01: Find the closest element

Total time limit: 1000ms memory limit: 65536kB
described
in a non-descending sequence, and find the closest value given element.

Input
The first line contains an integer n, a non-drop sequence length. 1 <= n <= 100000.
The second line contains n integers, the non-descending sequence of elements. All the elements are in the size between 0-1,000,000,000.
The third row contains an integer m, to ask for the number of a given value. 1 <= m <= 10000.
Next m lines, each an integer, the closest element to ask for the given value. All values are given size between 0-1,000,000,000.
Output
m lines, each an integer value closest to the corresponding element of a given value, holding the input order. If multiple values satisfy the condition, a minimum output.
Sample input
. 3
2. 8. 5
2
10
. 5
sample output
. 8
. 5

#include<iostream>
#include<cstring>
#include<algorithm>//排序、二分查找算法使用的头文件
using namespace std;
int main(){
	int n;
	cin>>n;
	long long a[n+5];
	for(int i=0;i<n;i++){
		cin>>a[i];
	}
	//二分查找需要先排序 
	sort(a,a+n);
	
	int m;
	cin>>m;
	long long target[m+5];
	for(int i=0;i<m;i++){
		cin>>target[i];
		long long *temp=lower_bound(a,a+n,target[i]);
		if(*temp==target[i]){
			//查找的值正好在输入的数据里面，输出 
			cout<<target[i]<<endl;
		}else{
			//目标查找值不在数据里面，比较查找值前后的两个值与查找值哪个更接近 
			if(temp-a>0&&temp-a<n){
			//查找的位置不在起始位置，也不再结束位置 
				if((target[i]-a[temp-a-1])<=(*temp-target[i])){
					cout<<a[temp-a-1]<<endl;
				}else{
					cout<<*temp<<endl;
				}	
			}else if(temp-a==0)
			//查找的位置在起始位置
			cout<<*temp<<endl;
			else if(temp-a==n)
			//查找的位置在结束位置
			cout<<a[temp-a-1]<<endl;
			
		}
	}
	
	return 0;
}