Codeforces Round #204 (Div. 2) C

写了一记忆化 TLE了  把double换成long long就过了 double 这么耗时间啊 

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<stdlib.h>
 5 #include<algorithm>
 6 #include<cmath>
 7 using namespace std;
 8 #define N 4010
 9 #define INF 0xfffffff
10 #define LL long long
11 double a[N];
12 LL dp[N][N],b[N][2];
13 int n;
14 LL dfs(int i,int r1,int r2,LL ss)
15 {
16     if(i>2*n)
17     return dp[r1][r2] = 0;
18     if(r1<=0&&r2<=0)
19     return 0;
20     if(dp[r1][r2]!=INF)
21     return dp[r1][r2];
22     LL bs = ss,as = ss,s1,s2;
23     if(r1)
24     {
25         s1 = (b[i][1]+dfs(i+1,r1-1,r2,ss+b[i][1]));
26         as += (b[i][1]+dfs(i+1,r1-1,r2,ss+b[i][1]));
27     }
28     if(r2)
29     {
30         s2 = b[i][0]+dfs(i+1,r1,r2-1,ss+b[i][0]);
31         bs+=b[i][0]+dfs(i+1,r1,r2-1,ss+b[i][0]);
32     }
33     if(r1&&r2)
34     {
35         if(abs(bs)<abs(as))
36         return dp[r1][r2] =s2 ;
37         else
38         return dp[r1][r2] = s1;
39     }
40     else if(r1)
41     return dp[r1][r2] = s1;
42     else
43     return dp[r1][r2] = s2;
44 }
45 int main()
46 {
47     int i,j;
48     scanf("%d",&n);
49     for(i = 0; i <= 2*n ; i++)
50     for(j =0 ; j <= 2*n ; j++)
51     dp[i][j] = INF;
52     for(i = 1; i <= 2*n ; i++)
53     {
54         scanf("%lf",&a[i]);
55         b[i][1] = (floor(a[i])-a[i])*10000000;
56         b[i][0] = (ceil(a[i])-a[i])*10000000;
57     }
58     /*for(i = 2*n-1 ;i >=1 ; i--)
59         for(j = 2*n; j >= 1 ; j--)
60         {
61             double
62         }*/
63     LL s = dfs(1,n,n,0);
64     printf("%.3lf\n",(double)fabs(s)/10000000.0);
65     return 0;
66 }

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