http://codeforces.com/problemset/problem/351/A
题意：
2*n个数，选n个数上取整，n个数下取整
最小化 abs（取整之后数的和-原来数的和）
先使所有的数都下取整,累积更改的sum
那么选1个小数上取整，就会使sum-1
整数上下取整不会产生影响
所以有1个整数就可以使上取整的小数少1个
所以ans=min（abs（i-sum）） i∈[n-整数个数,n]
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
const double eps=1e-7;
int main()
{
int n;
scanf("%d",&n);
int m=n<<1;
int cnt=0;
double x;
double sum=0;
for(int i=1;i<=m;++i)
{
scanf("%lf",&x);
sum+=x-floor(x);
if(x-floor(x)<eps) cnt++;
}
double ans=1e9;
for(int i=n;i>=n-cnt;--i) ans=min(ans,abs(i-sum));
printf("%.3lf",ans);
}
A. 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
30.000 0.500 0.750 1.000 2.000 3.000
output
0.250
input
34469.000 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.