给2n个数,对其中n个向上取整,n个向下取整,问前后的数的和的最小差值。
把所有数的整数部分去掉。
若
0<x<1
,
δ=−x/1−x
,此时和
=n−∑xi
若
x=0
,
δ=0
因此答案只和有几个0向上取整相关
#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <functional>
#include <cstdlib>
#include <queue>
#include <stack>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
int main() {
// freopen("B.in","r",stdin);
int T=read();
while(T--) {
int n=read();
int p=0;double ans=0;
For(i,n*2) {
double x;
scanf("%lf",&x);
x-=(int)x;
// cout<<x<<endl;
if (x<1e-8) p++;
else ans+=x;
}
double tot=fabs(n-ans);
Rep(i,1+min(p,n)) {
int j=p-i;
if (j<=n) {
tot=min(tot,fabs(ans-(n-j)));
}
}
printf("%.3lf\n",tot);
}
return 0;
}