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

2016-11-02 16:08 309 查看

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, ..., 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.

Examples

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。

思路：我们假设对全部的数字向下取整，用sum表示初始sum与取整后的sum的差值，然后找出整数的个数k（这部分不需要取整），然后遍历，找出把i个数向下取整的差值，取两个的最小即为答案。

#include<bits/stdc++.h>
using namespace std;
int main() {
int n, k = 0;
double sum = 0, ans = 1000000, a[40010];
cin >> n;

for(int i = 1; i <= 2 * n; i++) {
cin >> a[i];
sum += (a[i] - floor(a[i]));
if(a[i] - floor(a[i]) == 0)
k++;
}
for(int i = max(n - k, 0); i <= min(n, 2 * n - k); i++)
ans = min(ans, fabs(sum - i));

printf("%.3f\n", ans);
return 0;

}

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