Codeforces Round #204 (Div. 2) C. Jeff and Rounding

C. Jeff and Rounding

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Jeff got 2n real numbers a₁, a₂, ..., a_2n as a birthday present. The boy hates non-integer numbers, so he decided to slightly "adjust" the numbers he's got. Namely, Jeff consecutively executes n operations, each of them goes as follows:

• choose indexes i and j (i ≠ j) that haven't been chosen yet;
• round element a_i to the nearest integer that isn't more than a_i (assign to a_i: ⌊ a_i ⌋);
• round element a_j to the nearest integer that isn't less than a_j (assign to a_j: ⌈ a_j ⌉).

Nevertheless, Jeff doesn't want to hurt the feelings of the person who gave him the sequence. That's why the boy wants to perform the operations so as to make the absolute value of the difference between the sum of elements before performing the operations and the sum of elements after performing the operations as small as possible. Help Jeff find the minimum absolute value of the difference.

Input

The first line contains integer n (1 ≤ n ≤ 2000). The next line contains 2n real numbers a₁, a₂, ..., a_2n (0 ≤ a_i ≤ 10000), given with exactly three digits after the decimal point. The numbers are separated by spaces.

Output

In a single line print a single real number — the required difference with exactly three digits after the decimal point.

Sample test(s)

input

3
0.000 0.500 0.750 1.000 2.000 3.000

output

0.250

input

3
4469.000 6526.000 4864.000 9356.383 7490.000 995.896

output

0.279

Note

In the first test case you need to perform the operations as follows: (i = 1, j = 4), (i = 2, j = 3), (i = 5, j = 6). In this case, the difference will equal |(0 + 0.5 + 0.75 + 1 + 2 + 3) - (0 + 0 + 1 + 1 + 2 + 3)| = 0.25.

思路：

2n个数（设小数部分都看做x），n个向下取整，相对原来的数-x，n个向下取整，相对原来的数+（1-x），从两个式子中可以看出x的前面都是负号，所以可以统一处理，那么剩下来的就是统计1的个数cnt了。如果不出现a.000的情况的话，那么cnt=n，因为a.000如果向上取整的话是不需要+1的，所以统计出现a.000的次数num，那么+1的个数就在（max(0,n-num)，n）之间了，取最佳ans就够了。

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
//#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 100005
#define MAXN 100005
#define mod 1000000007
#define INF 0x3f3f3f3f
using namespace std;

typedef long long ll;
int n,m,cnt;
double ans,sum;

int main()
{
    int i,j,mi;
    ll t;
    double s,tmp;
    while(~scanf("%d",&n))
    {
        m=2*n;
        cnt=0;
        sum=0;
        for(i=1;i<=m;i++)
        {
            scanf("%lf",&s);
            t=ll(s);
            if(t==s) cnt++;
            else
            {
                tmp=s-t;
                sum+=tmp;
            }
        }
        mi=min(n,cnt);
        ans=INF;
        for(i=0;i<=mi;i++)
        {
            ans=min(ans,fabs(n-i-sum));
        }
        printf("%.3f\n",ans);
    }
    return 0;
}