ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.

For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.

Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.

Input

The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).

The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.

The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.

The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.

Output

For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print  - 1, if it is impossible to get the corresponding sum.

Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1

因为只能取2种面值,所以枚举一种面值,然后搞个map找另一个面值跟它匹配

 #include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define mkp(a,b) make_pair(a,b)
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
LL x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int n,k,q,cnt;
int a[];
struct p{int v,w,k;}b[];
bool operator <(p a,p b){return a.w<b.w||a.w==b.w&&a.k<b.k;}
map<int,pa>mp,mp2;
int main()
{
mp.clear();mp2.clear();
n=read();k=read();
for (int i=;i<=n;i++)
{
a[i]=read();
for (int j=;j<=k;j++)
b[++cnt].v=i,b[cnt].k=j,b[cnt].w=j*a[i];
}
sort(b+,b+cnt+);
for (int i=;i<=cnt;i++)
{
int w=b[i].w,kk=b[i].k,v=b[i].v;
if (mp[w]==mkp(,))mp[w]=mkp(kk,v);
else
{
if (mp[w].first>kk)mp2[w]=mp[w],mp[w]=mkp(kk,v);
else if (mp2[w].first>kk)mp2[w]=mkp(kk,v);
}
}
q=read();
for (int i=;i<=q;i++)
{
int x=read(),ans=;
for (int j=;j<=n;j++)
{
for (int l=;l<=k;l++)
{
int now=l*a[j];
if (now>x)break;
if (now==x){ans=min(ans,l);break;}
pa s=mp[x-now];
if (s.second==j)s=mp2[x-now];
if (s==mkp(,))continue;
ans=min(ans,s.first+l);
}
}
printf("%d\n",ans>k?-:ans);
}
}

cf524C

cf524C The Art of Dealing with ATM的更多相关文章

  1. Codeforces Round VK Cup 2015 - Round 1 (unofficial online mirror, Div. 1 only)E. The Art of Dealing with ATM 暴力出奇迹!

    VK Cup 2015 - Round 1 (unofficial online mirror, Div. 1 only)E. The Art of Dealing with ATM Time Lim ...

  2. codeforce The Art of Dealing with ATM

    题目大意 ATM取款机有n种不同的钱币kind[i],每次取款允许吐出不超过k张钱币,且钱币的种类数不能超过2(一开始没理解2的意思),现在有q次取款,钱数为ques,问ATM能否凑出这样的钱,若能的 ...

  3. CodeForces 524C The Art of Dealing with ATM (二分)

    题意:给定 n 种不同的钞票,然后用q个询问,问你用最多k张,最多两种不同的钞票能不能组成一个值. 析:首先如果要求的值小点,就可以用DP,但是太大了,所以我们考虑一共最多有n * k种钞票,如果每次 ...

  4. Codeforces 524C.The Art of Dealing with ATM(暴力)

    我先采用了智障解法(n * n枚举...刚开始把n看成1000了还以为能过) 理所当然的t了,不过我怀疑优化一下能过?(感觉数据不太行的亚子 然后就是O(n * k * k)的解法,看到好多人快乐二分 ...

  5. [转] Loren on the Art of MATLAB

    http://blogs.mathworks.com/loren/2007/03/01/creating-sparse-finite-element-matrices-in-matlab/ Loren ...

  6. 1179: [Apio2009]Atm

    1179: [Apio2009]Atm Time Limit: 15 Sec  Memory Limit: 162 MBSubmit: 1629  Solved: 615[Submit][Status ...

  7. The Art of Deception

    前言 一些黑客毁坏别人的文件甚至整个硬盘,他们被称为电脑狂人(crackers)或计算机破坏者(vandals).另一些新手省去学习技术的麻烦,直接下载黑客工具侵入别人的计算机,这些人被称为脚本小子( ...

  8. 设计模式(十二):通过ATM取款机来认识“状态模式”(State Pattern)

    说到状态模式,如果你看过之前发布的重构系列的文章中的<代码重构(六):代码重构完整案例>这篇博客的话,那么你应该对“状态模式”并不陌生,因为我们之前使用到了状态模式进行重构.上一篇博客我们 ...

  9. How the problem solved about " Dealing with non-fast-forward errors"

    Recently ,I got confused When I use  git to push one of my project. The problem is below: And I Foun ...

随机推荐

  1. javaweb基础(23)_jsp自定义标签

    一.自定义标签的作用 自定义标签主要用于移除Jsp页面中的java代码. 二.自定义标签开发和使用 2.1.自定义标签开发步骤 1.编写一个实现Tag接口的Java类(标签处理器类) 1 packag ...

  2. iOS项目工程及目录结构

    做过一些iOS的项目,不同项目的沉淀没有积累到一起,目录的管理都在后期随着人员的增加越来越混乱,因此在这里做一些梳理,希望达到两个目的. 一套相对通用的目录结构,作为后续项目的模版. 积累相应的基础库 ...

  3. OpenCV3.42+VS2017配置+模块计算机类型“X86”与目标计算机类型“x64”冲突”的问题解决

    目录 OpenCV3.42+VS2017配置 Visual Studio 2017 第三方依赖设置,附加依赖项和附加库目录 "fatal error LNK1112: 模块计算机类型&quo ...

  4. 【费用流】bzoj1834: [ZJOI2010]network 网络扩容

    还是稍微记一下这个拆点模型吧 Description 给定一张有向图,每条边都有一个容量C和一个扩容费用W.这里扩容费用是指将容量扩大1所需的费用. 求:  1.在不扩容的情况下,1到N的最大流:  ...

  5. Latex使用笔记,中文,字号等

    中文 编译器选择的xelatex 或者lualatex 我试过的两种方式(都是基于ctex的). 直接使用ctex的基本documentclass \documentclass[UTF8]{ctexa ...

  6. 转 Hystrix入门指南 Introduction

    https://www.cnblogs.com/gaoyanqing/p/7470085.html

  7. JAVA 修改环境变量不重启电脑生效方法

     1. 在安装JDK1.6(高版本)时(本机先安装jdk1.6再安装的jdk1.5),自动将java.exe.javaw.exe.javaws.exe三个可执行文件复制到了C:\Windows\Sys ...

  8. 【mysql】mysql存储过程实例

    ```mysql DELIMITER $$   DROP PROCEDURE IF EXISTS `system_number_update` $$   CREATE DEFINER=`root`@` ...

  9. python爬虫基础14-selenium大全8/8-常见问题

    Selenium笔记(8)常见的坑 本文集链接:https://www.jianshu.com/nb/25338984 用Xpath查找数据时无法直接获取节点属性 通常在我们使用xpath时,可以使用 ...

  10. Python 变量作用域 LEGB (上)—— Local,Global,Builtin

    Python 变量作用域的规则是 LEGB LEGB含义解释:L —— Local(function):函数内的名字空间E —— Enclosing function locals:外部嵌套函数的名字 ...