Investment (Dynamic Programming DP)

　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 3000	400 250

　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

1
10000 4
2
4000 400
3000 250

Sample Output

14050

Meaning of the questions:

Known number of existing money, and plans to save time, several deposit scheme (the deposit amount and profit), how can we ask a few years after the biggest profits

Code:

There is such a fine short I do not want to come out of the code tql:

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <string.h>
 5 
 6 using namespace std;
 7 
 8 int main()
 9 {
10     int t, n, year, mon, tail;
11     int i, j, k;
12     int dp[100005], cost[100005], earn[100005];
13     scanf("%d", &t);
14     while(t--)
15     {
16         scanf("%d %d", &mon, &year);
17         scanf("%d", &n);
18         for(i=0;i<n;i++)
19         {
20             scanf("%d %d", &cost[i], &earn[i]);
21             cost[i]/=1000;
22         }
23         memset(dp, 0, sizeof(dp));
24         for(i=0 ; I <year; I ++ )
 25          {
 26 is              tail = Mon / 1000 ;    // addition thousand good memory array to the post 
27              for (J = 0 ; J <n-; J ++)      // n-types of deposit 
28              {
 29                  for (cost = K [J]; K <= tail; K ++)   // either a profit money, money can not profit or not, taking the optimal 
30                  {
 31 is                      DP [K] = max (DP [K], DP [K-cost [J]] + earn [J]);
 32                  }
 33 is              }
 34 is              Mon = + DP [tail];     // maximum profit added in the amount of the deposit, the deposit on the basis of the year
35         }
36         printf("%d\n", mon);
37     }
38     return 0;
39 }