在python3中实现查找数组中最接近与某值的元素操作

import datetime
 
def find_close(arr, e):
 start_time = datetime.datetime.now()
 
 size = len(arr)
 idx = 0
 val = abs(e - arr[idx])
 
 for i in range(1, size):
  val1 = abs(e - arr[i])
  if val1 < val:
   idx = i
   val = val1
 
 use_time = datetime.datetime.now() - start_time
 
 return arr[idx], use_time.seconds * 1000 + use_time.microseconds / 1000
 
def find_close_fast(arr, e):
 start_time = datetime.datetime.now()
  
 low = 0
 high = len(arr) - 1
 idx = -1
 
 while low <= high:
  mid = int((low + high) / 2)
  if e == arr[mid] or mid == low:
   idx = mid
   break
  elif e   arr[mid]:
   low = mid
  elif e < arr[mid]:
   high = mid
 
 if idx + 1 < len(arr) and abs(e - arr[idx])   abs(e - arr[idx + 1]):
  idx += 1
  
 use_time = datetime.datetime.now() - start_time
 
 return arr[idx], use_time.seconds * 1000 + use_time.microseconds / 1000
 
if __name__ == "__main__":
 arr = []
 
 f = open("1Mints.txt")
 for line in f:
  arr.append(int(line))
 f.close()
 
 arr.sort()
 
 while 1:
  e = int(input("input a number:"))
  print("find_close ", find_close(arr, e))
  print ("find_close_fast ", find_close_fast(arr, e))

#include <iostream 
#include <map 
#include <cmath  
using namespace std;
map <long long ,int  a;
int main()
{
	a.clear() ;
	int N,t;
	long long int x;
	cin    N;
	while(N--)
	{
		cin    t    x;
		if(t==1)
			a[x]=1;
		else
		{
			if(a.empty() )//判断集合是否为空 
				cout << "Empty!\n" ;
			else
			{
				if(a.size() == 1 )//若只有一个元素，则直接输出 
					cout << a.begin()- first << endl;
				else
				{
					map <long long ,int ::iterator it,m,n;
					it=a.find(x);
					if(it!=a.end() )
					{
						cout << x <<endl;
						continue;
					}
					a[x]=1;
					it=a.find(x);
					if(it == a.begin() )
					{
						it++;
						cout << it -  first << endl;
					} 
					else if(it == a.end() )
					{
						it--;
						cout << it -  first << endl; 
					}
					else
					{
						m=--it;//m和n分别指向it的左右两侧 
						it++;
						n=++it;
						if(abs(m -  first - x)   abs(n -  first - x))
							cout << n -  first << endl;
						else if(abs(m -  first - x) == abs(n -  first - x))	
							cout << m -  first << " " << n -  first << endl;
						else
							cout << m -  first << endl;		
					}
					a.erase(a.find(x) ); 	
				}	 
			}	
		}	
	}
	return 0;
}