• poj2063 Investment


    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

    这道题用完全背包做,主要注意一点,就是一开始有的总钱数和每次买股票的钱数都是1000的倍数,所以可以把背包压缩,都除以1000,然后每年的本利用val保存,下一年计算的时候,背包容量为val/1000,因为val%1000这个数对买股票没有意义,因为股票的价钱都是1000的倍数。

    #include<stdio.h>
    #include<string.h>
    int max(int a,int b){
    	return a>b?a:b;
    }
    int dp[50000];
    int main()
    {
    	int n,m,i,j,T,num,val,s;
    	int w[15],v[16];
    	scanf("%d",&T);
    	while(T--)
    	{
    		scanf("%d%d",&m,&n);
    		scanf("%d",&num);
    		for(i=1;i<=num;i++){
    			scanf("%d%d",&w[i],&v[i]);
    			w[i]/=1000;
    		}
    		val=m;
    		while(n--)
    		{
    			memset(dp,0,sizeof(dp));
    			s=val/1000;
    			for(i=1;i<=num;i++){
    				for(j=w[i];j<=s;j++){
    					dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
    				}
    			}
    			val=val+dp[s];
    			//printf("%d
    ",m);
    		}
    		printf("%d
    ",val);
    	}
    	return 0;
    }


  • 相关阅读:
    .NET中使用嵌入的资源
    C#操作注册表
    .Net中大数加减乘除运算
    CYQ.Data 轻量数据层之路 V4.5 版本发布[更好的使用体验,更优的缓存机制]
    关于性能比较的应用误区
    秋色园QBlog技术原理解析:性能优化篇:打印页面SQL,全局的SQL语句优化(十三)
    CYQ.DBImport 数据库反向工程及批量导数据库工具 V1.0 发布
    框架设计之菜鸟漫漫江湖路系列 一:菜鸟入门
    MySql折腾小记二:text/blog类型不允许设置默认值,不允许存在两个CURRENT_TIMESTAMP
    CYQ.Data.Xml XmlHelper 助你更方便快捷的操作Xml/Html
  • 原文地址:https://www.cnblogs.com/herumw/p/9464726.html
Copyright © 2020-2023  润新知