POJ 2063 —— Investment【完全背包】

​​题目传送门​​

求投资k年获得最大投资，每年都选最大利息的方案进行投资k年后就可以得到最多的人民币。

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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:

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.

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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.

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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.

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Sample Input

1
10000 4
2
4000 400
3000 250

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Sample Output

14050

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题意：

给你钱数，还有年数year，还有多少钱一年多少利息，问几年后最多的钱数（利息算到下一年的本金中）。

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分析：

• 求year次完全背包
• 注意因为投资的钱都是1000的倍数，所以可以÷1000缩小背包数

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内存问题：

注意题目中的本金很大但是最开始最多的本金不超过$1 000 000，而且每种计算利息所需的本金都是$1 000的倍数，所以为了避免超内存，每次dp时先把本金除以$1 000，相应的cost也除以$1 000。那么dp数组到底开多大才合适呢，题中说最多存40年，开始时本金不超过$1 000 000而每次的利息又不会超过本金的 10% ，所以 数组的大小应该是 $1 000 000 / 1000 * ( 1.1 ) ⁴⁰而 1.1⁴⁰ 大概为 45.259256，就取 50 吧。从而数组开 1 000 * 50 = 50 000

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AC代码：

#include <iostream>
#include <vector>
#include <utility>
#include <cstring>
#include <algorithm>
#include <map>
#include <queue>
#include <stack>
#include <cstdio>
#include <fstream>
#include <set>
using namespace std;
typedef long long ll;
#define

const int MAXN = 1e6;
#define


ll dp[MAXN];
ll w[MAXN];
ll v[MAXN];

int main() {

  int T;
  while (cin >> T) {
    while (T--) {
      memset(dp, 0, sizeof dp);
      int V, year;
      cin >> V >> year;
      int n;
      cin >> n;
      for (int i = 1; i <= n; i++) {
        cin >> w[i] >> v[i];
        w[i] /= 1000;
      }
      while (year--) {
        int tmp = V / 1000;
        for (int i = 1; i <= n; i++) {
          for (int j =0; j <= tmp; j++) {
            if (j >= w[i])
              dp[j] = max(dp[j], dp[j - w[i]] + v[i]);
          }
        }
        V += dp[tmp];

      }
      cout << V << endl;
    }
  }

  return 0;
}