[HDU 1963] Investment

Investment

Time Limit:10000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

 

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	400
3000	250

With a capital of 10000 one could buy two bonds of 4000, giving a yearly interest of 800. Buying two bonds of 3000, and one of 4000 is a better idea, as it gives a yearly interest of 900. After two years the capital has grown to 11800, and it makes sense to sell a 3000 one and buy a 4000 one, so the annual interest grows to 1050. This is where this story grows unlikely: the bank does not charge for buying and selling bonds. Next year the total sum is 12850, which allows for three times 4000, giving a yearly interest of 1200.

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 1000000), 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 1000. The interest of a bond is never more than 10% of its value.

 

Output

For each test case, output �C on a separate line �C 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

 

由于HDU上这个题目显示有问题，不好复制，所以从虚拟OJ上复制了

多重背包模板题

进行N次（N表示多少年）多重背包即可，注意除以1000压缩空间、见代码 - -

#include<iostream>

#include<cstdio>

#include<cstring>

using namespace std;

#define N 100010



int w[N];

int v[N];

int dp[N];



int main()

{

    int T,n,s,time,i,j;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d%d%d",&s,&time,&n);

        for(i=1;i<=n;i++)

        {

            scanf("%d%d",&w[i],&v[i]);

            w[i]/=1000; //由于都是1000的倍数，所以同时除以1000压缩空间

        }

        int val=s;

        while(time--) //time次完全背包即可

        {

            s/=1000; //为什么可以这样?因为就算是整数除法，截去了百位数，但是由于价钱都是1000的倍数，所以几百元也买不起

            memset(dp,0,sizeof(dp));

            for(i=1;i<=n;i++)

            {

                for(j=w[i];j<=s;j++)

                {

                    dp[j]=max(dp[j],dp[j-w[i]]+v[i]);

                }

            }

            val+=dp[s]; //答案加上这一年的利息

            s=val;    //更新本金

        }

        cout<<val<<endl;

    }

    return 0;

}