POJ 1276 Cash Machine 【多重背包-二进制优化】
2015-09-12 16:30
525 查看
Cash Machine
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 29995 | Accepted: 10805 |
Description
A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machine uses exactly N distinct bill denominations, say Dk, k=1,N, and for each denominationDk the machine has a supply of nk bills. For example,
N=3, n1=10, D1=100, n2=4, D2=50, n3=5, D3=10
means the machine has a supply of 10 bills of @100 each, 4 bills of @50 each, and 5 bills of @10 each.
Call cash the requested amount of cash the machine should deliver and write a program that computes the maximum amount of cash less than or equal to cash that can be effectively delivered according to the available bill supply of the machine.
Notes:
@ is the symbol of the currency delivered by the machine. For instance, @ may stand for dollar, euro, pound etc.
Input
The program input is from standard input. Each data set in the input stands for a particular transaction and has the format:cash N n1 D1 n2 D2 ... nN DN
where 0 <= cash <= 100000 is the amount of cash requested, 0 <=N <= 10 is the number of bill denominations and 0 <= nk <= 1000 is the number of available bills for the Dk denomination, 1 <= Dk <= 1000, k=1,N. White spaces can occur freely between the numbers
in the input. The input data are correct.
Output
For each set of data the program prints the result to the standard output on a separate line as shown in the examples below.Sample Input
735 3 4 125 6 5 3 350 633 4 500 30 6 100 1 5 0 1 735 0 0 3 10 100 10 50 10 10
Sample Output
735
630
0
0
Hint
The first data set designates a transaction where the amount of cash requested is @735. The machine contains 3 bill denominations: 4 bills of @125, 6 bills of @5, and 3 bills of @350. The machine can deliver the exact amount of requested cash.
In the second case the bill supply of the machine does not fit the exact amount of cash requested. The maximum cash that can be delivered is @630. Notice that there can be several possibilities to combine the bills in the machine for matching the delivered
cash.
In the third case the machine is empty and no cash is delivered. In the fourth case the amount of cash requested is @0 and, therefore, the machine delivers no cash.
Source
Southeastern Europe 2002t
For each set of data the program prints the result to the standard output on a separate line as shown in the examples below.
Sample Input
735 3 4 125 6 5 3 350 633 4 500 30 6 100 1 5 0 1 735 0 0 3 10 100 10 50 10 10
Sample Output
735 630 0 0
Hint
The first data set designates a transaction where the amount of cash requested is @735. The machine contains 3 bill denominations: 4 bills of @125, 6 bills of @5, and 3 bills of @350. The machine can deliver the exact amount ofrequested cash.
In the second case the bill supply of the machine does not fit the exact amount of cash requested. The maximum cash that can be delivered is @630. Notice that there can be several possibilities to combine the bills in the machine for matching the delivered
cash.
In the third case the machine is empty and no cash is delivered. In the fourth case the amount of cash requested is @0 and, therefore, the machine delivers no cash.
Source
Southeastern Europe 2002嗯,题目大意就是说,找现金的机器,有一些面额的现金一定数量,然后问用这些现金所能找到的最接近且小于等于给定值的现金值。
输入就是首先所要求的现金值(即给定值)然后是现金面额的种类数,之后就是面额的数量以及面额值。
多重背包,若直接转化成01背包式超时的,这里用的是《背包九讲》中的二进制优化(然而我不怎么懂。。。)
#include <iostream> #include<cstdio> #include<cstring> #define maxn 100000+10 using namespace std; int cash,f[maxn],c[12],p[12]; int max(int a,int b) { return a>b?a:b; } void CompletePack(int cost) { for(int i=cost;i<=cash;++i) f[i]=max(f[i],f[i-cost]+cost); } void ZeroOnePack(int cost) { for(int i=cash;i>=cost;--i) f[i]=max(f[i],f[i-cost]+cost); } void MultiplePack(int cost,int amount) { if(cost*amount>=cash) { CompletePack(cost); return; } int k=1; while(k<amount) { ZeroOnePack(k*cost); amount-=k; k=k*2; } ZeroOnePack(amount*cost); } int main() { int n; while(~scanf("%d%d",&cash,&n)) { for(int i=1;i<=n;++i) scanf("%d%d",&c[i],&p[i]); for(int i=0;i<=cash;++i) f[i]=0; for(int i=1;i<=n;++i) { MultiplePack(p[i],c[i]); } printf("%d\n",f[cash]); } return 0; }
相关文章推荐
- lintcode-单词搜索-123
- 在Spring、Hibernate中使用Ehcache缓存--Ehcache 整合Spring 使用页面、对象缓存
- 网络协议综述
- c++ 玩转 vector容器 用法
- Linux下Nginx 中文文件处理方法
- 第三十五天 模拟音乐播放器MediaPlayer
- 导入Android自带Sample工程出错解决(Error retrieving parent for item)
- 项目2:科幻信息面板
- 晨间日记模板 Web应用版 晨间日记软件 开源
- 重定向和转发有什么区别
- 注册页面的搭建
- ACM POJ2533 简单dp
- C++学习笔记 -- 虚析构函数与纯虚析构函数
- DrawerLayout拦截了应该传到子View的事件
- Baskets of Gold Coins
- C#中复制数组
- Linux和Windows的换行符
- TCP协议中的三次握手和四次挥手(图解)
- python学习——heapq模块
- IO端口和IO内存