HDU 1963 Investment （完全背包）

                                                         Investment

                                             Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                         Total Submission(s): 491    Accepted Submission(s): 205

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 $10 000 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

题意： 买证券 

                T 组案例；

               输入本金s， 再输入要买几年的证券

               再输入有哪几种证券，

              买每种证券需要多少的金额，还有每种证券的年利率。

让你求4年后，John  ”最多 “将有多少钱。

明显是一个完全背包问题，这一年买证券赚的钱+本金 作为 下一年买证券的本金，不过应该注意的是，由于本金比较大，直接开一个10的6次方的数组，不RE的，肯定会超内存，所以需要优化一下，因为题目中说了，每种证券的金额都是1000的倍数，所以这里我可以把本题中的 w[ i ], 除以1000， 而本金则也可以除以1000，但是应该注意c此处  和。同学可以自己体会一下为什么这样做。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int max(int a, int b)
{
    return a>b?a:b;
}

int main()
{
    int T, v, year, n, i, j, k, t;
    int *dp, w[110], c[110];

    scanf("%d", &T);
    while(T--)
    {
        scanf("%d %d", &v, &year);
        scanf("%d", &n);

        memset(w, 0, sizeof(w));
        memset(c, 0, sizeof(c));

        for(i=0; i<n; i++)
            scanf("%d %d", &w[i], &c[i]);

     for(i=0; i<n; i++)
     {
         w[i] = w[i]/1000;
     }
      for(k=0; k<year; k++)
       {
             t = v;
             v = v / 1000;
         dp = (int *)malloc((v+1)*sizeof(int));

         for(i=0; i<=v; i++)
           dp[i] = 0;

        for(i=0; i<n; i++)
           for(j=w[i]; j<=v; j++)
           {
               dp[j] = max(dp[j], dp[j-w[i]]+c[i]);
           }
           v = t+dp[v];
       }
       printf("%d
", v);
    }
    return 0;
}