POJ 2063

Investment

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 7959 Accepted: 2729

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: 
ValueAnnual
interest4000
3000400
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

110000 424000 4003000 250

Sample Output

14050

题意：有一个人要买债券，他一开始有一笔资金，债券有很多种类，有的花钱少利息低，有的花钱多利息高。每一年他买的债券都会有收益，这时他又可以把现在拥有的所有钱（包括获得的利息）取出来在重新安排如何购买债券。问在若干年内他的资金最多能变成多少。很关键的一点是，债券的金额都是1000的倍数（用来降低时间和空间福再度）。

思路：对每一年都进行一次重新规划，这个规划就是完全背包的思路。dp[i][j]表示在前i种债券中，拥有j的资金，将这些资金拿去买债券能达到的最大年利润。j看成背包容量，显然这个问题可以用完全背包很容易的解决。最后对每一年都进行一次这样的操作，操作完后更新他现有的资金。但是如果单纯这样做，既会超内存又会爆空间，这时就应该把每个债券的单价除以1000，在进行完全背包时资金也除以1000（比1000多一点的零头肯定买不到债券）。

下面是代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <ctype.h>#include <cstring>#include <string>#include <queue>#include <cmath>#define MAXN 200000#define INF 2100000000#define pu system("pause")#pragma comment(linker, "/STACK:16777216");using namespace std;int dp[MAXN];int n;int cap, tim;int bond[15];int cost[15];int main(){    //freopen("C:/Users/Admin/Desktop/in.txt", "r", stdin);    int t;    cin >> t;    while(t--)    {        memset(dp, 0, sizeof(dp));        cin >> cap >> tim;        cin >> n;        for(int i = 0; i < n; i++)        {            cin >> cost[i] >> bond[i];            cost[i] /= 1000;        }        for(int i = 0; i < tim; i++)        {            long long temp = cap/1000;            for(int j = 0; j < n; j++)            {                for(int k = cost[j]; k <= temp; k++)                {                    if(dp[k] < dp[k-cost[j]]+bond[j])                    dp[k] = dp[k-cost[j]]+bond[j];                }            }            cap += dp[temp];        }        cout << cap << endl;    }    return 0;}