CF1186 D. Vus the Cossack and Numbers

题目传送门：https://codeforces.com/problemset/problem/1186/D

题目大意：
给定(n)个浮点数(a_i)，满足(sumlimits_{i=1}^na_i=0)，令(b_i=lfloor a_i floor) 或 (b_i=lceil a_i ceil)，求构造一组(b_i)，满足(sumlimits_{i=1}^nb_i=0)

---

首先令(b_i=lfloor a_i floor)，这样(sumlimits_{i=1}^nb_i<0)，我们再根据(|sumlimits_{i=1}^nb_i|)的值，将一些(b_i)改为(b_i+1)即可

注意存在(lfloor a_i floor=lceil a_i ceil)的情况，这样(b_i)是不能改成(b_i+1)的

/*program from Wolfycz*/
#include<map>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Fi first
#define Se second
#define ll_inf 1e18
#define MK make_pair
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
#define int_inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
	int f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')    f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
template<typename T>inline T read(T x){
	int f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=1e5;
const double eps=1e-8;
double A[N+10];
int B[N+10];
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	int n=read(0),All=0;
	for (int i=1;i<=n;i++){
		scanf("%lf",&A[i]);
		All+=(B[i]=floor(A[i]));
	}
	for (int i=1;All;i++){
		if (abs(A[i]-B[i])<=eps)	continue;
		B[i]++,All++;
	}
	for (int i=1;i<=n;i++)	printf("%d
",B[i]);
	return 0;
}