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

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<stdlib.h>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 #define N 4010
 #define INF 0xfffffff
 #define LL long long
 double a[N];
 LL dp[N][N],b[N][];
 int n;
 LL dfs(int i,int r1,int r2,LL ss)
 {
     if(i>*n)
     return dp[r1][r2] = ;
     if(r1<=&&r2<=)
     return ;
     if(dp[r1][r2]!=INF)
     return dp[r1][r2];
     LL bs = ss,as = ss,s1,s2;
     if(r1)
     {
         s1 = (b[i][]+dfs(i+,r1-,r2,ss+b[i][]));
         as += (b[i][]+dfs(i+,r1-,r2,ss+b[i][]));
     }
     if(r2)
     {
         s2 = b[i][]+dfs(i+,r1,r2-,ss+b[i][]);
         bs+=b[i][]+dfs(i+,r1,r2-,ss+b[i][]);
     }
     if(r1&&r2)
     {
         if(abs(bs)<abs(as))
         return dp[r1][r2] =s2 ;
         else
         return dp[r1][r2] = s1;
     }
     else if(r1)
     return dp[r1][r2] = s1;
     else
     return dp[r1][r2] = s2;
 }
 int main()
 {
     int i,j;
     scanf("%d",&n);
     for(i = ; i <= *n ; i++)
     for(j = ; j <= *n ; j++)
     dp[i][j] = INF;
     for(i = ; i <= *n ; i++)
     {
         scanf("%lf",&a[i]);
         b[i][] = (floor(a[i])-a[i])*;
         b[i][] = (ceil(a[i])-a[i])*;
     }
     /*for(i = 2*n-1 ;i >=1 ; i--)
         for(j = 2*n; j >= 1 ; j--)
         {
             double
         }*/
     LL s = dfs(,n,n,);
     printf("%.3lf\n",(double)fabs(s)/10000000.0);
     return ;
 }