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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