HDU 1963 Investment 最大利息

原题： http://acm.hdu.edu.cn/showproblem.php?pid=1963

题目：

Investment

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 689 Accepted Submission(s): 290

Problem 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 10000onecouldbuytwobondsof4 000, giving a yearly interest of 800.Buyingtwobondsof3 000, and one of 4000isabetteridea,asitgivesayearlyinterestof900. After two years the capital has grown to 11800,anditmakessensetosella3 000 one and buy a 4000one,sotheannualinterestgrowsto1 050. This is where this story grows unlikely: the bank does not charge for buying and selling bonds. Next year the total sum is 12850,whichallowsforthreetimes4 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 1000000),andthenumberofyearsthecapitalmaygrow(atmost40).Thefollowinglinecontainsasinglenumber:thenumberd(1<=d<=10)ofavailablebonds.Thenextdlineseachcontainthedescriptionofabond.Thedescriptionofabondconsistsoftwopositiveintegers:thevalueofthebond,andtheyearlyinterestforthatbond.Thevalueofabondisalwaysamultipleof1 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

思路：

有一笔钱，可以按不同的金额存到银行，每年都有相应的利息。一定年限后要求有最多钱。注意存入的钱只会以1000的倍数。

比如样例：
10000本金第一年买两次3000的，买一次4000的，就一共有10900元。
第二年因为900块用不出去，所以能用的钱和第一年一样，最后剩11800元。
第三年可以买两次4000的，买一次3000的，最后剩12850元。
第四年买三次4000的，最后剩14050元。

我们只需要写个函数调用4次就可以完成4年的操作。
因为每次都会以1000的倍数处理，所以我们可以把本金除以1000，大大减少了计算量和空间。
因为本金是1 000 000,10年满利息10%也不超过原来50倍，所以这里数组长度开50 000。

代码：

#include <iostream>
#include"string.h"
#include"cstdio"
#include"stdlib.h"
#include"algorithm"
using namespace std;

int m,n;
const int N=50000;
int dp[N];
int weight[50];
int value[50];
void init()
{
    memset(dp,0,sizeof(dp));
    memset(weight,0,sizeof(weight));
    memset(value,0,sizeof(value));
}

void pack()
{
    for(int i=1;i<=n;i++)
    {
        for(int j=weight[i];j<=m/1000;j++)
        {
            dp[j]=max(dp[j],dp[j-weight[i]]+value[i]);
        }
    }
    m=m+dp[m/1000];
    memset(dp,0,sizeof(dp));
}

int main()
{
    int tt;
    scanf("%D",&tt);
    while(tt--)
    {
        int t;
        scanf("%d %d",&m,&t);
        scanf("%d",&n);
        init();
        for(int i=1;i<=n;i++)
        {
            scanf("%d %d",&weight[i],&value[i]);
            weight[i]=weight[i]/1000;
        }
        for(int i=0;i<t;i++)
        {
            pack();
        }
        printf("%d\n",m);
    }

    return 0;
}