UVa11285 - Exchange Rates - c++编程基础

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

Sample Input

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 
   
    
#include 
    
      #include 
     
       #include 
      
        #include 
       
         #include 
        
          #include 
         
           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; }