POJ 2063 Investment

快速链接：http://poj.org/problem?id=2063

CSUST 2012年暑假8月组队后第八次个人赛：http://acm.hust.edu.cn:8080/judge/contest/view.action?cid=11473#problem/D

 

Investment

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 5370		Accepted: 1848

 

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

Source

Northwestern Europe 2004

重点:关于内存的处理,比赛时一直错就是内存没处理好。

        后来看了学长的题解( http://www.shabiyuan.com/?category欢迎访问)

        才清楚。具体分析见下面的代码区~~~~~~

//AC 948k 94ms C++ POJ 2063 Investment
/*思路:DPz之完全背包(如果不懂完全背包的推荐看我转载的《背包九讲》的博客，推荐一道完全背包入门题目hdu 1114)
       完全背包:即每种物品可以放N次。
       状态转移方程:if(dp[j]<dp[j-value[i]]+interest[i])
                        dp[j]=dp[j-value[i]]+interest[i];
       需要注意的是，这道题目是背包体积不断增大的完全背包。 
       开始背包的体积是本金，每隔一年都会得到一定的利息，所以每年背包的体积都会变大。
       重点:关于内存的处理,比赛时一直错就是内存没处理好,后来看了学长的题解(http://www.shabiyuan.com/?category欢迎访问)才清楚。
            由于本金的数值很大，而数组又不能开的特别大。所以用dp处理时,要把每次的本金除以1000减少内存，相应的每种债券所需要的钱也除以1000
*/
//400k 63ms (当数组开到maxn=60000时)

#include<cstdio> 
#include<cstring>
const int maxn=200000;
int dp[maxn];
int main()
{
    int test;
    int start,year;
    int d;
    int i,j;
    int value[15],interest[15];
    int ans,max;
    scanf("%d",&test);
    while(test--)
    {
        scanf("%d%d",&start,&year);
        ans=max=start;
        scanf("%d",&d);
        for(i=0;i<d;i++)
        {
            scanf("%d%d",&value[i],&interest[i]);
            value[i]/=1000;//每种债券除以1000减少内存。
        }
        while(year--)
        {
            memset(dp,0,sizeof(dp));
            max/=1000; //相应的每次的本金除以1000减少内存
            for(i=0;i<d;i++)
                for(j=value[i];j<=max;j++)
                    if(dp[j]<dp[j-value[i]]+interest[i])
                        dp[j]=dp[j-value[i]]+interest[i];
            ans+=dp[max];//本金+利息(注意利息开始没有除以也不用除以1000,所以ans即为所求)
            max=ans;
        }
        printf("%d\n",ans);
    }
    return 0;
}