百练 3420: Exchange Rates

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本文介绍了一种基于预测汇率变动进行货币投机的算法。该算法通过在最佳时机转换加元与美元来最大化最终收益，并考虑了每次交易需扣除3%佣金及舍去美分以下金额的影响。

百练 3420: Exchange Rates

原题OJ链接

描述

Now that the Loonie is hovering about par with the Greenback, you have decided to use your $1000 entrance scholarship to engage in currency speculation. So you gaze into a crystal ball which predicts the closing exchange rate between Canadian and U.S. dollars for each of the next several days. On any given day, you can switch all of your money from Canadian to U.S. dollars, or vice versa, at the prevailing exchange rate, less a 3% commission, less any fraction of a cent.
Assuming your crystal ball is correct, what’s the maximum amount of money you can have, in Canadian dollars, when you’re done?

输入

The input contains a number of test cases, followed by a line containing 0. Each test case begins with 0<d<=365, the number of days that your crystal ball can predict. d lines follow, giving the price of a U.S. dollar in Canadian dollars, as a real number.

输出

For each test case, output a line giving the maximum amount of money, in Canadian dollars and cents, that it is possible to have at the end of the last prediction, assuming you may exchange money on any subset of the predicted days, in order.

样例输入

样例输出

1001.60
1000.00

解题思路

这道题的难点在于英文阅读hhh。有几个坑：1）题目中的$1000指的是 Loonie，即 Canadian dollars 。2）less any fraction of a cent，这个是说小于美分的部分都舍去，所以代码中将初始的 1000 定义为100000，然后用int 取整。注意格式化输出，保留小数点后两位，用printf("%.2f\n",CA/100.0) 。
代码中的CA表示加拿大元，US表示美元。题目的意思是要输出最终经过汇率转化后的CA的最大值，每次转化都要付出3%的额外费用。

源代码

#include<iostream>
#include<algorithm>
using namespace std;
int main(){
    int d;

    while(cin>>d && d!=0){
        int CA=100000, US=0;
        double rate;
        for(int i=0;i<d;i++){
            cin>>rate;
            CA=max(CA,(int)(US*rate*0.97));
            US=max(US,(int)(CA/rate*0.97));
        }
        printf("%.2f\n",CA/100.0);
    }
    return 0;
}