poj 1276 背包啊
来源:互联网 发布:大数据平台开发面试 编辑:程序博客网 时间:2024/06/05 07:08
Cash Machine
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 22011 Accepted: 7723
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.
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.
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 350633 4 500 30 6 100 1 5 0 1735 00 3 10 100 10 50 10 10
Sample Output
73563000
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.
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
背包问题啊,可以看作01背包啊,不过会超时啊,这里要加上个记数的数组正序来就可以了
#include<iostream>#include<cstdio>#include<cstring>using namespace std;int cash,f[100005],n,m,a,c[100005];int main(){ while(scanf("%d",&cash)!=EOF) { memset(f,0,4*(cash+1));//人工计算位 scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d%d",&m,&a); memset(c,0,4*(cash+1)); for(int j=a;j<=cash;j++) { if(f[j-a]+a>f[j]&&c[j-a]<m) { f[j]=f[j-a]+a; c[j]=c[j-a]+1; } } } cout<<f[cash]<<endl; } return 0;}
- poj 1276 背包啊
- poj 1276 多重背包..
- POJ--1276:多重背包
- POJ 1276 多重背包
- POJ 1276 多重背包
- poj 1276 多重背包
- poj 1276 多重背包
- poj 1276 完全背包
- poj 1276 多重背包
- POJ 1276 背包
- POJ 1276 多重背包
- POJ 1276 多重背包
- POJ 1276(多重背包)
- poj 1276 多重背包
- POJ 1276 混合背包
- poj 1276 多重背包
- POJ 1276-CashMachine 背包问题
- POJ 1276 多重背包问题
- 从callback的角度来理解.NET/C# 中的 委托 (delegate)与 事件 (event)
- win7下打开CHM,索引选项出错
- 清除文件夹下SVN信息
- 编译DataNucleus Tutorial for JDO using Ant成功
- 分治、时间空间的权衡:最大合的连续字串问题 (PAT 1007)
- poj 1276 背包啊
- poj 3461
- 解决 “无法安装 Visual Studio 2010 Service Pack 1,因为此计算机的状态不支持”
- 这25条鲜活信息,有助你简单了解当下的中国互联网
- mips 异常处理
- 查询优化管道分析思路
- CSRF | XSRF 跨站请求伪造
- android adb shell 的ls命令出现奇怪的字符
- IOS中运行出现如下错误的解决办法