POJ 2063 Investment（完全背包）

 

Investment

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 5362		Accepted: 1843

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 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

Source

Northwestern Europe 2004

 

 

 

 

 

这题数据范围很大，比估算的要大很多。

 The value of a bond is always a multiple of $1 000.这句话告诉我们算完全背包的时候，可以把总的钱除于1000，把每种债券的价格也除于1000，这样可以减少时间。

而且注意的是原来的钱不一定是1000的整数倍的。这点有点坑，而且得到的利息也不是1000的整数倍。

而且必须把原来的价格和得到的利息记录下来。

每年做一次完全背包。。

 

代码1：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
int f[5000000];
int c[20],w[20];
int  fun(int V,int n)
{
    V/=1000;
    for(int i=0;i<n;i++)
      for(int j=c[i];j<=V;j++)
         f[j]=max(f[j],f[j-c[i]]+w[i]);
    return f[V];

}
int main()
{
    int V,p;
    int T;
    int n;
    scanf("%d",&T);
    while(T--)
    {
        memset(f,0,sizeof(f));
        scanf("%d%d",&V,&p);
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&c[i],&w[i]);
            c[i]/=1000;
        }
        int ans=V;
        for(int i=0;i<p;i++)ans+=fun(ans,n);
        printf("%d\n",ans);

    }
    return 0;
}

 

 

 

代码2：

 

 

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
int f[5010000];
int V;
int c[200],w[200];
void CompletePack(int cost,int weight)
{
    for(int v=cost;v<=V;v++)
      f[v]=max(f[v],f[v-cost]+weight);
}
int main()
{
   //freopen("in.txt","r",stdin);
   // freopen("out.txt","w",stdout);
    int T;
    int n;
    scanf("%d",&T);
    int p;
    while(T--)
    {
        scanf("%d%d",&V,&p);
        int tempV=V;
        V/=1000;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&c[i],&w[i]);
            c[i]/=1000;
        }
        memset(f,0,sizeof(f));
        for(int i=0;i<n;i++)
           CompletePack(c[i],w[i]);
        int sumtemp=f[V];
        int temp=f[V];
        int temp2=(tempV+temp)/1000;
        //printf("%d %d  %d\n",temp,V,temp2);
       int tt=temp2-V;
        for(int i=1;i<p;i++)
        {
            if(tt>0)
            {
                for(int j=0;j<n;j++)
                   for(int v=c[j];v<=temp2;v++)
                       f[v]=max(f[v],f[v-c[j]]+w[j]);
            }
            temp+=f[temp2];
            tt=(tempV+temp)/1000-temp2;
            temp2=(tempV+temp)/1000;
          //  printf("%d    %d   %d\n",temp,temp2,V);
        }
        if(p==0)printf("%d\n",tempV);
        else
        printf("%d\n",tempV+temp);
    }
    return 0;
}