Description

John never knew he had a grand-uncle, until he received the notary's letter. He learned that his late grand-uncle had gathered a lot of money, somewhere in South-America, and that John was the only inheritor.
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 contains a single positive integer N which is the number of test cases. The test cases follow. 
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

For each test case, output – on a separate line – the capital at the end of the period, after an optimal schedule of buying and selling.

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

  1. 【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 ...

  2. 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 ...

  3. HDU2159 二维完全背包

    FATE Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submis ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. POJ1717 Dominoes[背包DP]

    Dominoes Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6731   Accepted: 2234 Descript ...

  8. 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) ...

  9. 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

  1. 【待整理】PS切图基础教程

    http://www.w3cfuns.com/article-442-1-1.html http://www.w3cfuns.com/article-443-1-1.html 其他专题研究: floa ...

  2. Java 集合系列 05 Vector详细介绍(源码解析)和使用示例

    java 集合系列目录: Java 集合系列 01 总体框架 Java 集合系列 02 Collection架构 Java 集合系列 03 ArrayList详细介绍(源码解析)和使用示例 Java ...

  3. 【堆栈应用一】一个数divided=几个最小质因数的乘积

    /******************************************堆栈:一个数divided几个质因数(质因数的乘积为N)***************************** ...

  4. A-Making the Grade(POJ 3666)

    Making the Grade Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4656   Accepted: 2206 ...

  5. Computer Science Courses – Yan Yan

    CS: Compilers / Programming Languages Course Title Fundamentals of C++ Language Programming Textbook ...

  6. CentOS 常用命令大全(2)

    系统 # uname -a               # 查看内核/操作系统/CPU信息 # head -n 1 /etc/issue   # 查看操作系统版本 # cat /proc/cpuinf ...

  7. js中Array自定义contains, indexOf, delete方法.

    Array.prototype.contains = function (elem) { for (var i = 0; i < this.length; i++) { if (this[i] ...

  8. Android Phonebook编写联系人UI加载及联系人保存流程(六)

    2014-01-07 11:18:08 将百度空间里的东西移过来. 1. Save contact 我们前面已经写了四篇文章,做了大量的铺垫,总算到了这一步,见证奇迹的时刻终于到了. 用户添加了所有需 ...

  9. Rhel6-lanmp架构配置文档

    l--操作系统:windows  linux  unix  mac OS a--网页发布软件:apache nginx iis m--数据库:mysql  pgsql  oracle... p--网页 ...

  10. Unity游戏数据用Json保存

    (一)关于路径 unity有几个关键的路径 (1).Application.dataPath 只读路径,就是工作目录的Assets路径 (2).Application.streamingAssetsP ...