題目描述

Despite their innate prudence, the cows took a beating in the home mortgage market and now are trying their hand at stocks. Happily, Bessie is prescient and knows not only today's S (2 <= S <= 50) stock prices but also the future stock prices for a total of D days (2 <= D <= 10).

Given the matrix of current and future stock prices on various days (1 <= PR_sd <= 1,000) and an initial M (1 <= M <= 200,000) units of money, determine an optimal buying and selling strategy in order to maximize the gain realized by selling stock on the final day. Shares must be purchased in integer multiples, and you need not spend all the money (or any money). It is guaranteed that you will not be able to earn a profit of more than 500,000 units of money.

Consider the example below of a bull (i.e., improving) market, the kind Bessie likes most. In this case, S=2 stocks and D=3 days. The cows have 10 units of money to invest.

Today's price

| Tomorrow's price

| | Two days hence Stock | | | 1 10 15 15

2 13 11 20

If money is to be made, the cows must purchase stock 1 on day 1. Selling stock 1 on day 2 and quickly buying stock 2 yields 4 money in the bank and one share of 2. Selling stock 2 on the final day brings in 20 money for a total of 24 money when the 20 is added to the bank.

有S種股票，我們已經(jīng)知道每一天每一種股票的價格。一共有d天，一開始擁有的錢為m,求最后總共能夠擁有多少錢。

輸入輸出格式

輸入格式：

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* Line 1: Three space-separated integers: S, D, and M

* Lines 2..S+1: Line s+1 contains the D prices for stock s on days 1..D: PR_sd

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輸出格式：

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* Line 1: The maximum amount of money possible to have after selling on day D.

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輸入輸出樣例

輸入樣例
2 3 10 10 15 15 13 11 20 輸出樣例
24

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Measure:

我們來好好推下這個DP--

　　

在第二天，我們能取得的最大收益：

第一種股票第一天10元買入，第二天15元賣出，收益5元

第二種股票第一天……買不起！只有10元

所以，第二天我們就有15元啦。（開心）

在第三天，我們能取得的最大收益：

第一種股票第二天再買沒錢賺了

第二種股票第二天則可以低價買入，高價賣出，收益9元

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其實我一開始有些疑惑，我本來以為最便宜的時候買入，在最貴的時候賣出才應(yīng)該最優(yōu)

而DP的思路則是有錢賺就賣

現(xiàn)在驗證，它是對的。

假如我們一開始有 m=10 元，一種股票三天價格如下

5 10 20

按我的想法：第一天花10元買入兩股，第三天20元賣出，收益：40-10=30

按DP的做法：第一天花10元買入兩股，第二天10元賣出，收益：20-10=10，m=20;第三天有錢賺，又第二天賣出兩股【就相當(dāng)于沒賣】，收益：+（40-20）= 30

并不影響

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code(開 O(2) 才過)

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int MX=510000; int n,d,m,f[MX],pri[110][20];int main() {scanf("%d%d%d",&n,&d,&m);for(int i=1;i<=n;++i) for(int j=1;j<=d;++j) scanf("%d",&pri[i][j]); for(int k=2;k<=d;++k) { //前 k 天 int mx=0;memset(f,0,sizeof(f));for(int i=1;i<=n;++i) { //第 i 種股票 for(int j=pri[i][k-1];j<=m;++j) { //花 j 元 f[j]=max(f[j],f[j-pri[i][k-1]]+pri[i][k]-pri[i][k-1]); mx=max(mx,f[j]);}}m+=mx; }printf("%d",m);return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/qseer/p/9777188.html

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