CodeForces 352C. Jeff and Rounding（贪心）

Jeff got 2n real numbers a1, a2, ..., a2n 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 ai to the nearest integer that isn't more than ai (assign to ai: ⌊ ai ⌋);
round element aj to the nearest integer that isn't less than aj (assign to aj: ⌈ aj ⌉).

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, ..., a2n (0 ≤ ai ≤ 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.

题意：给2n个实数，对其中n个数做向上取整操作，另外n个数向下取整操作。求操作后的2n个数的和与原来2n个数的和差的绝对值的最小值。

做法：全部向上取整操作和记作res，再选取n个数做向下取整，只要减去向上取整与向下取整的差。一个数向上取整与向下取整的差只能是0或者1，那么如果res大于0.5，就让res尽量减1，否则减0。

第一次感到自己想出解题方法的感觉真好。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

double a[4005];
double ceil_a[4005];
double floor_a[4005];
double cha[4005];

int main()
{
    //freopen("in.txt", "r", stdin);

    int n;
    scanf("%d", &n);

    double res = 0;
    for (int i = 0; i < 2 * n; ++i) {
        scanf("%lf", &a[i]);
        ceil_a[i] = ceil(a[i]);
        floor_a[i] = floor(a[i]);
        cha[i] = ceil_a[i] - floor_a[i];
        res += ceil_a[i] - a[i];
    }
    sort(cha, cha + 2 * n);
    int l = 0, r = 2 * n - 1;
    for (int i = 0; i < n; ++i) {
        if (res > 0.5) res -= cha[r--];
        else res -= cha[l++];
    }

    printf("%.3f
", abs(res));
    return 0;
}