Cash Machine
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 33444   Accepted: 12106

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 denomination Dk 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 2002

一般解决方法:多重背包=01背包+完全背包
此处01处理时加上二进制思想,不然会超时

 //524K    47MS    C++    742B
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std; int dp[];
int m[], d[];
int main(void)
{
int cash, n;
while(scanf("%d%d",&cash,&n)!=EOF)
{
memset(dp,,sizeof(dp));
for(int i=;i<n;i++){
scanf("%d%d",&m[i],&d[i]);
}
for(int i=;i<n;i++){
if(m[i]*d[i]>cash){
for(int j=d[i];j<=cash;j++)
dp[j] = max(dp[j], dp[j-d[i]] + d[i]);
}else{
int km=;
int k=m[i];
while(km<k){
int val = km*d[i];
for(int j=cash;j>=val;j--)
dp[j] = max(dp[j], dp[j-val] + val); k -= km;
km<<=;
} int val = k*d[i];
for(int j=cash;j>=val;j--)
dp[j] = max(dp[j], dp[j-val] + val);
}
}
printf("%d\n",dp[cash]);
}
return ;
}

poj 1276 Cash Machine(多重背包)的更多相关文章

  1. POJ 1276 Cash Machine(多重背包的二进制优化)

    题目网址:http://poj.org/problem?id=1276 思路: 很明显是多重背包,把总金额看作是背包的容量. 刚开始是想把单个金额当做一个物品,用三层循环来 转换成01背包来做.T了… ...

  2. Poj 1276 Cash Machine 多重背包

    Cash Machine Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 26172   Accepted: 9238 Des ...

  3. [poj 1276] Cash Machine 多重背包及优化

    Description A Bank plans to install a machine for cash withdrawal. The machine is able to deliver ap ...

  4. poj 1276 Cash Machine_多重背包

    题意:略 多重背包 #include <iostream> #include<cstring> #include<cstdio> using namespace s ...

  5. 【转载】poj 1276 Cash Machine 【凑钱数的问题】【枚举思路 或者 多重背包解决】

    转载地址:http://m.blog.csdn.net/blog/u010489766/9229011 题目链接:http://poj.org/problem?id=1276 题意:机器里面共有n种面 ...

  6. POJ 1276 Cash Machine(单调队列优化多重背包)

    Cash Machine Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 38986   Accepted: 14186 De ...

  7. POJ 1276:Cash Machine 多重背包

    Cash Machine Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 30006   Accepted: 10811 De ...

  8. Cash Machine (POJ 1276)(多重背包——二进制优化)

    链接:POJ - 1276 题意:给你一个最大金额m,现在有n种类型的纸票,这些纸票的个数各不相同,问能够用这些纸票再不超过m的前提下凑成最大的金额是多少? 题解:写了01背包直接暴力,结果T了,时间 ...

  9. Cash Machine(多重背包)

    http://poj.org/problem?id=1276 #include <stdio.h> #include <string.h> ; #define Max(a,b) ...

随机推荐

  1. 关于Windows 7启动后网络一直转的问题的一个解决方法

    前两天给自己的X220i换了个SSD,装了个windows 7 的系统,然后自然搞了一些杀毒软件之类的,为了开发需要还装了Oracle数据库,可是,不知道从什么时候开始,系统启动后右下角那个网络图标显 ...

  2. 慕课网-Java入门第一季-6-10 练习题

    来源:http://www.imooc.com/ceping/1596 以下关于二维数组的定义和访问正确的是() A int[ ][ ] num = new int[ ][ ]; B int[ ][ ...

  3. java 配置文件读取

    1.getResourceAsStream Class.getClassLoader.getResourceAsStream(String path) :默认则是从ClassPath根下获取,path ...

  4. 如何使用THashedStringList

    1.添加 uses system.IniFiles 2.实例代码: unit Unit1; interface uses Winapi.Windows, Winapi.Messages, System ...

  5. 【MongoDB】MongoDB 3.2 SCRAM-SHA-1验证方式

    新版本已取消addUser方法,改使用createUser方法 官方地址:https://docs.mongodb.com/manual/tutorial/create-users/ 官方地址:htt ...

  6. css定位和浮动

    1.css中一切元素皆为框.div.p.h1等为块框:span.strong等为行内框,(在文本中每一行会被自动默认为行框,行框和行内框是不一样的概念).通过display可以改变框的类型,行内框通过 ...

  7. windows脚本配置ip地址

    背景:工作上经常涉及到要调试设备,每次都要手动配置静态ip地址,配置完之后还要重新改回来,有时候为了连续调试多台设备,来回手动更改ip,实在麻烦. 思考:想到windows有脚本,可以利用脚本文件达到 ...

  8. Maximo子表中增加附件功能

    附件功能的实现(详见ewell.webclient.beans.warranty.WarrantysDateBean ,ewell.webclient.beans.doclinks.custom.Ad ...

  9. python 实现 斐波那契数列

    #!usr/bin/python#_*_coding=utf-8_*_ def fin(n): li=[0,1] for i in range(2,n): li.append(li[-1]+li[-2 ...

  10. flex+java+blazeds 多通道好文

    http://www.cnblogs.com/noam/archive/2010/08/05/1793504.html blazeds, spring3整合实现RPC服务和消息服务 环境: MyEcl ...