You are a rich person, and you think your wallet is too heavy and full now. So you want to give me some money by buying a lovely pusheen sticker which costs p dollars from me. To make your wallet lighter, you decide to pay exactly pp dollars by as many coins and/or banknotes as possible.

For example, if p = 1 and you have two 10 coins, four 5 coins, and eight 1 coins, you will pay it by two 5coins and seven 1 coins. But this task is incredibly hard since you are too rich and the sticker is too expensive and pusheen is too lovely, please write a program to calculate the best solution.

Input Format

The first line contains an integer Tindicating the total number of test cases.

Each test case is a line with 1 integers p, c1, c5, c10, c20, c50, c100, c200, c500, c1000, c2000, specifying the price of the pusheen sticker, and the number of coins and banknotes in each denomination. The number ci means how many coins/banknotes in denominations of ii dollars in your wallet.

1≤T≤20000

0≤p≤109

0≤ci​≤100000

Output Format

For each test case, please output the maximum number of coins and/or banknotes he can pay for exactly p dollars in a line. If you cannot pay for exactly p dollars, please simply output '-1'.

样例输入复制

3
17 8 4 2 0 0 0 0 0 0 0
100 99 0 0 0 0 0 0 0 0 0
2015 9 8 7 6 5 4 3 2 1 0

样例输出复制

9
-1
36

题目来源

ACM Changchun 2015

/*
某些面值的钱分别有若干个,用这些钱来恰好组成某一金额的钱,问最多需要的钱的数目
逆向思维 用总价减去需求=X
那么贪心来用最少的数目凑成X即可(也可能凑不出来)
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <string>
#include <deque>
using namespace std;
#define ll long long
#define N 1009
const ll inf =9e18;
#define gep(i,a,b) for(int i=a;i<=b;i++)
#define gepp(i,a,b) for(int i=a;i>=b;i--)
#define gep1(i,a,b) for(ll i=a;i<=b;i++)
#define gepp1(i,a,b) for(ll i=a;i>=b;i--)
#define mem(a,b) memset(a,b,sizeof(a))
ll val[]={,,,,,,,,,,};
ll num[],a[];
ll ans,p;
ll sum,ret;
int t;
void dfs(int x,ll sum,ll cnt)
{
if(!sum)
{
ans=min(ans,cnt);
return ;
}
if(x<) return ;
a[x]=min(sum/val[x],num[x]);
dfs(x-,sum-val[x]*a[x],cnt+a[x]);
if(a[x]>=)
{
a[x]--;
dfs(x-,sum-val[x]*a[x],cnt+a[x]);
}
/*
如 : 1
150 0 0 0 3 1 1 0 0 0 0
如果没有上面的三行就是无解 -1
但是 dfs(4,60,0) 因为 用了50,后面的就无法凑成10
但是后面的20*3可以凑成60,因此a[x]可一每次都减1,再去dfs
这样才能考虑到所有的情况!
*/
}
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%lld",&p);
mem(a,);
sum=;
ret=;
gep(i,,)
{
scanf("%lld",&num[i]);
ret+=num[i];
sum+=val[i]*num[i];
}
sum-=p;
if(sum<)
{
printf("-1\n");
continue;
}
ans=inf;
dfs(,sum,);
if(ans!=inf)
{
printf("%lld\n",ret-ans);
}
else{
printf("-1\n");
}
}
return ;
}

ACM Changchun 2015 A. Too Rich的更多相关文章

  1. ACM Changchun 2015 L . House Building

    Have you ever played the video game Minecraft? This game has been one of the world's most popular ga ...

  2. ACM Changchun 2015 J. Chip Factory

    John is a manager of a CPU chip factory, the factory produces lots of chips everyday. To manage larg ...

  3. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 G. Garden Gathering

    Problem G. Garden Gathering Input file: standard input Output file: standard output Time limit: 3 se ...

  4. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 D. Delay Time

    Problem D. Delay Time Input file: standard input Output file: standard output Time limit: 1 second M ...

  5. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 I. Illegal or Not?

    I. Illegal or Not? time limit per test 1 second memory limit per test 512 megabytes input standard i ...

  6. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 K. King’s Rout

    K. King's Rout time limit per test 4 seconds memory limit per test 512 megabytes input standard inpu ...

  7. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 H. Hashing

    H. Hashing time limit per test 1 second memory limit per test 512 megabytes input standard input out ...

  8. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 C. Colder-Hotter

    C. Colder-Hotter time limit per test 1 second memory limit per test 512 megabytes input standard inp ...

  9. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 A. Anagrams

    A. Anagrams time limit per test 1 second memory limit per test 512 megabytes input standard input ou ...

随机推荐

  1. html文本框和按钮这些点击时不显示边框的光晕

    直接在样式加:focus{outline:0;}这样子就可以了

  2. 《Openstack的搭建》RHEL6.5

    Openstack就是搭建一个较为完整的虚拟化平台,把一个完整的物理机划分成若干个虚拟机来跑,从而实现资源的充分利用. Openstack对硬件的要求很高,要是你的物理机内存是4G的话,虚拟机的内存给 ...

  3. .NET Core 1.0 CentOS7 尝试(一、安装)

    昨天宣布 ASP.NET Core RC2,据说差不多稳定了,以后不会有大改了. 参考:https://blogs.msdn.microsoft.com/webdev/2016/05/16/annou ...

  4. okhttp使用心得(https验证不通过)(一)

    之前项目使用的是okhttp3.4版本的,tls协议支持1.0 1.2 等等 后来换成okhttp3.8.1,发现握手失败 找了好多原因之后  发现吧tls加上1.0后,就可以握手成功了,但是tls1 ...

  5. 字符串、String等转换

    (1) String 转换为字符串 例:String s = "abcde";char[] a = s.toCharArray(); (2) 字符串转换为Stringchar[] ...

  6. java代码(生成long类型数字)

    package test; public class GenerateNum { public static void main(String[] args) { //定义为long类型,需在数值后面 ...

  7. vue使用echarts可视化图形插件

    1.安装echarts:  cnpm/npm i echarts -S 2.main.js中   import echarts from 'echart'    Vue.prototype.$echa ...

  8. vue+element ui项目总结点(四)零散细节概念巩固如vue父组件调用子组件的方法、拷贝数据、数组置空问题 等

    vue config下面的index.js配置host: '0.0.0.0',共享ip (假设你的电脑启动了这个服务我电脑一样可以启动)-------------------------------- ...

  9. 微软爆料新型系统,Windows7,Windows10强势来袭

    本系统是10月5日最新完整版本的Windows10 安装版镜像,win10正式版,更新了重要补丁,提升应用加载速度,微软和百度今天宣布达成合作,百度成为win10 Edge浏览器中国默认主页和搜索引擎 ...

  10. asp.net 页面嵌套(非iframe)方法

    前台 <div id="divUrlDetail" runat="server"> </div> 后台 protected void P ...