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.

 

算法比较简单~~

因为一定会有n个数向上取整，n个数向下取整，所以最后差值d:

 

 d = 求和 (1-di) - 求和(di)  di表示第i个数向下取整丢掉的小数  。。。。。（i从1到n）

当ai含非零小数位时，

d = n - 求和(di)   （i从1到2n）.

 那么可以求出小数部分的所有和，即上式右边的第二项，数出有多少个含非零小数的数；

 然后为了使|d|最小，即等式右边第一项最接近第二项，从n中选数向上取整即可（直接贪心）。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<map>

using namespace std;

#define LL long long
#define ULL unsigned long long
#define UINT unsigned int
#define MAX_INT 0x7fffffff
#define MAX_LL 0x7fffffffffffffff
#define MAX(X,Y) ((X) > (Y) ? (X) : (Y))
#define MIN(X,Y) ((X) < (Y) ? (X) : (Y))

#define MAXN 4444
#define EPS 1e-9
#define INF 1e9

int main(){
//    freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);
    int i, n;
    while(scanf(" %d", &n)==1){
        double ans=INF, a, t=0;
        int cnt=0;
        for(i=0; i<n<<1; i++){
            scanf(" %lf", &a);
            double del = a - (int)a;
            t = t + del;
            if(del>EPS) cnt++;
        }
        int minn = (cnt>=n ? cnt-n : 0),
            maxx = (cnt>=n ? n : cnt);
        for(i=minn; i<=maxx; i++)
            ans=MIN(fabs(i-t), ans);
        printf("%.3f\n", ans);
    }
    return 0;
}