UVa11285 - Exchange Rates

Problem A: Exchange Rates

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 some or 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?

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. dlines follow, giving the price of a U.S. dollar in Canadian dollars. 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.

Sample Input

3
1.05
0.93
0.99
2
1.05
1.10
0

Output for Sample Input

1001.60
1000.00

输入的数字为汇率表示 1美元 = x加元

用usaDP[i] 表示第I天最多的美元 canaDP[i] 表示第I天最多的加元

因为舍去一美分以下的钱，因此可以将cannadp[0] = 100000

输出时只要将答案除去100即可

usadp[i] = max(usadp[i-1],int(cannadp[i-1]/t*0.97));
cannadp[i] = max(cannadp[i-1],int(usadp[i-1]*t*0.97));

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
using namespace std;
int n;
int usadp[400],cannadp[400];
int main(){
    while(~scanf("%d",&n)&&n){
        memset(usadp,0,sizeof usadp);
        memset(cannadp,0,sizeof cannadp);
        cannadp[0] = 100000;
        for(int i = 1; i <= n; i++){
            double t;
            scanf("%lf",&t);
            usadp[i] = max(usadp[i-1],int(cannadp[i-1]/t*0.97));
            cannadp[i] = max(cannadp[i-1],int(usadp[i-1]*t*0.97));
        }
        printf("%.2lf\n",cannadp[n]/100.0);
    }
    return 0;
}