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 ;
}