Investment (水题)完全背包加状态压缩--不然会TlE
来源:互联网 发布:weka中的数据预处理 编辑:程序博客网 时间:2024/06/04 18:38
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 400
3000 250
With a capital of
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.
The first line of a test case contains two positive integers: the amount to start with (at most
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.
110000 424000 4003000 250
14050
题意:
john 现在有好多钱,他想把这些钱存在银行里,银行给出几种存钱福利,每种存钱的方式所需的资金不同,每年的利润也不同,现在给你N多money,让你求出在Y年之后的所能得到的最大值。
看完题目后就发现是完全背包,但是没想到状态压缩,其实就是让循环的次数变少,减少复杂度。仔细看看题目其实有暗示。
The value of a bond is always a multiple of $1 000.
代码:
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define maxn 1000000using namespace std;int money,y,p;int w[100],v[100];int f[maxn];int main(){ int T; scanf("%d",&T); while(T--) { scanf("%d%d",&money,&y); scanf("%d",&p); for(int i=0;i<p;i++) { scanf("%d%d",&w[i],&v[i]); w[i]/=1000; //状态压缩第一步 } int sum=money; for(int i=0;i<y;i++) { memset(f,0,sizeof f); //注意memset 的使用,每年都得清零 sum=money; sum/=1000; for(int i=0;i<p;i++) for(int j=w[i];j<=sum;j++) { f[j]=max(f[j],f[j-w[i]]+v[i]); } money+=f[sum]; } printf("%d\n",money); } return 0;}
- Investment (水题)完全背包加状态压缩--不然会TlE
- Investment(完全背包)
- POJ 2063 Investment(dp ,空间压缩,完全背包)
- (完全背包)Investment(P2063)
- hdu1963 Investment(完全背包)
- POJ2063 Investment (完全背包)
- poj2063 Investment(完全背包)
- POJ 2063 Investment (完全背包)
- POJ 2063 Investment DP(完全背包)
- POJ 2063 Investment(完全背包)
- HDU2063:Investment-wust9(完全背包典型)
- HDU 1963 Investment (完全背包)
- POJ 2063 Investment(完全背包)
- POJ 2063Investment(完全背包)
- POJ 2063 - Investment(完全背包)
- POJ 2063 Investment(完全背包)
- HDOJ 题目1963 Investment(完全背包)
- POJ 2063 Investment (完全背包)
- 711总结
- Spirng实战学习笔记(转载)
- C++笔记(四)
- MySQL数据库开启远程连接(云服务器ubuntu)
- 大数据时代数据可视化的好处
- Investment (水题)完全背包加状态压缩--不然会TlE
- 读书札记 二 《大型网站技术架构:核心原理与案例分析》
- JS基础
- 差分对设置
- 在线绘制基因表达热图
- 欢迎使用CSDN-markdown编辑器
- bzoj 2748: [HAOI2012]音量调节
- 【SQL server】透视及分组
- 关于斐波那契数列的一些例题