ZOJ 2224 Investment （完全背包）-蒲公英云

ZOJ 2224 Investment （完全背包）

水深无声 2024-02-17 21:26 48阅读 0赞

Investment

Time Limit: 10 Seconds Memory Limit: 32768 KB

John never knew he had a grand-uncle, until he received the notary��s letter. He learned thathis late grand-uncle had gathered a lot of money, somewhere in South-America, and thatJohn was the only inheritor.

John did not need that much money for the moment. But he realized that it would be agood idea to store this capital in a safe place, and have it grow until he decided to retire. Thebank 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, payedto the owner at the end of each year. The bond has no fixed term. Bonds are available indifferent sizes. The larger ones usually give a better interest. Soon John realized that theoptimal set of bonds to buy was not trivial to figure out. Moreover, after a few years hiscapital 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 10000 one could buy two bonds of 4000, giving a yearly interest of800. Buying two bonds of 3000, and one of 4000 is a better idea, as it gives a yearlyinterest of 900. After two years the capital has grown to 11800, and it makes sense to sella 3000 one and buy a 4000 one, so the annual interest grows to 1050. This is where thisstory grows unlikely: the bank does not charge for buying and selling bonds. Next year thetotal sum is 12850, which allows for three times 4000, giving a yearly interest of 1200.

Here is your problem: given an amount to begin with, a number of years, and a set ofbonds with their values and interests, find out how big the amount may grow in the givenperiod, 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 testcases follow.

The first line of a test case contains two positive integers: the amount to start with (atmost 1000000), 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 availablebonds.

The next d lines each contain the description of a bond. The description of a bond consistsof two positive integers: the value of the bond, and the yearly interest for that bond. Thevalue of a bond is always a multiple of 1000. The interest of a bond is never more than10% of its value.

Output

For each test case, output �C on a separate line �C the capital at the end of the period, after anoptimal schedule of buying and selling.

Sample Input

1
10000 4
2
4000 400
3000 250

Sample Output

14050

分析：完全背包问题，刚开始数组开为100w，结果seg了，然后开为1000w，结果ME了，然后看了讨论说，可以吧数组压缩，因为债券值是1000的整数倍，所以所拥有的本息和的零头小于1000 不可能买到债券，所以，将本息和与债券值同时除以1000对正确结果没影响，这样就达到了优化数组。

AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define INF 10000000
const int maxn=1e6+10;
int dp[maxn];
int bond[12],interes[12];
int main() {
    int T;
    scanf("%d",&T);
    while(T--){
        int start,year;
        scanf("%d%d",&start,&year);
        int d;
        scanf("%d",&d);
        for(int i=0;i<d;i++){
            scanf("%d%d",&bond[i],&interes[i]);
            bond[i]/=1000;
        }
        int money=start;
        while(year--){
            int t=money/1000;
            for(int i=1;i<=t;i++)
            dp[i]=-INF;
            dp[0]=0;
            for(int i=1;i<=t;i++){
                for(int j=0;j<d;j++){
                 if(i>=bond[j])
                 dp[i]=max(dp[i],dp[i-bond[j]]+interes[j]);
                }
            }
            int cnt=0;
            for(int i=t;i>=1;i--)
            if(cnt<dp[i]){
                cnt=dp[i];
            }
            money+=cnt;
        }
        printf("%d\n",money);
    }
    return 0;
}