Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 35387   Accepted: 12816

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

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN 100509
#define N 21
#define MOD 1000000
#define INF 1000000009
#define eps 0.00000001
/*
两种思路
1.看作多个0 1 背包问题
2.利用二进制分解 优化
*/
int value[N], dp[MAXN], num[N];
int aim, n, k;
void solve(int &Max, int num, int value)
{
for (int i = Max; i >= ; i--)
{
if (dp[i])
{
for (int k = ; k <= num; k++)
{
if (i + value*k>aim) continue;
dp[i + value*k] = ;
if (i + value*k > Max)
{
Max = i + value*k;
}
}
}
}
}
int main()
{
while (scanf("%d%d", &aim,&n) != EOF)
{
for (int i = ; i < n; i++)
scanf("%d%d", &num[i], &value[i]);
if (n == || aim == )
{
printf("0\n");
continue;
}
memset(dp, , sizeof(dp));
dp[] = ;
int ans = ;
for (int i = ; i < n; i++)
solve(ans, num[i], value[i]);
printf("%d\n", ans);
}
return ;
}

Cash Machine POJ 1276 多重背包的更多相关文章

  1. Cash Machine POJ - 1276 多重背包二进制优化

    题意:多重背包模型  n种物品 每个m个  问背包容量下最多拿多少 这里要用二进制优化不然会超时 #include<iostream> #include<cstdio> #in ...

  2. poj 1276 多重背包

    735 3 4 125 6 5 3 350 //735的最大额,3种,4个125,6个5,3个350 633 4 500 30 6 100 1 5 0 1 735 0 0 3 10 100 10 50 ...

  3. Cash Machine POJ - 1276

    解法 多重背包板子题 多重背包板子 如果上限的体积大于了给定的体积那么套完全背包 否则二进制优化成01背包 代码 #include <iostream> #include <cstr ...

  4. POJ 1276 (多重背包) Cash Machine

    题意: 有n种纸币,已知每种纸币的面值和数量,求所能凑成的不超过cash的最大总面值. 分析: 这道题自己写了一下TLE了,好可耻.. 找了份比较简洁的代码抄过来了..poj1276 #include ...

  5. poj 2392 多重背包

    题意:有几个砖,给出高度,能放的最大高度和数目,求这些砖能垒成的最大高度 依据lim排个序,按一层一层进行背包 #include<cstdio> #include<iostream& ...

  6. POJ 3260 多重背包+完全背包

    前几天刚回到家却发现家里没网线 && 路由器都被带走了,无奈之下只好铤而走险尝试蹭隔壁家的WiFi,不试不知道,一试吓一跳,用个手机软件简简单单就连上了,然后在浏览器输入192.168 ...

  7. poj 1742 多重背包

    题意:给出n种面值的硬币, 和这些硬币每一种的数量, 要求求出能组成的钱数(小于等于m) 思路:一开始直接用多重背包套上去超时了,然后就没辙了,然后参考网上的,说只需要判断是否能取到就行了,并不需要记 ...

  8. poj 1014多重背包

    题意:给出价值为1,2,3,4,5,6的6种物品数量,问是否能将物品分成两份,使两份的总价值相等. 思路:求出总价值除二,做多重背包,需要二进制优化. 代码: #include<iostream ...

  9. Dividing POJ - 1014 多重背包二进制优化

    多重背包模型  写的时候漏了一个等号找了半天 i<<=1 !!!!!! #include<iostream> #include<cstdio> #include&l ...

随机推荐

  1. error: undefined reference to 'property_set (转载)

    转自:http://blog.csdn.net/u011589606/article/details/23474241 in the cpp file, please include #include ...

  2. J20170916-hm

    スタイルシート 样式表 シール 封条 シート 纸片 マニフェスト 货单(Rails) ダイジェスト 消化,(Rails 附加哈希值) インタプリタ n. 解释者; 口译译员; [军事] 判读员; [自 ...

  3. 元素类型以及overflow,white-space等属性

    1:预格式化标签:<pre></pre>2:overflow属性="visible/hidden(隐藏)"/scroll/auto(自动)/inherit; ...

  4. day24 03 多继承

    day24 03 多继承 正常的代码中  单继承==减少了代码的重复 继承表达的是一种 子类是父类的关系 1.简单的多继承关系 A,B,C,D四个类,其中D类继承A,B,C三个父类,因此也叫多继承,子 ...

  5. SQL数据库还原的二种方式和区别

    1.数据库还原 在SQL中,直接选择选择“还原数据库”:选中.bak 文件即可. 2.生成脚本 新建同样的DB名字,在SQL打开脚本,执行脚本语言.数据库里面就会自动填充内容.

  6. spring boot打包文件后,报错\No such file or directory

    现象: 一段代码: ClassLoader loader = XXXUtil.class.getClassLoader(); String jsFileName = loader.getResourc ...

  7. fcc html5 css 练习2

    <form action="/submit-cat-photo" >action属性的值指定了表单提交到服务器的地址 <input type="text ...

  8. [Windows Server 2012] 安装PHP+MySQL方法

    ★ 欢迎来到[护卫神·V课堂],网站地址:http://v.huweishen.com★ 护卫神·V课堂 是护卫神旗下专业提供服务器教学视频的网站,每周更新视频.★ 本节我们将带领大家:PHP+MyS ...

  9. cmd 启动mysql环境变量配置

    win10系统:(其他系统类似,改环境变量就可以) 1.我的电脑,右键选择属性,进入系统页面 2.点击高级系统设置,进入系统属性页面 3.点击高级选项卡,点击环境变量,进入环境变量设置 4.选择系统变 ...

  10. Linux监控实时log

    https://jingyan.baidu.com/article/93f9803f5545a3e0e46f5596.html