Investment_完全背包
Description
John did not need that much money for the moment. But he realized that it would be a good idea to store this capital in a safe place, and have it grow until he decided to retire. The bank convinced him that a certain kind of bond was interesting for him.
This kind of bond has a fixed value, and gives a fixed amount of yearly interest, payed to the owner at the end of each year. The bond has no fixed term. Bonds are available in different sizes. The larger ones usually give a better interest. Soon John realized that the optimal set of bonds to buy was not trivial to figure out. Moreover, after a few years his capital would have grown, and the schedule had to be re-evaluated.
Assume the following bonds are available:
| Value | Annual interest |
| 4000 3000 |
400 250 |
With a capital of e10 000 one could buy two bonds of $4 000, giving a yearly interest of $800. Buying two bonds of $3 000, and one of $4 000 is a better idea, as it gives a yearly interest of $900. After two years the capital has grown to $11 800, and it makes sense to sell a $3 000 one and buy a $4 000 one, so the annual interest grows to $1 050. This is where this story grows unlikely: the bank does not charge for buying and selling bonds. Next year the total sum is $12 850, which allows for three times $4 000, giving a yearly interest of $1 200.
Here is your problem: given an amount to begin with, a number of years, and a set of bonds with their values and interests, find out how big the amount may grow in the given period, using the best schedule for buying and selling bonds.
Input
The first line of a test case contains two positive integers: the amount to start with (at most $1 000 000), and the number of years the capital may grow (at most 40).
The following line contains a single number: the number d (1 <= d <= 10) of available bonds.
The next d lines each contain the description of a bond. The description of a bond consists of two positive integers: the value of the bond, and the yearly interest for that bond. The value of a bond is always a multiple of $1 000. The interest of a bond is never more than 10% of its value.
Output
Sample Input
1
10000 4
2
4000 400
3000 250
Sample Output
14050
【题意】给出本金和年数,给出几种物品价格和每年的盈利,求最大本息和
【思路】完全背包,数比较大,需要/1000压缩一下
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int N=;
int dp[N];
struct node
{
int val,w;
}a[];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,m,y;
scanf("%d%d",&m,&y);
scanf("%d",&n); for(int i=;i<=n;i++)
{
scanf("%d%d",&a[i].val,&a[i].w);
a[i].val/=;//数太大,对背包大小进行压缩
}
for(int i=;i<=y;i++)
{
int tmp=m/;
memset(dp,,sizeof(dp));//在每一年都要清零
for(int j=;j<=n;j++)//完全背包
{
for(int k=a[j].val;k<=tmp;k++)
{
dp[k]=max(dp[k],dp[k-a[j].val]+a[j].w);
}
}
m+=dp[tmp];//本息和
}
printf("%d\n",m);
}
return ;
}
Investment_完全背包的更多相关文章
- 【USACO 3.1】Stamps (完全背包)
题意:给你n种价值不同的邮票,最大的不超过10000元,一次最多贴k张,求1到多少都能被表示出来?n≤50,k≤200. 题解:dp[i]表示i元最少可以用几张邮票表示,那么对于价值a的邮票,可以推出 ...
- HDU 3535 AreYouBusy (混合背包)
题意:给你n组物品和自己有的价值s,每组有l个物品和有一种类型: 0:此组中最少选择一个 1:此组中最多选择一个 2:此组随便选 每种物品有两个值:是需要价值ci,可获得乐趣gi 问在满足条件的情况下 ...
- HDU2159 二维完全背包
FATE Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submis ...
- CF2.D 并查集+背包
D. Arpa's weak amphitheater and Mehrdad's valuable Hoses time limit per test 1 second memory limit p ...
- UVALive 4870 Roller Coaster --01背包
题意:过山车有n个区域,一个人有两个值F,D,在每个区域有两种选择: 1.睁眼: F += f[i], D += d[i] 2.闭眼: F = F , D -= K 问在D小于等于一定限度的时 ...
- 洛谷P1782 旅行商的背包[多重背包]
题目描述 小S坚信任何问题都可以在多项式时间内解决,于是他准备亲自去当一回旅行商.在出发之前,他购进了一些物品.这些物品共有n种,第i种体积为Vi,价值为Wi,共有Di件.他的背包体积是C.怎样装才能 ...
- POJ1717 Dominoes[背包DP]
Dominoes Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6731 Accepted: 2234 Descript ...
- HDU3466 Proud Merchants[背包DP 条件限制]
Proud Merchants Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others) ...
- POJ1112 Team Them Up![二分图染色 补图 01背包]
Team Them Up! Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 7608 Accepted: 2041 S ...
随机推荐
- 批量Load/Store指令的寻址方式
批量Load/Store指令用于实现在一组寄存器和一块连续的内存单元之间传输数据.也称为多寄存器寻址方式,即一条指令可以完成多个寄存器值的传送.这种寻址方式可以用一条指令最多完成传送16个通用寄存器的 ...
- 转载解决:错误的语法:”XXXX“必须是批处理中仅有的语句
SQL Server 数据库提示“错误的语法:”XXXX“必须是批处理中仅有的语句”报错的原因分析 解析:批处理必须以 CREATE 语句开始.也就是说一个查询分析器里面只有一个批处理语句才是规范的语 ...
- uva 10723
10723 - Cyborg Genes Time limit: 3.000 seconds Problem F Cyborg Genes Time Limit 1 Second Septembe ...
- PHP函数——urlencode() 函数
urlencode($str)的作用是对字符串$str进行url编码,方便$str作为一个变量传递给下一页,一般情况下$str有两种, 第一种是数组类型,如果想将数组作为url的一个参数,即必须将数组 ...
- lucene索引文件格式
转自:http://blog.csdn.net/whuqin 本文介绍下lucene生成的索引有哪些文件组成,每个文件包含了什么信息.基于Lucene 4.10.0. 数据结构 索引(index)包含 ...
- C#识别验证码技术-Tesseract
相信大家在开发一些程序会有识别图片上文字(即所谓的OCR)的需求,比如识别车牌.识别图片格式的商品价格.识别图片格式的邮箱地址等等,当然需求最多的还是识别验证码.如果要完成这些OCR的工作,需要你掌握 ...
- Entity Framework系列
这个系列主要记录学习EF的过程和碰到的问题以及解决问题的方法. EF中的那些批量操作 EF的Model First
- 使用Lucene.Net管理索引实现搜索
之前使用一直是没有问题的,只到今天发现删除的时候无法删除,增加的时候却一直在增加,导致搜索的时候可以搜出来很多相同的结果. 小猪决定趁今天这个机会好好的把这个问题给解决了. private void ...
- java 面向对象编程 第18章——网络编程
1. TCP/IP协议模型 应用层:应用程序: 传输层:将数据套接端口,提供端到端的通信服务: 网络互联层:负责数据包装.寻址和路由,同时还包含网间控制报文协议: 网络接口层:提供TCP/IP协议的 ...
- 小记:获取系统时间的long值,格式化成可读时间。
/** * 返回的字符串形式是形如:2013年10月20日 20:58 * */ public static String formatTimeInMillis(long timeInMillis) ...