POJ 2063 Investment
2015-02-05 09:59
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Investment
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.
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
Sample Output
Source
Northwestern Europe 2004
题意:约翰获得了一笔遗产,但他现在不需要,他想先存起来让它升值等自己退休的时候再用,银行说服他买一种有利息的债券。给出约翰遗产金额和存款的年数,及债券价格和利息,求出约翰最后能得到多少钱
完全背包,因为债券价格都是1000的倍数,因此可以把每种债券的价格除以1000,总资产金额也除以1000,这样可以节约时间。但要注意的是总金额和利息都不一定是1000的倍数,因此要记录下每年的总金额和利息
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 8857 | Accepted: 3039 |
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 |
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
题意:约翰获得了一笔遗产,但他现在不需要,他想先存起来让它升值等自己退休的时候再用,银行说服他买一种有利息的债券。给出约翰遗产金额和存款的年数,及债券价格和利息,求出约翰最后能得到多少钱
完全背包,因为债券价格都是1000的倍数,因此可以把每种债券的价格除以1000,总资产金额也除以1000,这样可以节约时间。但要注意的是总金额和利息都不一定是1000的倍数,因此要记录下每年的总金额和利息
#include <iostream> #include <stdio.h> #include <math.h> #include <string.h> #include <algorithm> using namespace std; int T,V,nyear,nbond; int cost[50],weight[50]; int dp[7000000]; int CompletePack(int V) { V/=1000; for(int i=1;i<=nbond;i++) { for(int j=cost[i];j<=V;j++) { dp[j]=max(dp[j],dp[j-cost[i]]+weight[i]); } } return dp[V];//返回该年的利息 } int main() { scanf("%d",&T); while(T--) { scanf("%d%d",&V,&nyear); //v/=1000; scanf("%d",&nbond); for(int i=1;i<=nbond;i++) { scanf("%d%d",&cost[i],&weight[i]); cost[i]/=1000;//除以1000,节约时间 } memset(dp,0,sizeof(dp)); int s=V; for(int i=1;i<=nyear;i++) { s+=CompletePack(s);//记录每年的总资产 } printf("%d\n",s); } return 0; }
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