2504: Exchange Rates

 

Status	In/Out	TIME Limit	MEMORY Limit	Submit Times	Solved Users	JUDGE TYPE
	stdin/stdout	3s	16384K	203	51	Standard

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?

Input

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.

Output

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
1.0500
0.9300
0.9900
2
1.0500
1.1000
0

Sample Output

1001.60
1000.00

 

Problem Source: Waterloo 9,23,2007

 

This problem is used for contest: 115 

 

#include<stdio.h>  //采用取百化整的方法
#include<string.h>
double rate[366];
int coin[366][2];
int main()
{
 //freopen("in.txt","r",stdin);
 //freopen("out.txt","w",stdout);
 int i,n,j;
 double k;
 while((scanf("%d",&n),n)!=0)
 {
  memset(coin,0,sizeof(coin));
  memset(rate,0,sizeof(rate));
  for(i=1;i<=n;i++)
      scanf("%lf",&rate[i]);
  coin[1][0]=1000/rate[1]*0.97*100;
  coin[1][1]=1000*100;
  for(i=2;i<=n;i++)
    for(j=1;j<n;j++)
    {
   if(coin[j][1]/rate[i]*0.97>coin[i][0])
          coin[i][0]=coin[j][1]/rate[i]*0.97;
   if(coin[j][0]*rate[i]*0.97>coin[i][1])
          coin[i][1]=coin[j][0]*rate[i]*0.97;
    }
  k=1000.00;
  for(i=1;i<=n;i++)
  if(coin[i][1]*1.0/100>k)
       k=coin[i][1]*1.0/100; 
  printf("%.2lf/n",k);
 }
 return 0;
}

 

以上是N^2的DP代码

 

以下是N的DP代码

我的DP还是不错的啊，哈哈！

 

#include<stdio.h>  //采用取百化整的方法
#include<string.h>
double rate[366];
int coin[366][2];
int main()
{
 //freopen("in.txt","r",stdin);
 //freopen("out.txt","w",stdout);
 int i,n;
 double k;
 while((scanf("%d",&n),n)!=0)
 {
  memset(coin,0,sizeof(coin));
  memset(rate,0,sizeof(rate));
  for(i=1;i<=n;i++)
      scanf("%lf",&rate[i]);
  coin[1][0]=1000/rate[1]*0.97*100;
  coin[1][1]=1000*100;
  for(i=2;i<=n;i++)
  {
   if(coin[i-1][1]/rate[i]*0.97>coin[i-1][0])
          coin[i][0]=coin[i-1][1]/rate[i]*0.97;
   else
          coin[i][0]=coin[i-1][0];
   if(coin[i-1][0]*rate[i]*0.97>coin[i-1][1])
          coin[i][1]=coin[i-1][0]*rate[i]*0.97;
   else
          coin[i][1]=coin[i-1][1]; 
  }
  k=0;
  for(i=1;i<=n;i++)
  if(coin[i][1]*1.0/100>k)
       k=coin[i][1]*1.0/100; 
  printf("%.2lf/n",k);
 }
 return 0;
}