jdk1.8 将war包放到to ...

  4. ACM-ICPC 比赛环境的使用

    ACM-ICPC 现场赛不同的赛站可能比赛环境不同,不过一般都是 Ubuntu 系统.附带的软件可能略有不同,可能会有使用习惯的差异导致效率下降或者无法运行代码,但是在终端下编译运行代码都是相同的.本 ...

  5. Leetcode 跳跃游戏 II

    题目链接:https://leetcode-cn.com/problems/jump-game-ii/ 题目大意: 略. 分析: 贪心 + DP. 代码如下: class Solution { pub ...

  6. haproxy附加

    1.安装haproxy yum -y install haproxy 2.编写文件  vim /etc/haproxy/haproxy.cfg

  7. Tomcat启动脚本(1)startup.bat

    @echo off rem Licensed to the Apache Software Foundation (ASF) under one or more rem contributor lic ...

  8. Vue学习之路之登录注册实例代码

    Vue学习之路之登录注册实例代码:https://www.jb51.net/article/118003.htm vue项目中路由验证和相应拦截的使用:https://blog.csdn.net/wa ...

  9. Spring Cloud配置中心客户端读取配置

    微服务连接配置中心来实现外部配置的读取. 引入依赖 <dependencies> <dependency> <groupId>org.springframework ...

  10. mysql架构精选

    ◆主从架构1.安装服务(主从) yum -y install mysql* /etc/init.d/mysqld start2.修改配置文件:/etc/my.conf(主从) vi /etc/my.c ...